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ICSE Class X Board Exam 2024 : Chemistry

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2 CHAPTER Some Basic Concepts of Chemistry MATTER A substance which occupies space, possesses mass and can be felt by any one or more of the five senses is called matter. Physical Classification of Matter It is based on physical state under ordinary conditions of temperature and pressure. (a) Solid: A substance is said to be solid if it possesses a definite volume and a definite shape, e.g. sugar, iron, gold, wood etc. (b) Liquid: A substance is said to be liquid if it possesses a definite volume but not definite shape. They take up the shape of the container, e.g. water, milk, oil, mercury, alcohol etc. (c) Gas: A substance is said to be gas if it neither possesses a definite volume nor a definite shape. This is because they fill up the whole container, e.g. Hydrogen (H2), Oxygen (O2), Carbon dioxide (CO2) etc. Chemical Classification of Matter (a) Pure Substance: A material containing only one type of substance. Pure Substance can not be separated into simpler substance by physical method. e.g.: Element = Na, Mg, Ca ............................ etc. Compound = HCl, H2O, CO2, HNO3 ..........etc. Pure substance is classified into two types: (I) Element (II) Compound (I) Element: The pure substance containing only one kind of atoms. It is classified into 3 types (i) Metal Zn, Cu, Hg, Ac, Sn, Pb etc. (ii) Non-metal N2, O2, Cl2, Br2, F2, P4, S8 etc. (iii) Metalloids B, Si, As, Te etc. (II) Compound: It is defined as pure substance containing more than one kind of elements or atoms which are combined together in a fixed proportion by weight and which can be decomposed into simpler substance by the suitable chemical method. The properties of a compound are completely different from those of its constituent element, e.g. HCl, H2O, H2SO4, HClO4, HNO3 etc. (b) Mixture: A material which contain more than one type of substances and which are mixed in any ratio by weight is called as mixture. The property of the mixture is the property of its components. The mixture can be separated by simple physical method. Classification of Mixture (i) Homogeneous mixture: The mixture, in which all the components are present uniformly is called as homogeneous mixture. Components of mixture are present in a single phase, e.g. Water + Salt, Water + Sugar, Water + alcohol. Homogeneous substances are of two types: (a) Pure substances: Substances which have definite and constant chemical composition are known as pure substances. For example, all elements and compounds are pure substances. (b) Solutions: A homogenous mixture of two or more pure substances is known as a solution. For example, air, a mixture of NaCl and water, alcohol and water, etc. A solution does not have a definite composition. (ii) Heterogeneous mixture: The mixture in which all the components are present non-uniformly, e.g. Water + Sand, Water + Oil, blood, petrol etc. ATOMS AND MOLECULES Atom: The term atom was introduced by Dalton. Atom is the smallest particle of matter that takes part in a chemical reaction. Atom is also defined as the smallest particle of an element that retains all the properties of an element. Molecule: The term molecule was introduced by Avogadro. Molecule is the smallest particle of matter that exists independently and is formed by the combination of atoms. Molecule is also defined as the smallest particle of matter that can exist and retains all the properties of that substance. Physical and Chemical Properties: (a) Physical Property: The property which can be measured without changing the chemical composition of the substance is known as physical property like mass, volume, density, refractive index etc. (b) Chemical Property: The property which can be evaluated at the cost of matter itself is known as chemical property. For example combustible nature of hydrogen gas can be verified by burning of hydrogen. The sweet taste of sugar by consuming it. Physical Quantity and their Measurements in Chemistry Physical Quantities All quantities that can be measured are called physical quantities. eg. time, length, mass, force, work done, etc. Fundamental Quantities A set of physical quantities which are completely independent of each other but all other physical quantities can be expressed in terms of these physical quantities is called set of Fundamental Quantities. Fundamental units are those units which can neither be derived from one another nor they can be further resolved into any other units. The Fundamental Quantities that are currently being accepted by the scientific community are mass, time, length, current, temperature, luminous intensity and amount of substance. International System (SI) of Units Table: SI base quantities and their units S. No. Physical quantity unit Symbol metre m 1 Length 2 Mass kilogram kg 3 Time second s 4 Temperature kelvin K 5 Electric current ampere A 6 Luminous Intensity candela cd 7 Amount of substance mole mol SCIENTIFIC NOTATION If a number P can be expressed as P = A 10x where 1 A < 10, this is called Scientific Notation and x is called order of magnitude of the number. SI Prefixes: The magnitudes of physical quantities vary over a wide range. The mass of an electron is 9.1 10 31 kg and that of our earth is about 6 1024 kg. Standard prefixes for certain power of 10. Table shows these prefixes: Power of 10 12 9 6 3 2 1 1 48 Prefix tera giga mega kilo hecto deca deci Symbol T G M k h da d Power of 10 2 3 6 9 12 15 Prefix centi milli micro nano pico femto Symbol c m n p f SIGNIFICANT FIGURES OR DIGITS The significant figures (SF) in a measurement are the figures or digits that are known with certainty plus one that is uncertain. Larger the number of significant figures obtained in a measurement, greater is its accuracy and vice versa. 1. Rules to find out the number of significant figures: I Rule: All the non-zero digits are significant e.g. 1984 has 4 SF. II Rule: All the zeros between two non-zero digits are significant, e.g. 10806 has 5 SF. III Rule: All the zeros to the left of first non-zero digit are not significant, e.g.00108 has 3 SF. IV Rule: If the number is less than 1, zeros on the right of the decimal point but to the left of the first non-zero digit are not significant, e.g. 0.002308 has 4 SF. V Rule : The trailing zeros (zeros to the right of the last nonzero digit) in a number with a decimal point are significant, e.g. 01.080 has 4 SF. VI Rule: The trailing zeros in a number without a decimal point are not significant e.g. 010100 has 3 SF. VII Rule: When the number is expressed in exponential form, the exponential term does not affect the number of S.F. For example in x = 12.3 = 1.23 101 = .123 102 = 0.0123 103 = 123 10 1, each term has 3 SF only. (Note: It has 3 significant figure in each expression.) 2. Rules for arithmetical operations with significant figures: I Rule: In addition or subtraction the number of decimal places in the result should be equal to the number of decimal places of that term in the operation which contain lesser number of decimal places, e.g. 12.587 12.5 = 0.087 = 0.1 ( second term contain lesser i.e. one decimal place) II Rule: In multiplication or division, the number of SF in the product or quotient is same as the smallest number of SF in any of the factors, e.g. 5.0 0.125 = 0.625 = 0.62 To avoid the confusion regarding the trailing zeros of the numbers without the decimal point the best way is to report every measurement in scientific notation (in the power of 10). In this notation every number is expressed in the form a 10b , where a is the base number between 1 and 10 and b is any positive or negative exponent of 10. The base number a is written in decimal form with the decimal after the first digit. While counting the number of SF only base number is considered (Rule VII). The change in the unit of measurement of a quantity does not affect the number of SF. For example in 2.308 cm = 23.08 mm = 0.02308 m = 23080 m each term has 4 SF. P W JEE (XI) Module-1 CHEMISTRY ACCURACY AND PRECISION The accuracy of a measurement is a measure of how close the measured value is to the true value of the quantity. Precision tells us to what resolution or limit the quantity is measured. Illustration: Assertion: If the true value for a result is 2.00 g and a student A takes two measurements and reports the results as 1.95 g and 1.93 g. These values are precise as they are close to each other but are not accurate. Reason: Precision refers to the closeness of various measurements for the same quantity. Whereas, accuracy is the agreement of a particular value to the true value of the result. In the light of the above statement, choose the correct answer from the options given below: (a) Both Assertion and Reason are true and Reason is correct explanation of Assertion. (b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion. (c) Assertion is true but Reason is false. (d) Assertion is false but Reason is true. Sol. (a) Illustration: Assertion: 100 has only one significant figure, but 100.0 has three significant figures and 100.0 has four significant figures. Reason: Zeros at the end or right of a number are significant provided they are on the right side of the decimal point. In the light of the above statement, choose the correct answer from the options given below: (a) Both Assertion and Reason are true and Reason is correct explanation of Assertion. (b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion. (c) Assertion is true but Reason is false. (d) Assertion is false but Reason is true. Sol. (a) Illustration: Assertion: All the zeros to the left of first non-zero digit are not significant. Reason: Zeros at the end or right of a non-zero digit are significant provided they are on the right side of the decimal point. In the light of the above statement, choose the correct answer from the options given below: (a) Both Assertion and Reason are true and Reason is correct explanation of Assertion. (b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion. (c) Assertion is true but Reason is false. (d) Assertion is false but Reason is true. Sol. (b) Some Basic Concepts of Chemistry LAWS OF CHEMICAL COMBINATION The law of conservation of mass (Law of Indestructibility of matter) According to this law, the mass can neither be created nor be destroyed in a balanced chemical reaction but one form is changed into another form. In a chemical change total mass remains conserved i.e., total mass before the reaction is always equal to total mass after the reaction. 1 O (g) H2O ( ) H2(g) + 2 2 1 mole 1 mole 1 mole 2 1 Mass before the reaction = 1 2 + 32 = 18 gm 2 Mass after the reaction = 1 18 = 18 gm [Total mass of reactants = Total mass of products + Mass of unreacted reactants] Law of constant composition (definite proportions): All chemical compounds are found to have constant composition of elements irrespective of their method of preparation or sources. E.g. In H2O, hydrogen & oxygen combine in 2 : 1 molar ratio, this ratio remains constant whether it is tap water, river water or seawater or produced by any chemical reaction. Law of Multiple Proportions: When one element combines with the other element to form two or more different compounds, the mass of one element, which combines with a constant mass of the other bear a simple ratio to one another. Nitrogen and oxygen combine to form five oxides: weights of oxygen which combine with the fixed weight of nitrogen in these oxides are calculated as under: Oxide Ratio of weights of nitrogen and oxygen N2O 28 : 16 NO 28 : 32 N2O3 28 : 48 N2O4 28 : 64 N2O5 28 : 80 Number of parts by weight of oxygen which combine with 14 parts by weight of nitrogen from the above are 8, 16, 24, 32 and 40 respectively. Their ratio is 1 : 2 : 3 : 4 : 5, which is a simple whole number ratio. Hence, the law of multiple proportion is illustrated. ADVANCED LEARNING Law of Reciprocal Proportion: The ratio of the masses of two elements A and B which combine separately with a fixed mass of the third element C is either the same or some simple multiple of the ratio of the masses in which A and B combine directly with each other. H Like CH4, CO2 and H2O CH4 H2O CH4 C : H = 12 : 4 CO2 C : O = 12 : 32 H2O H : O = 2 : 16 C CO2 O 49 Gay-Lussac s Law of Combining Volume: Gases combine in a simple whole number ratio of their volumes provided all measurements should be done at the same temperature and pressure. H2 (g) + Cl2 (g) 2HCl 1 vol. 1 vol. 2 vol. Avogadro s hypothesis: Equal volume of all gases have equal number of molecules (not atoms) at same temperature and pressure condition. S.T.P. (Standard Temperature and Pressure) Temperature = 0 C or 273 K, Pressure = 1 atm = 760 mm of Hg. Volume of one mole of gas at STP is found to be experimentally equal to 22.4 litres which is known as molar volume. Measuring the volume is equivalent to counting the number of molecules of the gas. Example 1: A 15.9 g sample of sodium carbonate is added to a solution of acetic acid weighing 20.0 g. The two substances react, releasing carbon dioxide gas to the atmosphere. After reaction, the contents of the reaction vessel weigh 29.3 g. What is the mass of carbon dioxide given off during the reaction? Sol. The total mass of reactants taken = 15.9 + 20.0 = 35.9 gm. From the conservation of mass, the final mass of the contents of the vessel should also be 35.9 gm. But it is only 29.3 gm. The difference is due to the mass of released carbon dioxide gas. Hence, the mass of carbon dioxide gas released = 35.9 29.3 = 6.6 gm Example 2: The following are the results of analysis of two samples of the same or two different compounds of phosphorus and chlorine. From these results, decide whether the two samples are from the same or different compounds. Also state the law, which will be obeyed by the given samples. Amount of P Amount of Cl Compound A 1.156 gm 3.971 gm Compound B 1.542 gm 5.297 gm Sol. The mass ratio of phosphorus and chlorine in compound A, mP : mCl = 1.156:3.971 = 0.2911:1.000 The mass ratio of phosphorus and chlorine in compound B, mP : mCl = 1.542:5.297 = 0.2911:1.000 As the mass ratio is same, both the compounds are same and the samples obey the law of definite proportion. Example 3: 2.5 ml of a gaseous hydrocarbon exactly requires 12.5 ml oxygen for complete combustion and produces 7.5 ml carbon dioxide and 10.0 ml water vapour. All the volumes are measured at the same pressure and temperature. Show that the data illustrates Gay Lussac s law of volume combination. Sol. Vhydrocarbon : Voxygen : Vcarbon dioxide : Vwater vapour 50 = 2.5 : 12.5 : 7.5 : 10.0 = 1 : 5 : 3 : 4 (simple whole no. ratio) Hence, the data is according to the law of volume combination. 1. A sample of pure carbon dioxide, irrespective of its source contains 27.27% carbon and 72.73% oxygen. The data support: (a) Law of constant composition. (b) Law of conservation of mass. (c) Law of reciprocal proportions. (d) Law of multiple proportions. 2. The percentage of hydrogen in water and hydrogen peroxide is 11.1 and 5.9 respectively. These figures illustrate: (a) Law of multiple proportions. (b) Law of conservation of mass. (c) Law of constant proportions. (d) Law of combining volumes. 3. 1.0 g of an oxide of A contained 0.5 g of A. 4.0 g of another oxide of A contained 1.6 g of A. The data indicate the law of: (a) Reciprocal proportions. (b) Constant proportions. (c) Conservation of energy. (d) Multiple proportions. 4. Carbon is found to form two oxides which contain 42.9% & 27.3% of carbon respectively show that these figures shows the (a) Law of multiple proportion (b) Law of definite proportion (c) Law of mass conservation (d) All of these ATOMIC MASS & MOLECULAR MASS Relative Atomic Mass: One of the most important concept from Dalton s atomic theory was that of relative atomic mass or relative atomic weight. This is done by expressing mass of one atom with respect to a fixed standard. Dalton used hydrogen as the standard (H = 1). Later on oxygen (O = 16) replaced hydrogen as the reference. The present standard unit which was adopted internationally in 1961, is based on the mass of one carbon-12 isotopic atom, taken as exactly 12.000 u (amu). Relative atomic mass (R.A.M) Mass of one atom of an element = 1 Mass of one C12 atom 12 Atomic Mass Unit (or amu): The atomic mass unit (amu) is 1 equal to 12 th of mass of one atom of carbon-12 isotope. 1 mass of one C12 isotopic atom 12 ~ mass of one nucleon in C12 atom. 1 amu = 1.66 10 24 gm or 1.66 10 27 kg 1 amu = P W JEE (XI) Module-1 CHEMISTRY Today, amu has been replaced by u which is known as unified atomic mass One amu is also called One Dalton (Da). Atomic mass = R.A.M 1 amu Relative atomic mass indicates the number of nucleons present in the atom. Average / Mean Atomic Mass The weighted average of the isotopic masses of the element s naturally occurring isotopes. Mathematically, average atomic mass of X (Ax) a1 x1 + a 2 x 2 + ..... + a n x n = 100 Where: a1, a2, a3 ........... atomic mass of isotopes. and x1, x2, x3 ........ mole% or % of natural abundance of isotopes. Key Note Where: M1, M2, M3 ........... are molar masses. n1, n2, n3 ........... are moles of substances. Formula Mass The formula mass of a substance is defined as the sum of the atomic masses of constituent atoms in an ionic compound. This is generally used for ionic compounds which do not contain discrete molecules, but ions as their constituent units. For example: Formula mass of NaCl is: Formula mass = mass of sodium atom + mass of chlorine atom = (23 + 35.5) u = 58.5 u Illustration: The molar composition of polluted air is as follows: Gas At. wt. mole percentage composition Oxygen 16 16% Nitrogen 14 80% Carbon dioxide - 03% Sulphur dioxide - 01% What is the average molecular weight of the given polluted air? (Given, atomic weights of C and S are 12 and 32 respectively. j= n P P P Atomic weights of many elements are not whole numbers due to the presence of stable isotopes. The number of atoms of a particular isotope present in 100 atoms of a natural sample of that element is called its relative abundance which always remains constant for a given element. Shortcut for % determination if average atomic weight is given for X having isotopes XA & XB. % of XA = Average atomic weight wt of X B difference in weight of X A & X B Sol. Mavg = n M j =1 j= n j n j =1 j Here j= n n j =1 j = 100 j 16 32 + 80 28 + 44 3 + 64 1 100 512 + 2240 + 132 + 64 2948 = = = 29.48 Ans. 100 100 \ Mavg = 100 Illustration: Naturally occurring chlorine is 75% Cl35 which has an atomic mass of 35 amu and 25% Cl37 which has a mass of 37 amu. Calculate the average atomic mass of chlorine: (a) 35.5amu (b) 36.5amu (c) 71amu (d) 72amu Sol. (a) Average atomic mass = (% of I isotope) (its A.M) + (% of II isotope) (its A.M.) 100 = 75 35 + 25 37 = 35.5 amu 100 Example 4: Find the relative atomic mass of O atom and its atomic mass. Sol. The number of nucleons present in O atom is 16. Relative atomic mass of O atom = 16. Atomic mass = R.A.M 1 amu = 16 1 amu = 16 amu Example 5: The weight of one atom of uranium is 235 amu. Its actual weight in g is: Sol. 235 1.67 10 24 g = 3.95 10 22 g Relative molecular mass: = mass of one molecule of the substance 1 mass of one C12 atom 12 Molecular mass = Relative molecular mass 1 amu Mean Molar Mass or Molecular Mass The average molar mass of the different substance present in the n M + n 2 M 2 + .....n n M n . container = 1 1 n1 + n 2 + ....n n Some Basic Concepts of Chemistry 5. The Relative molecular mass of ammonia is: (a) 17 (b) 22 (c) 28 (d) 44 6. The mass of an atom of sodium is: (a) 23 amu (b) 23gm (c) 46 amu (d) 12 amu 7. The atomic mass & molecular mass of hydrogen is: (a) 1amu & 2amu (b) 2 amu & 4 amu (c) 3amu & 6amu (d) 4 amu & 8 amu 51 8. One u stands for the (a) mass of an atom of carbon-12 atom. (b) 1/12th of mass of carbon-12. (c) 1/12th of mass of hydrogen atom. (d) mass of one atom of any of the element. 9. Mass of 1 amu in g is equal to: (a) 1.66 1024 g (b) 1.66 10 24 g (c) 1.008 g (d) 9.1 10 28 g MOLE CONCEPT Mole Mole is a counting unit mostly used for microscopic particles and is defined as follows: A mole is the amount of a substance that contains as many entities (atoms, molecules or other particles) as there are atoms in exactly 0.012 kg (or 12 gm) of the carbon-12 isotope. From mass spectrometer we found that there are 6.022 1023 atoms present in 12 gm of C12 isotope. The number of entities in 1 mol is so important that it is given a separate name and symbol known as Avogadro constant denoted by NA. i.e., on the whole we can say that 1 mole is the collection of 6.022 1023 entities. Here entities may represent atoms, ions, molecules or even pens, chair, paper etc but as this number (NA) is very large therefore it is significant only for very micro-particles. Illustration: Chlorophyll, the green colouring material of plants contains 3.68 % of magnesium by mass. Calculate the number of magnesium atom in 5.00 g of the complex. Sol. Mass of magnesium in 5.0 g of complex 3.68 5.00 = 0.184 g = 100 Atomic mass of magnesium = 24 24 g of magnesium contain = 6.022 1023 atoms 6.022 1023 0.184 g of magnesium would contain = 0.184 24 = 4.617 1021 Therefore, 5.00 g of the given complex would contain 4.617 1021 atoms of magnesium. How Big is a Mole? Amount of water in world's oceans (litres) Avogadro's number Age of earth (seconds) Population of earth 602,200,000,000,000,000,000,000 Distance from earth to sun (centimeters) In modern practice gram-atom and gram-molecule are termed as mole. Gram Atomic Mass: The atomic mass of an element expressed in gram is called gram atomic mass of the element. or It is also defined as mass of 6.022 1023 atoms. or It is also defined as the mass of one mole atoms. 52 Eg: Element R.A.M. Atomic mass Gram Atomic (Relative Atomic (mass of one mass/weight Mass) atom) N 14 14 amu 14 gm He 4 4 amu 4 gm C 12 12 amu 12 gm For example, for oxygen atom: Atomic mass of O atom = mass of one O atom = 16 amu Gram atomic mass = mass of 6.022 1023 O atoms = 16 amu 6.022 1023 = 16 1.66 10 24 g 6.022 1023 = 16 g ( 1.66 10 24 6.022 1023 ~ 1) Gram Molecular Mass: The molecular mass of a compound expressed in gram is called the gram-molecular mass of the compound. or It is also defined as mass of 6.022 1023 molecules. or It is also defined as the mass of one mole molecules. For example, for O2 molecule: Molecular mass of O2 molecule = mass of one O2 molecule = 2 mass of one O atom = 2 16 amu = 32 amu Gram molecular mass = mass of 6.022 1023 O2 molecules = 32 amu 6.022 1023 = 32 1.66 10 24 gm 6.022 1023 = 32 gm RELATIONSHIP BETWEEN GRAM AND AMU 1 of wt. of one C12 atom. 12 For carbon (C12) atom, 1 mole C = 12 gm = 6.022 1023 atoms wt. of 6.022 1023 C12 atoms = 12 gm 12 gm wt. of 1 atom of C12 = NA (NA Avogadro s number = 6.022 1023) 1 amu = 1 amu = 1 of wt. of one C12 atom 12 1amu = = 1 12 gm 12 N A 1 gm NA METHODS OF CALCULATIONS OF MOLE (a) If no. of particles of some species is given, then no. of moles Given no. of particles = NA (b) If weight of a given species is given, then no. of moles = Given wt. (for atoms), Atomic wt. Given wt. or (for molecules) Molecular wt. P W JEE (XI) Module-1 CHEMISTRY (c) If volume of a gas is given along with its temperature (T) PV and pressure (P) use n = (assuming gas to be ideal) RT where R = 0.0821 lit-atm/mol K (when P is in atmosphere and V is in litre.) Y-map: Interconversion of mole - volume, mass and number of particles: STP L L or or ( ) ( ) ATOMICITY It is equal to number of atoms present in one molecule. For example atomicity of H2, CO2, O3, CCl4, respectively. DENSITY For Liquids and Solids, (a) Absolute density Mass Absolute density = Volume (b) Relative density Density of substance Relative density = Density of standard substance Density of substance e.g., Specific gravity = Density of H 2 O at 4 C For Gases: PM Absolute density (mass / volume) = RT where P is pressure of gas, M = mol. wt. of gas, R is the gas constant, T is the temperature. Vapour density: It is defined only for gas. It is a density of gas with respect to H2 gas at same temperature & pressure. PM gas / RT M gas d gas M V.D. = = = = P M H 2 / RT M H 2 d H2 2 V.D. = M 2 Example 6: Total number of atoms of all elements present in 1 mole of ammonium dichromate is? (a) 14 (b) 19 23 (c) 6 10 (d) 114 1023 Sol. (NH4)2Cr2O7 =19 6.022 1023 114 1023 atoms. Example 7: How many atoms of oxygen are there in 16 g of oxygen? (a) 2 NA (b) NA (c) 1.5 NA (d) 4 NA Sol. Let x atoms of oxygen are present So, 16 1.66 10 24 x = 16 g x = P Some Basic Concepts of Chemistry = NA Volume of hydrogen gas at STP = 10 22.4 L Example 9: The number of atoms contained in 11.2 L of SO2 at S.T.P. are: (a) 3/2 6.022 1023 (b) 2 6.022 1023 (c) 6.022 1023 (d) 4 6.022 1023 Sol. 22.4 litre gas has = 1 mole 1 1 1 litre gas has = 11.2 = mole of molecules 2 22.4 1 = 3 mole of atoms 2 3 = 6.022 1023 atoms 2 Example 10: 7.5 litre of the particular gas at S.T.P. weighs 16 gram. What is the V.D. of gas? 7.5 16 = Sol. Moles at S.T.P. = 22.4 M 48 M = 48 gram; V.D. = = 24. 2 Example 11: Find the density of CO2(g) with respect to N2O(g). Key Note 1 g O, 1 g O2, 1 g O3 each have same number of oxygen atoms. Density of liquid water at 4 C is 1 g/mL = 1 g/cc = 103 kg/m3 1.66 10 24 Example 8: Calculate the volume in litres of 20 g hydrogen gas at STP. (a) 2.24 L (b) 22.4 L (c) 224 L (d) 4.48 L Mass Sol. No. of moles of hydrogen gas = Molecular mass 20 gm = = 10 mol 2 gm Sol. V.D. = P 1 M.wt. of CO 2 44 = = 1 M.wt.of N 2 O 44 Example 12: Find the vapour density of N2O5. = Sol. V.D. Mol.wt. of N 2 O5 108 = = 54 . 2 2 53 10. The number of electrons present in 1 mol of methane molecule are: (a) 6.022 1025 (b) 6.022 1024 (c) 6.022 1023 (d) 6.022 1022 11. The mass of one molecule of water is approximately: (a) 3 10 23 g (b) 18 g (c) 1.5 10 23 g (d) 4.5 10 23 g 12. The molar mass of ferrous sulphate (FeSO4.7H2O) is: (a) 152 gm (b) 278 gm (c) 137 gm (d) None of these 13. The vapour density of carbon dioxide is: (a) 44 (b) 32 (c) 22 (d) 12 3 14. The density of air is 0.001293 g/cm at STP. Identify which of the following statement is correct? (a) Vapour density is 12.72. (b) Molecular weight is 28.96. (c) Vapour density is 0.001293 g/cm3. (d) Vapour density and molecular weight cannot be determined. The molecular formula is an integral multiple of the empirical formula. i.e. Molecular formula = Empirical formula n Molecular Formula Mass where n = Empirical Formula Mass If sum of mass percent of all elements is less than 100 then difference is due to oxygen. mCO 2 12 100 Mass % of C = 44 mcompound Mass For n mole of a compound (C3H7O2); Moles of C = 3n Moles of H = 7n Moles of O = 2n PERCENTAGE FORMULAE COMPOSITION % of element in a compound Atomic weight of element Number of atom of that element in one molecule 100 = Total molecular weight of compound Here we are going to find out the percentage of each element in the compound by knowing the molecular formula of compound. EMPIRICAL AND MOLECULAR FORMULA We have just seen that knowing the molecular formula of the compound, we can calculate percentage composition of the elements. Conversely, if we know the percentage composition of the elements initially, we can calculate the relative number of atoms of each element in the molecules of the compound. This gives us the empirical formula of the compound. Further if the molecular mass is known then the molecular formula can easily be determined. An empirical formula represents the simplest whole number ratio of various atoms present in a compound. The molecular formula gives the actual number of atoms of each element in a molecule. The molecular formula shows the exact number of different types of atoms present in a molecule of a compound. 54 m H2O 2 100 18 mcompound Example 13: Every molecule of ammonia always has formula NH3 irrespective of method of preparation or sources. i.e. 1 mole of ammonia always contains 1 mol of N and 3 mole of H. In other words 17 gm of NH3 always contains 14 gm of N and 3 gm of H. Now find out % of each element in the compound. Sol. Mass % of N in NH3 = ELEMENTAL ANALYSIS % of H = 14 Mass of N in 1 mole NH3 100 = 100 = 82.35 % 17 Mass of 1 mole of NH3 Mass % of H in NH3 = 3 Mass of H in 1 mole NH3 100 Mass of 1 mol e of NH3 3 = 100 = 17.65 % 17 Example 14: Acetylene and benzene both have the empirical formula CH. The molecular masses of acetylene and benzene are 26 and 78 respectively. Deduce their molecular formulae. Sol. Empirical Formula is CH Step-1: The empirical formula of the compound is CH Empirical formula mass = (1 12) + 1 = 13. Molecular mass = 26 Step-2: To calculate the value of n Molecular mass 26 = =2 Empirical formula mass 13 Step-3: To calculate the molecular formula of the Compound. Molecular formula = n (Empirical formula of the compound) = 2 CH = C2 H2 Thus the molecular formula is C2 H2 Similarly for benzene To calculate the value of n n = n = Molecular mass 78 = = 6. Empirical formula mass 13 Thus the molecular formula is 6 CH = C6H6 P W JEE (XI) Module-1 CHEMISTRY Example 15: An organic substance containing carbon, hydrogen and oxygen gave the following percentage composition. C = 40.684% ; H = 5.085% and O = 54.228% Sol. Element %/Atomic mass divide by Simple smallest ratio integer ratio The molecular weight of the compound is 118. Calculate the molecular formula of the compound. N 25.94 = 1.85 14 1.85 =1 1.85 2 (a) C4H6O4 (b) C2H3O2 O 74.06 = 4.63 16 4.63 = 2.5 1.85 5 (c) C2H4O2 (d) C4H6O8 Sol. Step-1: To calculate the empirical formula of the compound. Empirical Formula is C2 H3 O2 Step-2: To calculate the empirical formula mass. The empirical formula mass of the compound is 12 2 + 3 1 + 16 2 = 59 . Step-3: To calculate the value of n n = Molecular mass 118 = =2 Empirical formula mass 59 Step-4: To calculate the molecular formula of the salt. Molecular formula = n (Empirical formula) = 2 C2 H3 O2 = C4 H6 O4 Thus the molecular formula is C4 H6 O4. Example 16: Acetylene & butene have empirical formula CH & CH2 respectively. The molecular mass of acetylene and butene are 26 & 56 respectively. Deduce their molecular formula. Sol. n= Molecular mass Empirical formula mass For Acetylene : n = 26 =2 13 Molecular formula = C2H2 56 For Butene: n= =4 14 Molecular formula = C4H8 Example 17: An oxide of nitrogen gave the following percentage composition by mass: N = 25.94 and O = 74.06 Calculate the empirical formula of the compound. (a) NO2 (b) N2O4 (c) N2O5 (d) N2O Some Basic Concepts of Chemistry So empirical formula is N2O5. 15. A compound contains 25% hydrogen and 75% carbon by mass. Determine the empirical formula of the compound. (a) CH4 (b) C2H6 (c) C3H8 (d) C2H2 16. The empirical formula of a compound of molecular mass 120 u is CH2O. The molecular formula of the compound is: (a) C2H4O2 (b) C4H8O4 (c) C3H5C3 (d) All of these 17. Calculate the molecular formula of compound which contains 20% Ca and 80% Br (by wt.) if molecular weight of compound is 200 u. (Atomic wt. Ca = 40. Br = 80) (a) Ca1/2Br (b) CaBr2 (c) CaBr (d) Ca2Br CONCENTRATION OF SOLUTION Concentration of a solution can be expressed in any of the following ways. (i) % by wt.: Amount of solute (in g) dissolved in 100 gm of solution. 4.9% H2SO4 by wt. 100 gm of solution contains 4.9 gm of H2SO4. (ii) % by volume: Volume of solute (in ml) dissolved in 100 ml of solution. x% H2SO4 by volume 100 ml of solution contains x ml H2SO4. (iii) % wt. by volume: wt. of solute (in g) present in 100 ml of solution. CONCENTRATION TERMS Molarity (M): No. of moles of solute present in 1000 ml of solution. Moles of solute Molarity (M) = Volume of solution (L) 55 Molality (m): No. of moles of solute present in 1000 gm of solvent. Moles of solute m= wt. of solvent in kg Mole fraction: The mole fraction of a particular component in a solution is defined as the number of moles of that component per mole of solution. If a solution has nA mole of A & nB mole of B. nA mole fraction of A (XA) = nA + nB nB mole fraction of B (XB) = n + n A B XA + XB = 1 Parts per million (ppm): Mass of solute Mass of solute 106 = 106 Mass of solvent Mass of solution CONVERSION OF CONCENTRATION TERMS 1. Molarity and % solute by mass: Let d = density of solution in g/mL and let it contains x% (w/w) solute by x d 10 mass. M = mA Specific gravity has no units and its numerical value equals density in g/mL. 2. Molality and mole fraction: Consider a binary solution consisting of two components A (Solute) and B (Solvent). Let XA & XB are the mole fraction of A & B respectively. nA nB XA = , XB = nA + nB nA + nB If molality of solution be m then: nA nA = m 1000 = 1000 mass of solvent nB MB where MB is the molecular wt. of the solvent B. mole fraction of A 1000 X A 1000 = m m= mole fraction of B M B XB MB m mole fraction of solute 1000 mole fraction of solvent molecular wt. of solvent 3. Mole fraction of solute into molarity of solution X 2 d 1000 M= X1 M1 + M 2 X 2 Mole fraction of solvent and solute are X1 and X2 so X1 + X2 = 1 Suppose total mole of solution is = 1 then mole of solute and solute and solvent are X2 & X1 respectively weight of solute = X2M2, weight of solvent = X1M1 & total wt of solution = X1M1 + X2M2 volume of solution = molarity (M) = 56 X1M1 + X 2 M 2 X M + X2M2 ml = 1 1 L d d 1000 X 2 d 1000 X1M1 + X 2 M 2 4. Molarity into mole fraction X2 = 1000M/ [1000d MM2] Molarity = M moles solute in 1000 ml of solution So, moles of solute = M & mass of solution = d 1000 wt. of solute = MM2 & wt. of solvent = 1000d MM2 Where M2 is molar mass of solute mole fraction of solute = 1000M / [1000d MM2] mM1 5. Molality into mole fraction X2 = 1000 + mM1 Molality = moles of solute in 1000 gm of solvent = m 1000 where M1 is molar mass of solvent moles of solvent = M1 mM1 m = mole fraction X2 = 1000 1000 + mM1 +m M1 md 1000 6. Molality into molarity M = 1000 + mM 2 Molality = m moles of solute in 1000 gm of solvent mole of solute = m & weight of solute = mM2 Weight of solution = 1000 + mM2 1000 + mM 2 volume of solution = mL = 1000 + mM 2 L d d 1000 m d 1000 molarity = 1000 + mM 2 M 1000 7. Molarity into Molality m = 1000d - MM 2 M1 and M2 are molar masses of solvent and solute. Molarity = M mole of solute in 1000 ml of solution moles of solute = M & weight of solute = MM2 weight of solution = 1000d mass of solvent = 1000d MM2 M 1000 molality = 1000d MM 2 M 1 d M + 2 = on simplifying m 1000 DILUTION & MIXING OF TWO LIQUIDS Upon dilution no. of moles of solute remains constant. If a particular solution having volume V1 mL and molarity M1 is diluted upto volume V2 mL. M1V1 = M2V2 M2 : final molarity If a solution having volume V1 and molarity M1 is mixed with another solution of same solute having volume V2 & molarity M2 then M1V1 + M2V2 = MR (V1 + V2) M V +M 2 V2 MR = Resultant molarity = 1 1 V1 +V2 Key Note P Molality is the most convenient method to express the concentration because it involves the mass of liquids rather than their volumes. It is also independent of the variation in temperature. P W JEE (XI) Module-1 CHEMISTRY P P P P P P No. of moles of solute = Molarity volume of solution (in L) No. of millimoles of solute = Molarity volume of solution (in mL) 1 mole = 1000 millimole All those concentration terms which does not involve volume terms e,g., ppm, mass %, molality, mole fraction are independent of temperature of the solution. For dilue aqueous solution, molarity = molality Concentration of solids or pure liquids is constant. (C = n/V = d/M) Example 21: 117 g NaCl is dissolved in 500 ml aqueous solution. Find the molarity of the solution. 117 / 58.5 = 4M. 500 / 1000 Example 22: Calculate the resultant molarity of following: Sol. Molarity = (a) 200 ml 1M HCl + 300 ml water (b) 1500 ml 1M HCl + 18.25 g HCl (c) 200 ml 1M HCl + 100 ml 0.5 M H2SO4 (d) 200 ml 1M HCl + 100 ml 0.5 M HCl 200 1 + 0 = 0.4M. 200 + 300 18.25 1000 1500 1 + 36.5 = 1.33M (b) Final molarity = 1500 200 1 + 100 0.5 2 (c) Final molarity of H+ = = 1M 200 + 100 Sol. (a) Final molarity = Example 18: 0.2 mole of HCl and 0.1 mole of barium chloride were dissolved in 0.5 L of water to produce solution. The molarity of the Cl ions is: (a) 0.06 M (b) 0.09 M (c) 0.12 M (d) 0.80 M Cl Sol. HCl 0.2 mole BaCl2 2 Cl 2 0.1 = 0.2 w 1000 Total moles of Cl = 0.4 M = m v 0.4 1000 w Molarity = = 0.8 = 0.4 m 500 Example 19: 149 g of potassium chloride (KCl) is dissolved in 10 L of an aqueous solution. Determine the molarity of the solution. (K = 39, Cl = 35.5) (a) 0.2 M (b) 0.4 M (c) 0.5 M (d) 2.2 M Sol. Molecular mass of KCl = 39 + 35.5 = 74.5 g 149g =2 Moles of KCl = 74.5g 2 Molarity of the solution = = 0.2M 10 Example 20: 255 g of an aqueous solution contains 5 g of urea. What is the concentration of the solution in terms of molality? (Mol. wt. of urea = 60) (a) 0.222 m (b) 0.333 m (c) 2.22 m (d) 3.33 m Sol. Mass of urea = 5 g Molecular mass of urea = 60 5 Number of moles of urea = = 0.083 60 Mass of solvent = (255 5) = 250 g Molality of the solution Number of moles of solute 1000 = Mass of solvent in gram 0.083 1000 = 0.332 m = 250 Some Basic Concepts of Chemistry 200 1 + 100 0.5 = 0.83M. 200 + 100 Example 23: 518 g of an aqueous solution contains 18 g of glucose (mol.wt. = 180). What is the molality of the solution? Sol. wt. of solvent = 518 18 = 500 g. (d) Final molarity = So molarity = 18 / 180 = 0.2M 500 / 1000 Example 24: 0.25 g of a substance is dissolved in 6.25 g of a solvent. Calculate the percentage amount of the substance in the solution. Sol. Wt. of solution = 0.25 + 6.25 = 6.50 So % (w/w) = 0.25 100 = 3.8% 6.50 Example 25: An aqueous solution is 1.33 molal in methanol. Determine the mole fraction of methanol & H2O. Sol. Molality = Mole fraction of solute 1000 Mole fraction of solvent mol.wt. of solvent 1.33= XA 1.33 18 X A 23.94 X A = 1000 ; = ; XB MB 1000 X B 1000 X B XA = 0.02394 XB, XA + XB = 1 1.02394 XB = 1 X B = 1 = 0.98, XA = 0.02 1.02394 2nd Method: Let wt. of solvent = 1000 gm, molality = 1.33 means 1.33 moles of solute are present in 1000 g solvent 57 mole fraction of solute moles of solute , = moles of solute + moles of solvent = m 1.33 = 1000 1.33 + (1000 / 18) m+ 18 Mole fraction of solute = 0.02 Mole fraction of solvent = 1 0.02 = 0.98 Example: When potassium chlorate (KClO3) is heated it gives potassium chloride (KCl) and oxygen (O2). KCl + O2 (unbalanced chemical equation ) KClO3 2KCl + 3O2 (balanced chemical equation) 2KClO3 Attributes of a balanced chemical equation: (a) It contains an equal number of atoms of each element on both sides of equation. (b) It should follow law of charge conservation on either side. (c) Physical states of all the reagents should be included in brackets. 18. If 500 ml of 1 M solution of glucose is mixed with 500 ml of 1 M solution of glucose, final molarity of solution will be: (a) 1 M (b) 0.5 M (c) 2 M (d) 1.5 M 19. The volume of water that must be added to a mixture of 250 ml of 0.6 M HCl and 750 ml of 0.2 M HCl to obtain 0.25 M solution of HCl is: (a) 750 ml (b) 100 ml (c) 200 ml (d) 300 ml 20. The molarity of Cl in an aqueous solution which was (w/v) 2% NaCl, 4% CaCl2 and 6% NH4Cl will be: (a) 0.342 (b) 0.721 (c) 1.12 (d) 2.18 21. 2M of 100 ml Na2SO4 is mixed with 3M of 100 ml NaCl solution and 1 M of 200 ml CaCl2 solution. Then the ratio of the concentration of cation and anion. (a) 1/2 (b) 2 (c) 1.5 (d) 1 22. Equal moles of H2O and NaCl are present in a solution. Hence, molality of NaCl solution is : (a) 0.55 (b) 55.5 (c) 1.00 (d) 0.18 23. Mole fraction of A in H2O is 0.2. The molality of A in H2O is: (a) 13.9 (b) 15.5 (c) 14.5 (d) 16.8 24. What is the molarity of H2SO4 solution that has a density of 1.84 g/cc and contains 98% by mass of H2SO4? (Given atomic mass of S = 32) (a) 4.18 M (b) 8.14M (c) 18.4 M (d) 18 M 25. The molarity of the solution containing 2.8% (mass/ volume) solution of KOH is : (Given atomic mass of K = 39) is : (a) 0.1 M (b) 0.5 M (c) 0.2 M (d) 1 M STOICHIOMETRY BASED CONCEPT (PROBLEMS BASED ON CHEMICAL REACTION) All chemical reaction are represented by chemical equations by using chemical formulae of reactants and products. Qualitatively a chemical equation simply describes what the reactants and products are. However, a balanced chemical equation gives us a lot of quantitative information mainly the molar ratio in which reactants combine and the molar ratio in which products are formed. 58 (d) All reagents should be written in their standard molecular forms (not as atoms) (e) The coefficients give the relative molar ratios of each reagent. Balancing a chemical equation Many chemical equations can be balanced by trial and error. Let us take the reactions of a few metals and non-metals with oxygen to give oxides For Example: Combustion reaction of C2H6 : C2H6 + O2 CO2 + H2O (skeleton equation) Balance carbon atoms: C2H6 + O2 2CO2 + H2O Now balance hydrogen atoms: C2H6 + O2 2CO2 + 3H2O 7 Now balance oxygen atoms: C2H6 + O2 2CO2 + 3H2O 2 Always remember that subscripts in formula of reactants and products cannot be changed to balance an equation. One of the most important aspects of a chemical equation is that when it is written in the balanced form, it gives quantitative relationships between the various reactants and products in terms of moles, masses, molecules and volumes. Mole - Mole Analysis This analysis is very much important for quantitative analysis point of view. Consider the decomposition of KClO3. 2KClO3 2KCl + 3O2 In very first step of mole-mole analysis you should read the balanced chemical equation like 2 moles KClO3 on decomposition gives you 2 moles KCl and 3 moles O2 and from the stoichiometry of reaction we can write Moles of KClO3 Moles of O 2 Moles of KCl = = 2 2 3 Now for any general balanced chemical equation like a A + b B c C + d D you can write. Moles of A reacted Moles of B reacted = a b = Moles of D produced Moles of C produced = d c P W JEE (XI) Module-1 CHEMISTRY Further, a balanced chemical equation along with the quantitative information conveyed by it is given below: CaCO3 + 2HCl CaCl2 + 1 Mole 2 Mole 1 Mole 40+12+3 16 2(1 + 35.5) 40+2 35.5 = 100 g = 73 g = 111 g H2O + CO2 1 Mole 1 Mole 2 1+16 12+2 16 = 18 g = 44 g or 22.4 L at STP Thus, (i) 1 mole of calcium carbonate reacts with 2 moles of hydrochloric acid to give 1 mole of calcium chloride, 1 mole of water and 1 mole of carbon dioxide. (ii) 100 g of calcium carbonate react with 73 g hydrochloric acid to give 111 g of calcium chloride, 18 g of water and 44 g (or 22.4 litres at STP) of carbon dioxide. Stoichiometry: + 3H2 2NH3 N2 1 mole 3 mole 2 mole 22.4 litre 3 22.4 litre 2 22.4 litre (at STP) 1 litre 3 litre 2 litre 1000 mL 3000 mL 2000 mL 1 mL 3 mL 2 mL 28 gm 6 gm 34 g (law of conservation of mass is followed). Mass can not be represented by stoichiometry. quantitative information conveyed by a chemical equation helps in a number of calculations. The problems involving these calculations may be classified into the following different types: The Type (I) Mass - Mass Relationships i.e. mass of one of the reactants or products is given and the mass of some other reactant or product is to be calculated. Mass - Mass Analysis Consider the reaction 2 KClO3 2KCl + 3O2 According to stoichiometry of the reaction mass-mass ratio = 2 122.5 : 2 74.5 : 3 32 or Mass of KClO3 Mass of KCl = 2 122.5 2 74.5 Mass of KClO3 2 122.5 = Mass of O 2 3 32 Illustration: Calculate the weight of iron which will be converted into its oxide by the action of 36 g of steam. (Given : 3Fe + 4H2O Fe3O4 + H2) Sol. Mole ratio of reaction suggests, Mole of Fe 3 = Mole of H 2 O 4 3 36 3 3 \ Mole of Fe = mol of H2O = = 4 18 2 4 3 wt. of Fe = 56 = 84 g 2 Some Basic Concepts of Chemistry Illustration: What amount of silver chloride is formed by the action of 5.850 g of sodium chloride on an excess of silver nitrate? Sol. Writing the balanced equation for the reaction NaCl 1 mol n= NaCl + AgNO3 1 mol AgCl 1 mol + NaNO3 1 mol Weight 5.85 = = 0.1 mol Mw 58.5 mole (NaCl) 1 = mole (AgCl) 1 Weight Weight = Mw 143.5 weight = 0.1 143.5 g = 14.35 g. Illustration: How much iron can be theoretically obtained in the reduction of 1 kg of Fe2O3? mole (AgCl) = 0.1 = Sol. Writing the balanced equation for the decomposition 3 reaction. Fe 2 O3 2Fe + O 2 2 Weight 1000 = mol Mw 160 mole (Fe 2 O3 ) 1 mole (Fe) = 2 = n Fe2O3 2 1000 =12.5 mol 160 Weight Weight = = Atomic weight 56 moles of Fe = Weight of iron obtained = 12.5 56 g = 700 g Type (II) Mass - Volume Relationships i.e. mass/volume of one of the reactants or products is given and the volume/mass of the other is to be calculated. 2KClO3 2KCl + 3O2 Mass volume ratio = 2 122.5 g : 2 74.5 g : 3 22.4 L at STP we can use two relation for volume of oxygen and Mass of KClO3 2 122.5g = (i) Volume of O 2 at STP 3 22.4 L 2 74.5g Mass of KCl = (ii) 3 22.4 L Volume of O 2 at STP Illustration: How much marble of 90.5% purity would be required to prepare 10 litres of CO2 at STP when the marble is acted upon by dilute HCl? Sol. CaCO3 + 2HCl CaCl2 + H2O + CO2 100 g 22.4litre 22.4 L of CO2 at STP will be obtained from 100 g of CaCO3 \ 10 L of CO 2 at STP will be obtained from pure 100 10 = 44.64 g CaCO3 = 22.4 100 \ Impure marble required = 44.64 = 49.326 g 90.5 59 Illustration: At 100 C for complete combustion of 3g ethane the required volume of O2 & produced volume of CO2 at STP will be? Sol. 2C2H6 + 7O2(g) 4CO2(g) + 6H2O(g) 2 7 4 6 Weight 3 1 n C2 H 6 = = = = 0.1 mol Mw 30 10 Step-I: Calculation of volume of CO2 from x litre of propane. C3H8 + 5O2 3CO2 + 4H2O x litre 3x litre Step-II: Calculation of volume of CO2 from (3 x) litre of butane. The combustion equation for butane is: 7 0.1 = 0.35 mol. 2 Required Volume of O2 at STP = 0.35 22.4 = 7.84 L. 13 C4H10 + 2 O2 4CO2 + 5H2O (3 x) litre 4(3 x) litre And produced moles of CO2 = Step-III: Calculation of composition of the mixture. Required moles of O2 = 4 0.1 = 0.2 mol. 2 Volume of CO2 obtained at STP = 0.2 22.4 = 4.48 L. Illustration: In the following reaction, if 10 g of H2 is reacted with N2, what will be the volume of NH3 produced at STP? N2 + 3H2 2NH3 Sol. N2 + 3H2 2NH3 = n H2 Weight 10 = = 5 mol. Mw 2 2 10 5 = . 3 3 10 22.4 = 74.67 L Volume of NH3 produced at STP = 3 Type (III) Volume - Volume Relationships i.e. volume of one of the reactants or the products is given and the volume of the other is to be calculated. Illustration: At 100 C for complete combustion of 1.12 litre of butane (C4H10), the produced volume of H2O(g) & CO2 at STP will be. Produced moles of NH3 = Sol. 1 13 4 5 2 C4H10(g) + 13 O2(g) 4CO2(g) + 5H2O(g) 2 1.12 litre Volume of H2O(g) at STP = 5 1.12 = 5.6 litre Volume of CO2(g) at STP = 4 1.12 = 4.48 litre Illustration: At 25 C for complete combustion of 5 mole propane (C3H8), the required volume of O2 at STP will be? Sol. For C3H8 , the combustion reaction is C3H8(g) + 5O2(g) 3CO2(g) + 4H2O(g) 5 mol V . 22.4 Volume of O2 gas at STP (V) = 25 22.4 = 560 L. Illustration: 3 litre of mixture of propane (C3H8) & butane (C4H10) on complete combustion give 10 litre CO2. Find the composition of the mixture. Sol. Let the volume of propane in the mixture = x litre, The volume of butane in the mixture = (3 x) litre Now let us calculate the volume of CO2 evolved with the help of chemical equation. Required moles of O2 = 5 5 = 25 mol = 60 Total volume of CO2 formed in the step (I) and step (II) = [3x + 4(3 x)] litre But the volume of CO2 actually formed = 10 litre 3x + 4(3 x) = 10 or 3x + 12 4x = 10 or x = 2 litre Volume of propane = x litre = 2 litre Volume of butane = (3 x) litre = (3 2) = 1 litre Example 26: Write a balanced chemical equation for the following reaction: When ammonia (NH3) decompose into nitrogen (N2) gas & hydrogen (H2) gas. 1 3 Sol. NH3 N 2 + H 2 or 2NH3 N2 + 3H2 2 2 Example 27: When 170 g NH3 (M =17) decomposes, how many grams of N2 & H2 is produced? 1 3 Sol. NH3 N 2 + H 2 2 2 moles of NH3 moles of N 2 moles of H 2 = = 1 1/ 2 3/ 2 1 170 So, moles of N2 = 5 = 2 17 So, wt. of N2 = 5 28 = 140 g 3 170 Similarly moles of H2 = 15 = 2 17 So, wt. of H2 = 15 2 = 30 g. Example 28: When 340 g NH3 (M = 17) decomposes, how many litres of nitrogen gas is produced at STP? (a) 2.24 L (b) 22.4 L (c) 224 L (d) None 1 3 Sol. NH3 N 2 + H 2 2 2 340 Moles of NH3 = = 20 17 1 So moles of N2 = 20 = 10 2 Vol. of N2 at STP = 10 22.4 = 224 L. P W JEE (XI) Module-1 CHEMISTRY PERCENTAGE YIELD The percentage yield of product 26. If 1.5 moles of dioxygen combine with Al to form Al2O3, the weight of Al used in the reaction is: (a) 27 g (b) 40.5 g (c) 54 g (d) 81 g 27. How many litres of CO2 at STP will be formed when 0.01 mol of H2SO4 reacts with excess of Na2CO3? Na2CO3 + H2SO4 Na2SO4 + CO2 + H2O (a) 22.4 L (b) 2.24 L (c) 0.224 L (d) 1.12L 28. How many moles of potassium chlorate need to be heated to produce 11.2 litre oxygen at S.T.P.? 3 KClO3 KCl + O 2 2 1 1 (a) mol (b) mol 2 3 1 2 (c) mol (d) mol 4 3 LIMITING REAGENT (L.R.) CONCEPT Quite often one of the reactants is present in larger amount than the other as required according to the balanced equation. The amount of the product formed then depends upon the reactant which has reacted completely. This reactant is called the limiting reactant. The excess of the other is left unreacted. Limiting Reagent (L.R.): The reactant which is completely consumed in a reaction is called as L.R. Calculation of Limiting Reagent: (a) By calculating the required amount by the equation and comparing it with given amount. [Useful when only two reactant are there] = Actual yield 100 theoretical maximum yield The actual amount of any limiting reagent consumed is given by (% yield given moles of limiting reagent). C + 3D , in this reaction Example 29: A + 5B 10mole 10mole which is a L.R.? Sol. For A For B 10 =2 5 10 = 10 1 2 < 10 So, B is L.R. 1 Example 30: H 2 (g) + O 2 (g) H 2 O(g) ; in the above 2 4g 32g reaction, what is the volume of water vapour produced at STP? (a) 4.48 L (b) 44.8 L (c) 2.24 L (d) 11.2 L Sol. H 2 (g) + 4g 1 O 2 (g) H 2 O(g) 2 32g For H2 For O2 4 n = = 2mol 2 2 = 2 1 n= 1 (b) By calculating amount of any one product obtained taking each reactant one by one irrespective of other reactants. The one giving least product is limiting reagent. For L.R. (c) Divide given moles of each reactant by their stoichiometric coefficient, the one with least ratio is limiting reagent. [Useful when number of reactants are more than two.] Ex. A + 2B C + 2D Given moles 3 9 0 0 3 3 9 6 0 3 3 6 A is L.R. Formula for checking L.R. = Both H2 & O2 are L.R. Given value (moles, volume, or molecules) Stoichiometry Coefficient Least value indicate the L.R. Ex. A B 9 3 = 4.5 =3 1 2 3 < 4.5 So, A is L.R Some Basic Concepts of Chemistry 1 2 32 = 1mol 32 = 2 mol Moles of H2O(g) produced = 2 mol = V 22.4 Volume of H2O(g) produced at STP = 22.4 2 = 44.8 litre Example 31: At STP, In a container 100 mL N2 and 100 mL of H2 are mixed together. Then find out the produced volume of NH3. (a) 6.66 mL (b) 66.6 mL (c) 3.33 mL (d) 5.55 mL Sol. Balanced equation will be N2 + 3H2 2NH3. Given 100mL 100mL For determination of Limiting reagent. Divide the given quantities by stoichiometry coefficients 100 100 = 100 = 33.3 (Limiting reagent) 3 1 61 In this reaction H2 is limiting reagent so reaction will proceed according to H2. According to stoichiometry from 3 mL of H2 produced volume of NH3 = 2 ml That is from 100 ml of H2 produced volume of 2 NH3 = 100 = 66.6 mL 3 Example 32: Number of moles of NH3 produced if 140 gm of N2 reacts with 40 gm of hydrogen. (Given % yield of reaction is 50%) (a) 12 (b) 10 (c) 5 (d) 6 50% Sol. N2 + 3H2 2NH3 140 gm 40 gm or 5 mol 20 mol Number of moles of NH3 produced = 5 2 0.5 = 5 mole 29. 4 mole of MgCO3 is reacted with 6 moles of HCl solution. Find the volume of CO 2 gas (in litres) produced at STP, the reaction is: MgCO3 + 2HCl MgCl2 + CO2 + H2O. (a) 11.2 (b) 22.4 (c) 67.2 (d) 44.8 30. For a reaction, N2(g) + 3H2(g) 2NH3(g); which of the following reaction mixtures has dihydrogen (H2) as a limiting reagent, : (a) 14g of N2 + 4g of H2 (b) 35g of N2 + 8g of H2 (c) 28g of N2 + 6g of H2 (d) 56g of N2 + 10g of H2 31. The percent yield for the following reaction carried out in carbon tetrachloride (CCl4) solution is 80%. Br2 + Cl2 2BrCl How many moles of BrCl is formed from the reaction of 0.025 mol Br2 and 0.025 mol Cl2? (a) 0.04 (b) 0.08] (c) 0.02 (d) 0.01 32. If 240 g of carbon is taken in a container to convert it completely to CO2 but in industry it has been found that 280 g of CO was also formed along with CO2. Find the mole percentage yield of CO2. The reactions occurring are: 1 C + O 2 CO 2 ;C + O 2 CO 2 (a) 25 62 (b) 50 (c) 75 (d) 35 ADVANCED LEARNING SEQUENTIAL REACTION Here, we solve problems in which the products of one reaction are used up in one or more subsequent reactions. In order to attempt such problems the following solving strategy has to be used. (i) The balanced and molecular equations are written for all reactions involved separately. (ii) Later the equations are multiplied, as a whole, by suitable factors, so that products of one reaction which are utilized in subsequent reactions are cancelled out. (iii) The final reaction obtained is used to find out the required quantities. Illustration: How many kilograms of pure H2SO4 could be obtained from one kilogram of pure iron pyrites (FeS2) according to the following reactions? 2Fe2O3 + 8SO2 4FeS2 + 11O2 2SO3 2SO2 + O2 SO2 + H2O H2SO4 (a) 0.184kg (b) 1.633kg (c) 2.643kg (d) 3.234kg Sol. Gram molecular weight of FeS2 = 120g Gram molecular weight of H2SO4 = 98 g Let us multiply the equations with suitable factors. 4FeS2 + 11 O 2 2Fe 2 O 3 + 8 SO 2 2SO 2 + O 2 2SO3 4 SO3 + H 2 O H 2SO 4 8 4FeS2 + 11O 2 2Fe 2 O3 + 8SO 2 8SO3 8SO 2 + 4O 2 8H 2SO 4 8SO3 + 8H 2 O 2Fe 2 O3 + 8H 2SO 4 4FeS + 15O 2 + 8H 2 O 2 From the above, it is clear that, 4 moles of FeS2 produces 8 moles of H2SO4 (or) produces 1 mole of FeS2 2 moles of H2SO4 produces 120g (1 mole) of FeS2 2 98 g (2 moles) of H2SO4 produces 1000g (l kg) of FeS2 1000 2 98 = 1633.33 g 120 or 1.633 kg of H2SO4 Therefore, 1.633 kg of H2SO4 is produced from 1 kg of iron pyrites. Illustration: 20 g of KC1O3 on heating give enough oxygen to react completely with hydrogen produced by the action of dil. H2SO4 on zinc. Find the weight of zinc required for the purpose. The reactions are as follows: (K = 39, Zn = 65, Cl = 35.5) P W JEE (XI) Module-1 CHEMISTRY 2KCl + 3O2 (i) 2KC1O3 ZnSO4 + H2 (ii) Zn + H2SO4 H2O (iii) H2 + O2 (a) 13.84g (b) 21.48g (c) 31.84g (d) 43.48g Sol. Gram molecular weight of KClO3 = 122.5 g Gram atomic weight of zinc = 65 g Let us multiply the equations with suitable factors. Zn + H 2SO 4 ZnSO 4 + H 2 6 H 2 + 1 O 2 H2O 6 2 2KClO3 2KCl + 3O 2 2KClO3 2KCl + 3O 2 6Zn + 6H 2SO 4 6ZnSO 4 + 6H 2 6H 2 + 3O 2 6H 2 O 2KClO3 + 6Zn + 6H 2SO 4 2KCl + 6ZnSO 4 + 6H 2 O From the above, it is clear that, 2 moles of KClO3 requires 6 moles of Zn (or) requires 1 mole of KClO3 3 moles of Zn requires 122.5 g (l mole) of KClO3 3 65 g (3 moles) of Zn 20 3 65 requires 20 g of KClO3 = 31.84 g of Zn 122.5 Therefore, 31.84 g of Zn is required for the given purpose. POAC POAC is based upon law of conservation of mass. Atoms are conserved, hence moles of atoms shall also be conserved in a chemical reaction (but not in nuclear reactions.) Consider the reaction: KClO3(s) KCl(s) + O2(g) (unbalanced chemical reaction) Apply POAC for K atoms. Moles of K atoms in reactant (KClO3) = moles of K atoms in product (KCl) Moles of K atoms in KClO3 = 1 moles of KClO3 and moles of K atoms in KCl = 1 moles of KCl. moles of KClO3 = moles of KCl wt. of KClO3 in g wt. of KCl in g or = mol. wt. of KClO3 mol. wt. of KCl Again, applying the POAC for O atoms, moles of O in KClO3 = 3 moles of KClO3 moles of O in O2 = 2 moles of O2 3 moles of KClO3 = 2 moles of O2 wt. of KClO3 vol. of O 2 at STP = 2 or 3 mol. wt. of KClO3 22.4 L Some Basic Concepts of Chemistry Illustration: 0.32 mole of LiAlH4 in ether solution was placed in a flask and 74 g (1 moles) of t-butyl alcohol was added. The product is LiAlHC12H27O3. Find the weight of the product if lithium atoms are conserved. [Li = 7, Al = 27, H = 1, C = 12, O = 16] Sol. Applying POAC for lithium atoms, 1 moles of LiAlH4 = 1 moles of LiAlH C12H27O3 weight of LiAlH C12 H 27 O3 254 wt. of LiAlH C12H27O3 = 81.28 g. 0.32 = 1 Illustration: 27.6 g K2CO3 was treated by a series of reagents so as to convert all of its carbon to K2Zn3 [Fe(CN)6]2. Calculate the weight of the product. [mol. wt. of K2CO3 = 138 and mol. wt. of K2Zn3 [Fe(CN)6]2 = 698] Sol. Here we have not knowledge about series of chemical reactions but we known about initial reactant and final product accordingly Several K2Zn3 [Fe(CN)6]2 K2CO3 Steps Since C atoms are conserved, applying POAC for C atoms, moles of C in K2CO3 = moles of C in K2Zn3 [Fe(CN)6]2 1 moles of K2CO3 = 12 moles of K2Zn3 [Fe(CN)6]2 wt.of K 2 CO3 wt.of the product = 12 mol.wt.of K 2 CO3 mol.wt.of product wt. of K2Zn3 [Fe(CN)6]2 = 27.6 698 = 11.6 g 138 12 STRENGTH (LABELLING) OF OLEUM Oleum is SO3 dissolved in 100% H2SO4. Sometimes, oleum is reported as more than 100% by weight, say y% (where y > 100). This means that (y - 100) grams of water, when added to 100 g of given oleum sample, will combine with all the free SO3 in the oleum to give 100% sulphuric acid. Hence, weight % of free SO3 in oleum = 80 (y 100)/18. Example: If in a sample of oleum, mole fraction of SO3 is 0.5. Label the oleum sample. Sol. Total moles = 1 Moles of SO3 = mole of H2SO4 = 0.5 Total Mass of SO3 & H2SO4 = 40 + 49 = 89 gm SO3 + H2O H2SO4 0.5 0.5 Mass of H2O required = 0.5 18 = 9 gm 89 gm require 9 gm H2O 9 100 = 10.11 gm 100 gm require = 89 % Labelling = (100 + 10.11) = 110.11% 63 AARAMBH (SOLVED EXAMPLES) 1. 1.80 g of a certain metal burnt in oxygen gave 3.0 g of its oxide. 1.50 g of the same metal heated in steam gave 2.50 g of its oxide. The law shown by above data is: (a) Law of constant proportion (b) Law of multiple proportion (c) Law of reciprocal proportion (d) All of the above Sol. In the first sample of oxide, Weight of metal = 1.80 g; Weight of oxygen = (3.0 1.80) g = 1.2 g 5.23 gm glucose has = = 1.75 1022 molecules Therefore, option (b) is the correct answer. 5. A sample of (C 2H 6) ethane has the same mass as 107 molecules of methane. How many C2H6 molecules does the sample contain? (a) 5.34 106 (b) 1.26 108 6 (c) 4.26 10 (d) 6.022 106 wt of metal 1.80g == 1.5 wt of oxygen 1.2g Sol. Moles of CH4 = In the second sample of the oxide, Weight of metal = 1.50 g; Weight of oxygen = (2.50 - 1.50) g = 1 g Mass of CH4 = Thus, in both samples of the oxide, the proportions of the weights of the metal and oxygen are fixed. Hence the results follow the law of constant proportion. Therefore, option (a) is the correct answer. 2. Calculate the total charge present on 4.2 gm of N3 . (a) 8.67 104 C (b) 9.05 104 C 3 (c) 8.67 10 C (d) 7.67 104 C wt.in gm 4.2 Sol. Mole = = = 0.3 Ionic wt. 14 Total no. of ions = 0.3 NA ions. Total charge = 0.3 NA 3 1.6 10 19 C = 0.3 6.022 1023 3 1.6 10 19 = 8.67 104 C (a) 6.022 1023 carbon atoms. (b) 1.26 1023 carbon atoms. (c) 1.26 1024 carbon atoms. (d) 6.022 1024 carbon atoms. Sol. 1 mol of C6H12O6 has 6 NA atoms of C 0.35 mol of C6H12O6 has 6 0.35 NA atoms of C = 2 .1 NA atoms = 2.1 6.022 1023 = 1.26 1024 carbon atoms. Therefore, option (c) is the correct answer. 4. How many molecules are present in 5.23 gm of glucose (C6H12O6)? (c) 1.75 1021 (d) None of these Sol. 180 gm glucose has = NA molecules 64 107 16 = mass of C2H6 NA 107 16 N A 30 So no. of molecules of C2H6= 107 16 NA= 5.34 106. N A 30 Therefore, option (a) is the correct answer. 6. From 160 g of SO2 (g) sample, 1.2046 1024 molecules of SO2 are removed then find out the volume of left over SO2 (g) at STP. (a) 11.2 L (b) 12.5 L (c) 9.5 L (d) 10.8 L 160 Sol. Given moles = = 2.5. 64 Removed moles = Therefore, option (a) is the correct answer. 3. How many carbon atoms are present in 0.35 mol of C6H12O6? (a) 1.65 1022 (b) 1.75 1022 107 NA So Moles of C2H6 = wt of metal = 1.5 wt of oxygen 5.23 6.022 1023 180 1.2046 1024 So left moles = 0.5. 6.022 1023 = 2. Volume left at STP = 0.5 22.4 = 11.2 L. Therefore, option (a) is the correct answer. 7. 14 g of Nitrogen gas and 22 g of CO2 gas are mixed together. Find the volume of gaseous mixture at STP. (a) 10.2 L (b) 12.2 L (c) 15.5 L (d) 22.4 L 14 Sol. Moles of N2 = = 0.5. 28 22 Moles of CO2 = = 0.5. 44 So total moles = 0.5 + 0.5 = 1. So vol. at STP = 1 22.4 = 22.4 L. Therefore, option (d) is the correct answer. 8. How many years it would take to spend Avogadro s number of rupees at the rate of 1 million rupees per second? (a) 19.098 1019 years (b) 19.098 years (c) 19.098 109 years (d) None of these P W JEE (XI) Module-1 CHEMISTRY Sol. 106 rupees are spent in 1 sec. 6.022 1023 rupees are spent in 1 6.022 1023 sec 106 1 6.022 1023 or 6 years = 19.098 109 years 10 60 60 24 365 Therefore, option (c) is the correct answer. 9. The density of O2 at STP is 1.429g/litre. Calculate the standard molar volume of gas. (a) 22.4 lit. (b) 11.2 lit (c) 33.6 lit (d) 5.6 lit. Sol. 1.429 gm of O2 gas occupies volume = 1 litre. 32 32 gm of O2 gas occupies = = 22.4 litre/mol. 1.429 Therefore, option (a) is the correct answer. 10. Calculate the weight of lime (CaO) obtained by heating 200 kg of 95% pure lime stone (CaCO3). (a) 104.4 kg (b) 105.4 kg (c) 212.8 kg (d) 106.4 kg Sol. 100 kg impure sample has pure CaCO3 = 95 kg 200 kg impure sample has pure CaCO3 95 200 = = 190 kg. 100 CaCO3 CaO + CO2 100 kg CaCO3 gives CaO = 56 kg. 56 190 190 kg CaCO3 gives CaO = = 106.4 kg. 100 Therefore, option (d) is the correct answer. 11. A compound containing beryllium has the following composition, Be = 6.1%, N = 37.8%, Cl=48%, H = 8.1%. One mole of the compound has mass of 148g and average atomic mass of beryllium is 9. The molecular formula of the compound is: (a) BeN4H12Cl2 (b) BeN2H10Cl (c) BeN4H2Cl3 (d) Be2N4H10Cl2 Sol. Element % %/A Simplest ratio Be 6.1 6.1/9 = 0.677 1 N 37.8 37.8/14 = 2.7 4 Cl 48 48/35.5 = 1.35 2 H 8.1 8.1/1 = 8.1 12 Empirical formula = BeN4Cl2H12 = 9 + 56 + 71 +12 = 148 n = 1 Molecular formula = BeN4Cl2H12 Therefore, option (a) is the correct answer. 12. One litre of a mixture of CO and CO2 is passed through red hot charcoal in tube. The new volume becomes 1.4 litre. Find out % composition of mixture by volume. All measurements are made at same P and T. Some Basic Concepts of Chemistry (a) CO2 40%, CO 60% (b) CO2 60%, CO 40% (c) CO2 25%, CO 75% (d) CO2 30%, CO 70% Sol. On passing through charcoal only CO2 reduces to CO. CO + C No reaction Volume a CO2 + C 2 CO Volume before reaction b 0 Volume after reaction 0 2b As given a + b = 1 and a + 2b = 1.4 0.4 b = 0.4 litre % of b = 100 = 40 % 1 0.6 a = 0.6 litre % of a = 100 = 60 % 1 Therefore, option (a) is the correct answer. 13. Calculate the molarity of H+ ion in the resulting solution when 200 ml 0.5M HCl is mixed with 200 ml 0.5M H2SO4 Sol. n H+ = 0.1 (from HCl) & n H+ = 0.2 (from H2SO4) Total H+ = n H+ (from HCl) + n H+ (from H2SO4) = 0.1 + 0.2 = 0.3 Total volume = 200 + 200 = 400 mL = 0.4 L n H+ 0.3 = = 0.75 M Ans. MR = Resultant molarity = Vsolution 0.4 Therefore, [0.75] is the correct answer. 14. What are the final concentration of all the ions when following are mixed 50 ml of 0.12 M Fe(NO3)3 + 100 ml of 0.1 M FeCl3 + 100 ml of 0.26 M Mg(NO3)2. 50 0.12 3 + 100 0.26 2 18 + 52 70 Sol. [NO3 ] = = = = 0.28 250 250 250 [Cl ] = 0.12 M; [Mg++] = 0.104 M; [Fe3+] = 0.064 M Therefore, [0.064] is the correct answer. 15. A sample of 3 g containing Na2CO3 and NaHCO3 loses 0.248 g when heated to 300 C, the temperature at which NaHCO3 decomposes to Na2CO3, CO2 and H2O. What is the percentage of Na2CO3 in the given mixture? Sol. The loss in weight is due to removal of CO2 and H2O which escape out on heating. wt. of Na2CO3 in the product = 3.00 0.248 = 2.752 g Let wt. of Na2CO3 in the mixture be x g \ wt. of NaHCO3 = (3.00 x) g Since Na2CO3 in the products contains x g of unchanged reactant Na2CO3 and rest produced from NaHCO3. The wt. of Na2CO3 produced by NaHCO3 = (2.752 x)g Na2CO3 + (H2O + CO2) NaHCO3 (3.0 x) (2.752 x) Applying POAC for Na atom 1 moles of NaHCO3 = 2 moles of Na2CO3 (2.752 x ) (3 x) =2 106 84 \ x = 2.328 g 2.328 100 = 77.6 % \ % of Na2CO3 = 3 Therefore, [77.6] is the correct answer. 65 SCHOOL LEVEL PROBLEMS SINGLE CORRECT TYPE QUESTIONS 1. 6.02 1020 molecules of urea are present in 100 mL of its solution. The concentration of the solution is (a) 0.02 M (b) 0.01 M (c) 0.001 M (d) 0.1 M 2. 10 mol of Zn react with 10 mol of HCl. Calculate the number of moles of H2 produced. (a) 5 mol (b) 10 mol (c) 20 mol (d) 2.5 mol 3. One mole of any substance contains 6.022 1023 atoms/ molecules. Number of molecules of H2SO4 present in 100 mL of 0.02M H2SO4 Solution is (a) 12.044 1020 molecules (b) 6.022 1023 molecules (c) 1 1023 molecules (d) 12.044 1023 molecules 4. What is the mass percent of carbon in carbon dioxide? (a) 0.034% (b) 27.27% (c) 3.4% (d) 28.7% 5. Which of the following statements about a compound is incorrect? 8. A gaseous hydrocarbons gives upon combustion, 0.72 g of water and 3.08 g of CO2. The empirical formula of the hydrocarbon is: (a) C6H5 (b) C7H8 (c) C2H4 (d) C3H4 9. Assertion: No. of moles of H2 in 0.224 L of hydrogen is 0.01 mole. Reason: 22.4 L of H2 at STP contain 6.022 1023 moles. (a) Both Assertion and Reason are true and Reason is correct explanation of Assertion. (b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion. (c) Assertion is true but Reason is false. (d) Assertion is false but Reason is true. 10. Assertion: The empirical mass of ethene is half of its molecular mass. Reason: The empirical formula represents the whole number ratio of various atoms present in a compound. (a) Both Assertion and Reason are true and Reason is correct explanation of Assertion. (a) A molecule of a compound has atoms of different elements. (b) Both Assertion and Reason are true but Reason is not a correct explanation of Assertion. (b) A compound cannot be separated into its constituent elements by physical methods of separation. (c) Assertion is true but Reason is false. (c) A compound retains the physical properties of its constituent elements. (d) The ratio of atoms of different elements in a compound is fixed. 6. Which of the following statements is correct about the reaction given below: 4Fe(s) + 3O2(g) 2Fe2O3(g) (a) Total mass of iron and oxygen in reactants = total mass of iron and oxygen in product therefore it follows law of conservation of mass. (b) Total mass of reactants = total mass of product; therefore, law of multiple proportions is followed. (c) Amount of Fe2O3 can be increased by taking any one of the reactants (iron or oxygen) in excess. (d) Amount of Fe2O3 produced will decrease if the amount of any one of the reactants (iron or oxygen) is taken in excess. 7. Number of atoms of He in 100 u of He ( Atomic mass of He is 4 u) (a) 25 (b) 50 (c) 100 (d) 400 66 (d) Assertion is false but Reason is true. MATCH THE COLUMN TYPE QUESTIONS 11. Match the following: List-I List-II (i) Molarity p. For very dilute solution (ii) Molality q. No units (iii) mole fraction r. Mol L 1 (iv) ppm s. independent of temperature 12. Match the following: List-I List-II (i) 88 g of CO2 6 . 0 2 2 1 0 23 (ii) molecules of H2O 5.6 litres of O2 at (iii) STP (iv) 96 g of O2 (v) 1 mol of any gas p. 0.25 mol q. 2 mol r. 1 mol s. 6.022 1023 molecules 3 mol P W JEE (XI) Module-1 CHEMISTRY SHORT ANSWER TYPE QUESTIONS 13. Volume of a solution changes with change in temperature, then, will be molality of the solution be affected by temperature? Give reason for your answer. 14. If 4 g of NaOH dissolves in 36 g of H2O, calculate the mole fraction of each component in the solution. Also, determine the molarity of solution (specific gravity of solution is 1g mL 1). 15. A solution is prepared by adding 2g of a substance A to 18 g of water. Calculate the mass per cent of the solute. 16. How many atoms and molecules are present in 124 gm of phosphorus (P4). 17. The cost of table salt ( NaCl ) is Rs. 10 per Kg. Calculate its cost per mole. (Molar mass of NaCl is 58.5 g mol 1) 18. Calculate the mole fraction of the solute in a 1.00 molal aqueous solution. LONG ANSWER TYPE QUESTIONS 19. A vessel contains 1.6 g of dioxygen at STP (273.15 K, 1 atm pressure). The gas is now transferred to another vessel at constant temperature, where pressure becomes half of the original pressure. Calculate (a) volume of the new vessel. (b) number of molecules of dioxygen. 20. Calcium carbonate reacts with aqueous HCl to give CaCl2 and CO2 according to the reaction given below: CaCO3(s) + 2HCl(aq) CaCl2(aq) + CO2(g) + H2O(l) What mass of CaCl2 will be formed when 250 mL of 0.76 M HCl reacts with 1000 g of CaCO3? Name the limiting reagent. Calculate the number of moles of CaCl2 formed in the reaction. 21. In a compound CxHyOz, the mass % of C and H is 6 : 1 and the amount of oxygen present is equal to the half of the oxygen required to react completely CxHy. Find the empirical formula of the compound. 22. An LPG cylinder weighs 14.8 Kg when empty. When full, it weighs 29.0 kg and shows a pressure of 2.5 atm. In the course of use at 27 C, the weight of cylinder is reduced to 23.2 Kg. Find the volume of n-butane in cubic meters used up at 27 C and 1 atm (Molecular weight of n-butane = 58). 23. A box contains some identical red coloured balls, labelled as A, each weighing 2 grams. Another box contains identical Some Basic Concepts of Chemistry bule coloured balls, labelled as B, each weighing 5 grams. Consider the combinations AB, AB2, A2B and A2B3 and show that law of multiple. CASE STUDY BASED QUESTIONS 24. The ideas underlying our modern understanding of thermodynamics and kinetic theory were developed during the nineteenth century. Central to these developments was the discovery that matter reacting chemically does not do so simply between equal masses of the samples involved. We now call the study of this phenomenon stoichiometry , defined as: the relationship between the amounts of substance that react together, and the products that are formed . Another development during the nineteenth century that was central to our modern understanding of the chemical nature of matter was the observation by Avogadro that equal volumes of ideal or perfect gases, at the same temperature and pressure, contain the same number of particles, or molecules . This is now known as Avogadro s law. It provides the motivation to formulate expressions for the quantity of a sample that reacts with another sample. The most notable example of such a formulation is the gram-molecule, which has been used to refer to both a unit and a quantity. The following questions are multiple choice questions. Choose the most appropriate answer: I. The concept of stoichiometry mentioned in the study is based on the (a) formation of chemical bonds. (b) amount of reactant and product involved in a chemical reaction. (c) idea of temperature and pressure required for the reaction to occur. (d) oxidation states of reactant and product involved. II. How much gram-molecules of H 2O are produced on combustion of 32 g of methane in excess oxygen? (a) 72 (b) 4 (c) 2 (d) 36 III. When an antacid tablet is used, Ca(OH)2 reacts with HCl in the stomach to form inert CaCl2 and H2O. If the molar mass of Ca(OH)2 is 75 g/mol, how many moles of HCl are required to fully react with 150 g of Ca(OH)2? (a) 4 (b) 1 (c) 8 (d) 2 IV. What must be held constant when applying Avogadro s law? (a) pressure and temperature (b) volume and temperature (c) moles and temperature (d) pressure and volume 67 PRARAMBH (TOPICWISE) FUNDAMENTAL QUANTITIES, LAWS OF CHEMICAL COMBINATION 1. Express the result of (0.582 + 324.65) to the appropriate number of significant figures: (a) 325.24 (b) 325.23 (c) 325.2 (d) 325.232 2. The correctly reported answer of the area of rectangle which is 12.34 cm long and 1.23 cm wide is : (a) 15.2 m2 (b) 15.2 cm2 (c) 15.1 cm 2 (d) 15.17 cm2 3. If an object has a mass of 0.2876 g, then find the mass of nine such objects: (a) 2.5884 g (b) 2.5886 g (c) 2.588 g (d) 2.5 g 4. Two elements X and Y combine in gaseous state to form XY in the ratio 1:35.5 by mass. The mass of Y that will be required to react with 2 g of X is: (a) 7.1 g (b) 3.55 g (c) 71 g (d) 35.5 g 5. 4.4 g of an oxide of nitrogen gives 2.24 L of nitrogen and 60 g of another oxide of nitrogen gives 22.4 L of nitrogen at S.T.P. The data illustrates: (a) Law of conservation of mass (b) Law of constant proportions (c) Law of multiple proportions (d) Law of reciprocal proportions 6. Two elements X and Y combine to form compounds A, B and C. The ratio of different masses of Y which combine with a fixed mass of X in A, B and C is 1 : 3 : 5. If 32 parts by mass of X combines with 84 parts by mass of Y in B, then in C, 16 parts by mass of X will combine with; (a) 14 parts by mass of Y (b) 42 parts by mass of Y (c) 70 parts by mass of Y (d) 84 parts by mass of Y ATOMIC MASS & MOLECULAR MASS, MOLE CONCEPT AND APPLICATIONS 7. 1 amu is equal to: 1 of mass of C12 atom (a) 12 1 of mass of O16 atom 14 (c) 1 g of H2 (d) 1.66 10 23 kg (b) 8. 1 mol of CH4 contains: (a) 6.02 1023 atoms of H (b) 4 g-atom of Hydrogen (c) 1.81 1023 molecules of CH4 (d) 3.0 g of carbon 68 9. 7.5 grams of a gas occupy 5.6 litres of volume at STP, the gas is: (a) NO (b) N2O (c) CO (d) CO2 10. The number of atoms in 4.25 g of NH3 is approximately: (a) 1 1023 (b) 2 1023 (c) 4 1023 (d) 6 1023 11. One litre of a gas at STP weighs 1.16 g. The possible gas is: (b) CO (a) C2H2 (d) CH4 (c) O2 12. The mass of a molecule of water is approximately: (b) 3 10 25 kg (a) 3 10 26 kg (d) 2.5 10 26 kg (c) 1.5 10 26 kg 13. If N A is Avogadro s number, then number of valence electrons in 4.2 g of nitride ions (N3 ) is: (b) 4.2 NA (a) 2.4 NA (c) 1.6 NA (d) 3.2 NA 14. The number of molecules at STP in 1 ml of an ideal gas will be: (b) 2.69 1019 (a) 6 1023 23 (c) 2.69 10 (d) None of these 15. Volume of a gas at STP is 1.12 10 7 cc. The number of molecules in it are: (b) 3.01 1012 (a) 3.01 1020 23 (c) 3.01 10 (d) 3.01 1024 16. 4.4 g of an unknown gas occupies 2.24 L of volume at standard temperature and pressure. The gas may be: (a) Carbon dioxide (b) Carbon monoxide (c) Oxygen (d) Sulphur dioxide 17. The number of oxygen atoms in 4.4 g of CO2 is approx.: (b) 6 1022 (a) 1.2 1023 (c) 6 1023 (d) 12 1023 18. The total number of protons in 10 g of calcium carbonate is: (NA = 6.022 1023) (b) 2.0478 1024 (a) 1.5057 1024 (c) 3.0115 1024 (d) 4.0956 1024 19. Number of molecules in 100 ml each of O2,NH3 and CO2 at STP are: (a) In the order: CO2 < O2 < NH3 (b) In the order: NH3 < O2 < CO2 (c) The same in all (d) In the order: NH3 < CO2 < O2 20. The number of water molecules in 1 litre of water is: (a) 18 (b) 18 1000 (d) 55.55 NA (c) NA 21. 2 g of oxygen contains number of atoms equal to that in: (a) 0.5 g of hydrogen (b) 4 g of sulphur (c) 7 g of nitrogen (d) 2.3 g of sodium P W JEE (XI) Module-1 CHEMISTRY PERCENTAGE COMPOSITION, EMPIRICAL FORMULA & MOLECULAR FORMULA 22. Caffeine has a molecular weight of 194. If it contains 28.9% by mass of nitrogen, number of atoms of nitrogen in one molecule of caffeine is: (a) 4 (b) 6 (c) 2 (d) 3 23. The percentage of oxygen in NaOH is: (a) 40 (b) 60 (c) 8 (d) 10 24. A compound (60 g) on analysis gave C = 24 g, H = 4 g, O = 32 g. Its empirical formula is: (a) C2H2O2 (b) C2H2O (c) CH2O2 (d) CH2O 25. What is the % of H2O in Fe(CNS)3.3H2O? (a) 45 (b) 30 (c) 19 (d) 25 26. A hydrocarbon contains 86% carbon, Then, the hydrocarbon is an: (a) Alkane (b) Alkene (c) Alkyne (d) Arene 27. The simplest formula of a compound containing 50% of element X (atomic mass 10) and 50% of element Y (atomic mass 20) is: (a) XY (b) X2Y (c) XY3 (d) X2Y3 28. Haemoglobin contains 0.33% of iron by weight. The molecular weight of haemoglobin is approximately 67,200. The number of iron atoms (At. wt. of Fe = 56) present in one molecule of haemoglobin is: (a) 6 (b) 1 (c) 4 (d) 2 CONCENTRATION TERMS 29. What is the molarity of NaOH solution if 250 mL of it contains 1 mg of NaOH? (a) 10 1 M (b) 10 2 M (c) 10 4 M (d) 10 3 M 30. Molarity of H2SO4 (density 1.8 g/mL) is 18 M. The molality of this H2SO4 is: (a) 36 (b) 200 (c) 500 (d) 18 31. The percentage of sodium in a breakfast cereal be labeled as 110 mg of sodium per 100 g of cereal is (a) 11% (b) 1.10% (c) 0.11% (d) 1.10% 32. 200 ml, 3M NaOH is mixed with 300 ml 2 M NaOH forming a solution of density 1200 Kg/m3 . Then molality of final solution is : (a) 2.2 m (b) 3 m (c) 2.4 m (d) 2 m 33. Calculate molality of 0.2 M urea solution having density 1.2 gm/ml. 1 1 1 1 (a) m (b) m (c) m (d) m 6 3 9 12 34. Commercial HNO2(aq) has density 1.41 g/ml and is 10M. Then mass percentage of solution is : 100 % (b) 50% (a) (c) 75% (d) 25% 3 35. 0.02 gm of an unknown substance is dissolved in 3.9 gm of benzene. The molality of solution is 0.08 m.Calculate the molecular mass of unknown substance: Some Basic Concepts of Chemistry (a) 60 g/mol (c) 264 g/mol (b) 164 g/mol (d) 64.1 g/mol 36. A 500 g solution has 0.1 g fluoride concentration. Then fluoride concentration in term of ppm will be: (a) 200 ppm (b) 100 ppm (c) 400 ppm (d) 50 ppm STOICHIOMETRY, EQUATION BASED CALCULATIONS 37. In the reaction; 4NH3(g) + 5O2 (g) 4NO(g) + 6H2O(g), when 1 mole of ammonia and 1 mole of O 2 are made to react to completion: (a) 1.0 mole of H2O is produced. (b) 1.0 mole of NO will be produced. (c) All the oxygen will be consumed. (d) All the ammonia will be consumed. 38. H2 evolved at STP on complete reaction of 27 g of aluminium with excess of aqueous NaOH would be: 3 Al + H 2 O + NaOH NaAlO 2 + H 2 2 (a) 22.4 litres (b) 44.8 litres (c) 67.2 litres (d) 33.6 litres 39. 12 g of Mg (at. mass 24) will react completely with acid to give: (a) One mole of H2 (b) 1/2 mole of H2 (c) 2/3 mole of O2 (d) Both 1/2 mol of H2 and 1/2 mol of O2 40. 100 g CaCO3 reacts with 1 litre 1 N HCl. On completion of reaction, how much weight of CO2 will be obtained? (a) 5.5 g (b) 11 g (c) 22 g (d) 33 g 41. What weight of HNO3 is needed to convert 5 g of iodine into iodic acid according to the reaction, I2 + 10HNO3 2HIO3 + 10NO2 + 4H2O (a) 12.4 g (b) 24.8 g (c) 0.248 g (d) 49.6 g 42. How much Cl2 at STP is liberated when 1 mole KMnO4 reacts with HCl? (a) 11.2 L (b) 22.4 L (c) 44.8 L (d) 56 L 43. 27 g of Al will react completely with how many grams of oxygen? (a) 8 g (b) 16 g (c) 32 g (d) 24 g 44. If 0.50 mole of BaC12 is mixed with 0.20 mol of Na3PO4, the maximum number of moles of Ba3(PO4)2 that can be formed is: (a) 0.70 (b) 0.50 (c) 0.20 (d) 0.10 45. 0.5 mole of H2SO4 is mixed with 0.2 mole of Ca(OH)2. The maximum number of moles of CaSO4 formed is: (a) 0.2 (b) 0.5 (c) 0.4 (d) 1.5 69 PRABAL (JEE MAIN LEVEL) 1. A sample of calcium phosphate Ca3(PO4)2 contains 8 mol of O atoms. The number of mole of Ca atoms in the sample is: (a) 4 (b) 1.5 (c) 3 (d) 8 2. Ratio of masses of H2SO4 and Al2(SO4)3 each containing 32 grams of S is __________. (a) 0.86 (b) 1.72 (c) 0.43 (d) 2.15 3. Which has maximum number of atoms of oxygen? (a) 10 ml H2O(l) (b) 0.1 mole of V2O5 (c) 12 gm O3(g) (d) 12.044 1022 molecules of CO2 4. Mass of one atom of the element A is 3.9854 10 23g. How many atoms are contained in 1g of the element A? (a) 2.509 1022 (b) 6.022 1023 (c) 12.044 1023 (d) None of these 5. The number of atoms present in 0.5 g-atoms of nitrogen is same as the atoms in: (a) 12 g of C (b) 32 g of S (c) 8 g of oxygen (d) 24 g of Mg 6. How many moles of magnesium phosphate Mg3(PO4)2 will contain 0.25 mole of oxygen atoms? (a) 0.02 (b) 3.125 10 2 (c) 1.25 10 2 (d) 2.5 10 2 7. 64 g of an organic compound has 24 g carbon and 8 g hydrogen and the rest is oxygen. The empirical formula of the compound is: (a) CH4O (b) CH2O (c) C2H4O (d) None of these 8. Two elements X (atomic mass = 75) and Y (atomic mass = 16) combine to give a compound having 75.8% of X. The formula of the compound is: (a) X2Y3 (b) X2Y (c) X2Y2 (d) XY 9. A definite amount of gaseous hydrocarbon was burnt with just sufficient amount of O 2. The volume of all reactants was 600 ml, after the explosion the volume of the products [CO2(g) and H2O(g)] was found to be 700 ml under the similar conditions. The possible molecular formula of the compound is: (a) C3H8 (b) C3H6 (c) C3H4 (d) C4H10 10. Mole fraction of ethyl alcohol in aqueous ethyl alcohol (C2H5OH) solution is 0.25. Hence, percentage of ethyl alcohol by weight is: (a) 54% (b) 25% (c) 75% (d) 46% 11. 74 gm of sample on complete combustion gives 132 gm CO2 and 54 gm of H2O. The molecular formula of the compound may be: 70 (a) C5H12 (b) C4H10O (c) C3H6O2 (d) C3H7O2 12. Weight of oxygen in Fe2O3 and FeO in the simple ratio for the same amount of iron, is: (a) 3 : 2 (b) 1 : 2 (c) 2 : 1 (d) 3 : 1 13. A person needs on average of 2.0 mg of riboflavin (vitamin B2) per day. How many gm of butter should be taken by the person per day if it is the only source of riboflavin? Butter contains 5.5 microgram riboflavin per gm. (a) 363.6 gm (b) 2.75 mg (c) 11 gm (d) 19.8 gm 14. The oxide of a metal contains 30% oxygen by weight. If the atomic ratio of metal and oxygen is 2 : 3, determine the atomic weight of metal. (a) 12 u (b) 56 u (c) 27 u (d) 52 u 15. When a mixture of 10 mole of SO2 and 15 mole of O2 was passed over catalyst, 8 mole of SO3 was formed. How many mole of SO2and O2 did not enter into combination? (a) 2 moles of SO2, 11 moles of O2 (b) 3 moles of SO2, 11.5 moles of O2 (c) 2 moles of SO2, 4 moles of O2 (d) 8 moles of SO2, 4 moles of O2 16. C6H5OH(g) + O2(g) CO2(g) + H2O(l) Magnitude of volume change if 30 ml of C6H5OH (g) is burnt with excess amount of oxygen, is: (a) 30 ml (b) 60 ml (c) 20 ml (d) 10 ml 17. Mass of sucrose C12H22O11 produced by mixing 84 gm of carbon, 12 gm of dihydrogen and 56 lit. O2 at 1 atm & 273 K according to given reaction, is: C(s) + H2(g) + O2(g) C12H22O11(s) (a) 138.5 (c) 172.5 (b) 155.5 (d) 199.5 18. What volume (in ml) of 0.2 M H2SO4 solution should be mixed with the 40 ml of 0.1 M NaOH solution such that the 6 M. resulting solution has the concentration of H2SO4 as 55 (a) 70 (b) 45 (c) 30 (d) 58 19. For the reaction; 2x + 3y + 4z 5w, initially 1 mol of x, 3 mol of y and 4 mol of z is taken. If 1.25 mol of w is obtained then % yield of this reaction is: (a) 50% (b) 60% (c) 70% (d) 40% P W JEE (XI) Module-1 CHEMISTRY 20. If 10 g of Ag reacts with 1 g of sulphur , the amount of Ag2S formed will be: (a) 7.75 g (b) 0.775 g (c) 11 g (d) 10 g 29. What volume of a 0.8 M solution contains 100 milli moles of the solute? 21. A solution of A (MM = 20) and B (MM = 10), [Mole fraction XB = 0.6] having density 0.7 gm/ml then molarity and molality of B in this solution will be ________ and ________ respectively. (a) 30 M, 75 m (b) 40 M, 75 m (c) 30 M, 65 m (d) 50 M, 55 m 30. 4.4 g of CO2 and 2.24 litre of H2 at STP are mixed in a container. The total number of molecules present in the container will be: 22. 125 ml of 8% w/w NaOH solution (sp. gravity = 1) is added to 125 ml of 10 % w/v HCl solution. The nature of resultant solution would be ________. (a) Acidic (b) Basic (c) Neutral (d) None of these 23. 36.5 % (w/w) HCl has density equal to 1.20 g mL 1. The molarity (M) and molality (m), respectively, are: (a) 15.7, 15.7 (b) 12, 12 (c) 15.7, 12 (d) 12, 15.7 24. An aqueous solution of ethanol has density 1.025 g/mL and it is 2 M. What is the molality of this solution? (a) 1.79 (b) 2.143 (c) 1.951 (d) None of these. 25. 500 mL of a glucose solution contains 6.02 1022 molecules of glucose. The concentration of the solution is: (a) 0.1 M (b) 1.0 M (c) 0.2 M (d) 2.0 M 26. Equal moles of H2O and NaCl are present in a solution. Hence, molality of NaCl solution is: (a) 0.55 (b) 55.5 (c) 1.00 (d) 0.18 27. Decreasing order of mass of pure NaOH in each of the aqueous solution. I. 50 g of 40% (w/w) NaOH II. 50 ml of 50% (w/v) NaOH (dsol = 1.2 g/ml). III. 50 g of 15 M NaOH (dsol = 1 g/ml). (a) I, II, III (b) III, II, I (c) II, III, I (d) III = II = I M , its molarity for Cl ion will 28. A solution of FeCl3 is 30 be: M M (b) (a) 90 30 M M (c) (d) 10 5 Some Basic Concepts of Chemistry (a) 100 mL (b) 125 mL (c) 500 mL (d) 62.5 mL (a) 6.022 1023 (b) 1.2046 1023 (c) 6.022 1022 (d) 6.022 1024 31. 10 g of CaCO3 on heating gives 5 g of the residue (as CaO). The percent yield of the reaction is approximately: (a) 50% (b) 72% (c) 89% (d) 100% 32. 33.6 g of an impure sample of sodium bicarbonate when heated strongly gave 4.4 g of CO2. The percentage purity of NaHCO3 would be: (a) 25% (b) 50% (c) 75% (d) 100% 33. A gaseous mixture contains oxygen and nitrogen in the ratio of 1 : 4 by weight. Therefore the ratio of their number of molecules is: (a) 1 : 4 (b) 1 : 8 (c) 7 : 32 (d) 3 : 16 34. 2.76 g of silver carbonate on being strongly heated yields a residue weighing: (a) 2.16 g (b) 2.48 g (c) 2.32 g (d) 2.64 g 35. 12 g of alkaline earth metal gives 14.8 g of its nitride. Atomic weight of metal is: (a) 12 (b) 20 (c) 40 (d) 14.8 36. How many liters of CO2 at STP will be formed when 0.01 mol of H2SO4 reacts with excess of Na2CO3. Na2CO3 + H2SO4 Na2SO4 + CO2 + H2O (a) 22.4 L (c) 0.224 L (b) 2.24 L (d) 1.12 L 37. Equal weight of X (At. wt. = 36) and Y (At. wt. = 24) are reacted to form the compound X2Y3. Then (a) X is the limiting reagent (b) Y is the limiting reagent (c) No reactant is left over and mass of X2Y3 formed is double the mass of X taken (d) None of these 38. A person adds 1.71 gram of sugar (C12H22O11) in order to sweeten his tea. The number of carbon atoms added are (mol. mass of sugar = 342) (a) 3.6 1022 (b) 7.2 1021 (c) 0.05 1020 (d) 6.6 1022 71 39. A compound contains 38.8 % C, 16.0 % H and 45.2 % N. The formula of the compound would be (a) CH3NH2 (b) CH3CN (c) C2H5CN (d) CH2(NH)2 How much CS2 (in kg) can be produced from 440 kg of waste SO2 with 60 kg of coke if the SO2 conversion is 80%? 49. Pure iron pyrite, FeS2, is burnt with 50% excess air than required for complete oxidation of FeS2, in a closed vessel. 40. When 100 g of ethylene polymerises entirely to polyethene, the weight of polyethene formed as per the equation n(C2H4) ( CH2 CH2 )n is: (a) (n/2)g (b) 100g (c) (100/n)g (d) 100ng 2Fe2O3(s) + 8SO2(g) 4FeS2(s) + 11O2(g) 41. What is the concentration of nitrate ions if equal volumes of 0.1 M AgNO3 and 0.1 M NaCl are mixed together? (a) 0.1 N (b) 0.2 M (c) 0.05 M (d) 0.25 M 50. An element A forms both a dichloride (ACl 2 ) and a tetrachloride (ACl4). Treatment of 27.8 g ACl2 with excess chlorine forms 34.9 g ACl4. Then atomic mass (in g/mol) of A is: 42. 0.16 g of dibasic acid required 25 ml of M/10 NaOH for complete neutralization. Molecular weight of acid is: (a) 32 (b) 64 (c) 128 (d) 256 51. Mixture of 10 moles of Fe2S3, 20 moles of H2O and 30 mole of O2 react with 30% yield of given reaction: 43. An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 ml. The volume of 0.1 M NaOH required to completely neutralize 10 ml of this solution is: (a) 40 ml (b) 20 ml (c) 10 ml (d) 4 ml 44. Which of the following contains the greatest number of oxygen atoms? (a) 1g of O (b) 1g of O2 (c) 1g of O3 (d) All same 45. For sequential reaction : B+C A ...(i) 2B C + 2D ...(ii) If % yield of (i) and (ii) reactions are 90% and 80% respectively then the overall % yield is expected to be: (a) 90% (b) 80% (c) 72% (d) 10% 46. If 32 g of O2 contains 6.022 1023 molecules at STP then 32 g of S, under the same conditions, will contain, (a) 6.022 1023 S atoms (b) 3.011 1023 S atoms (c) 12.044 1023 S atoms (d) 1 1023 S atoms INTEGER TYPE QUESTIONS 47. The specific gravity of a solution is 1.8, having 62% by weight of acid. It is to be diluted to specific gravity of 1.2. What volume of water (in mL) should be added to 100 ml of this solution? 48. Carbon disulphide, CS2, can be made from by-product SO2. The overall reaction is CS2 + 4CO 5C + 2SO2 72 Air contains 20% O2 and 80% N2, by volume. The mole percent of N2 gas in the gaseous mixture, after complete reaction, is. Fe2S3 + H2O + O2 Fe(OH)3 + S Calculate moles of Fe(OH)3 that can be produced in above reaction. 52. In order to obtain NaBr following set of reactions are involved. Fe + Br2 FeBr2 I FeBr2 + Br2 Fe3Br8 II Fe3Br8 + Na2CO3 NaBr + CO2 + Fe3O4 III If % yield of reaction I, II & III are 60%, 20% & 30% respectively then calculate mass of iron required (in g) to obtain 20.6 kg of NaBr. 53. A solution of A (MM=20) and B (MM= 10), [Mole fraction XB = 0.6] has density 0.7 gm/ml then molarity and molality of B in this solution will be _____M and ____m respectively. 54. A piece of aluminium weighing 2.7 g is heated with 75.0 ml of H2SO4 (sp. gr. 1.2 containing 25% H2SO4 by mass). After the metal is completely dissolved, the solution is diluted to 400ml. What is the molarity of the free H2SO4 in the resulting solution (Multiply final answer by 10)? 55. The odour of skunk is caused by chemical compounds called thiols (C4H10S). These can be deodorized by reaction with household bleach (NaOCl) according to following unbalanced reaction: C4H10S + NaOCl(aq) C8H18S2 + NaCl(aq) + H2O(aq) How many gram of thiol can be deodorized by 74.5 gm of NaOCl? P W JEE (XI) Module-1 CHEMISTRY PARIKSHIT (JEE ADVANCED LEVEL) SINGLE CORRECT TYPE QUESTIONS 1. (i) 2Al + 6HCl 2AlCl3 + 3H2 (ii) AlCl3 + 3NaOH Al(OH)3 + 3NaCl (iii) Al(OH)3 + NaOH NaAlO2 + 2H2O Above series of reactions are carried out starting with 18 g of Al and 109.5 g of HCl in first step and further 100 g of NaOH is added for step (ii) and (iii). Find out limiting reagent in each step and calculate the maximum amount of NaAlO2 that can be produced in step (iii). (Assume reactions are taken in sequence and also that each reaction goes to 100% completion) L.R. in step (I) L.R. in step (II) L.R. in step (III) Moles of NaAlO2 (a) Al AlCl3 Al(OH)3 0.66 (b) Al Na(OH) Al(OH)3 0.5 (c) Al AlCl3 NaOH 0.5 (d) HCl AlCl3 NaOH 0.5 2. A mixture of CH4 and C2H2 was completely burnt in an excess of oxygen yielding equal volumes of CO2 and steam, measured at the same temperature and pressure. The mole percent of CH4 in the original mixture is (a) 25% (b) 30% (c) 75% (d) 50% 3. The strength of 10 2 M Na2CO3 solution in terms of molality will be (density of solution = 1.10 g mL 1). (Molecular weight of Na2CO3 = 106 g mol 1) (a) 9.00 10 3 (b) 1.5 10 2 (c) 5.1 10 3 (d) 11.2 10 3 4. 10 moles of X, 12 mole of Y and 20 moles of Z are mixed to produce a final product P, according to the given balanced reactions: I X + 2Y Y+P I + Z then the maximum moles of P, which can be produced assuming that the products formed can also be reused in the reaction? (a) 6 mole (b) 9 mole (c) 10 mole (d) 12 mole 14 5. A compound contains 4% oxygen, % nitrogen, 3 4% sulphur. Then empirical formula of compound contains __________ number of nitrogen atoms. (a) 6 (b) 8 (c) 3 (d) Can not be determined 6. 56 gm of N2 and 9 gm of H2 are made to react completely to produce a mixture of NH3 and N2H4. The ratio of moles of NH3 and N2H4 is: Some Basic Concepts of Chemistry (a) 1 : 1 (c) 2 : 3 (b) 3 : 2 (d) None of these 7. 100 gm of an oleum sample (labelled as 109%) is mixed with 300 gm of another oleum sample (labelled as 118%). The new labelling of resulting oleum sample becomes (a) 115.75% (b) 106.75% (c) 163% (d) 15.75% 8. Iodobenzene is prepared from aniline (C6H5NH2) in a twostep process as shown here. C6H5NH2 + HNO2 + HC1 C6H5N+2Cl + 2H2O C6H5N+2 Cl + KI C6H5I + N2 + KCl In an actual preparation, 9.30 g of aniline was converted to 16.32 g of iodobenzene. The percentage yield of iodobenzene is (I = 127) (a) 8% (b) 50% (c) 75% (d) 80% 9. A protein isolated from a bovine preparation, was subjected to amino acid analysis. The amino acid present in the smallest amount was lysine, C6H14N2O2 and the amount of lysine was found to be 365 mg per 100 g protein. What is the minimum molecular mass (in g/mol) of the protein? (a) 40,000,000 (b) 40,000 (c) 40 (d) 4,00,000 MULTIPLE CORRECT TYPE QUESTIONS 10. Which is/are correct statements about 1.7 g of NH3? (a) It contain 0.3 mol H atoms. (b) It contain 2.408 1023 atoms. (c) Mass % of hydrogen is 17.65%. (d) It contains 0.3 mol N-atoms. 11. If 27 g of carbon is mixed with 88 g of oxygen and is allowed to burn to produce CO2, then: (a) (b) (c) (d) Oxygen is the limiting reagent. Volume of CO2 gas produced at STP is 50.4 L. C and O combine in mass ratio of 3.8. Volume of unreacted O2 at STP is 11.2 L. 12. For the following reaction : Na2CO3 + 2HCl 2NaCl + CO2 + H2O 106.0 g of Na2CO3 reacts with 109.5 g of HCl. Which of the following is/are correct? (a) The HCl is in excess amount. (b) 117.0 g of NaCl is formed. (c) The volume of CO2 produced at NTP is 22.4 L. (d) None of these 13. A sample of a mixture of CaCl 2 and NaCl weighing 4.44 g was treated to precipitate all the Ca as CaCO3, which was then heated and quantitatively converted to 1.12 g of CaO. (At . wt. Ca = 40, Na = 23, Cl = 35.5) 73 (a) Mixture contains 50% NaCl by mass (b) Mixture contains 60% CaCl2 by mass (c) Mass of CaCl2 is 2.22 g in the mixture (d) Mass of CaCl2 is 1.11 g in the mixture 14. A + B A3B2 (unbalanced) A3B2 + C A3B2C2 (unbalanced) Above two reactions are carried out by taking 3 moles each of A and B and one mole of C. Then, which option is/are correct? (a) 1 mole of A3B2C2 is formed. (b) 1/2 mole of A3B2C2 is formed. (c) 1/2 mole of A3B2 is formed. (d) 1/2 mole of A3B2 is left finally. 15. The incorrect statement(s) regarding 2 M MgCl2 aqueous solution is/are: (dsolution = 1.09 gm/ml) (a) Molality of Cl is 4.44 m. (b) Mole fraction of MgCl2 is approximately 0.035. (c) The conc. of MgCl2 is 19% w/v. (approx) (d) The conc. of MgCl2 is 19 104 ppm. 19. What is the % by mass of oxygen in the compound? (a) 53.33% (b) 88.88% (c) 33.33% (d) None of these 20. What is the empirical formula of the compound? (b) CH2O (a) CH3O (c) C2H2O (d) CH3O2 21. Which of the following could be molecular formula of compound? (a) C6H6O6 (b) C6H12O6 (c) C6H14O12 (d) C6H14O6 Comprehension (Q. 22 to 24): Oleum is considered as a solution of SO3 in H2SO4, which is obtained by passing SO3 in solution of H2SO4. When 100 g sample of oleum is diluted with desired weight of H2O then the total mass of H2SO4 obtained after dilution is known as % labelling in oleum. For example, a oleum bottle labelled as 109% H2SO4 means the 109g total mass of pure H2SO4 will be formed when 100 g of oleum is diluted by 9 g of H2O which combines with all the free SO3 to form H2SO4 as H2SO4 SO3 + H2O 16. Solution containing 23 g HCOOH is/are: w (a) 46 g of 70% v HCOOH (dsolution = 1.40 g/mL) (b) 50 g of 10 M HCOOH (dsolution = 1 g/mL) w (c) 50 g of 25% w HCOOH (d) 46 g of 5 M HCOOH (dsolution = 1 g/mL) 17. A student was carrying out the following chemical reaction in Lab 6C (balanced reaction) 2A + 3B He used 30 mole of A and 30 mole of B and at the end of reaction he found 40 moles of C were formed. Identify the correct statement(s). (a) Total mass before and after the reaction will remain same. (b) B is limiting reagent (c) % yield is 60. (d) At 50% yield 30 moles of C will be formed. 18. Which of the following contain same number of entities? (a) Number of atoms in 1 mole CuSO4 5H2O (b) Number of neutrons in 3.5 mole of CH4 (c) Number of atoms in 2 mole of FeCr2O4 (d) Number of electrons in 2.1 mole of NH+4 COMPREHENSION BASED QUESTIONS Comprehension (Q. 19 to 21): A chemist decided to determine the molecular formula of an unknown compound. He collects following informations: I. Compound contain 2 : 1 H to O atoms (number of atoms). II. Compound has 40% C by mass. III. Approximate molecular mass of the compound is 178 g. IV. Compound contains C, H and O only. 74 22. What is the % of free SO3 in an oleum that is labelled as 104.5% H2SO4 ? (a) 10 (b) 20 (c) 40 (d) None of these 23. If excess water is added into a 100 g bottle sample labelled as 112% H2SO4 and is reacted with 5.3g Na2CO3, then find the volume of CO2 evolved at 1 atm pressure and 300 k temperature after the completion of the reaction: [R = 0.0821 L atom mol 1 K 1] Na2SO4 + H2O + CO2 H2SO4 + Na2CO3 (a) 2.46 L (b) 24.6 L (c) 1.23 L (d) 123 24. 1g of oleum sample is diluted with water. The solution required 54 ml of 0.4 N NaOH for compete neutralisation. The % of free SO3 in the sample is: (a) 74 (b) 26 (c) 20 (d) None of these MATCH THE COLUMN TYPE QUESTIONS 25. A. B. C. D. Column-I 100 mL of 0.2 M AlCl3 solution + 400 ml of 0.1 M HCl solution 50 mL of 0.4 M KCl + 50 ml H2O 30 mL of 0.2 M K2SO4 + 70 ml H2O 200 mL 24.5% (w/v) H2SO4 q. Column-II Total concentration of cation(s) = 0.12 M [SO42 ] = 0.06 M r. [SO42 ] = 2.5 M s. [Cl ] = 0.2 M p. (a) A r,s; B s; C p,q; D r (b) A p,s; B s; C p,q; D r (c) A p,s; B s; C s,q; D r (d) A p,s; B p,s; C p,q; D r P W JEE (XI) Module-1 CHEMISTRY NUMERICAL TYPE QUESTIONS (UPTO ONE DECIMAL) 26. Nitric acid can be produced from ammonia in three step process. 4NO(g) + 6H2O(g) ...(i) 4NH3(g) + 5O2(g) 2NO2(g) ...(ii) 2NO(g) + O2(g) 2HNO3(aq.) + NO(g) ...(iii) 3NO2(g) + H2O(l) Calculate weight of NH3(g) (in kg) required to produce 1260 kg of HNO3. (When % yield of 1st, 2nd and 3rd reaction are respectively 69%, 60% and 68% respectively.) 27. The mineral Argyrodite is a stoichiometric compound that contain silver, sulphur ( 2) and an unknown element Y (+4). The mass-ratio of silver and Y in the compound is, m (Ag) : m (Y) = 11.88 Y forms a reddish brown lower sulphide on heating the mineral in stream of H2 (g), in which Y is in +2 state. The residue are Ag2S and H2S. To convert 10 g Argyrodite completely, 0.295 L of H2 (g) measured at 400 K and 1.0 atmosphere is required. Determine molar mass of Y = p and empirical formula of mineral = AgqYrSd. Find sum of p + q + r + s + d. INTEGER TYPE QUESTIONS 28. Sample of an element "X" consist of its three isotopes A1, A2 & A3 and population of A2 is three times the population of A3. If the average molar mass of sample is 1.25. Determine percentage population of A1 (Molar masses of isotopes A1, A2 & A3 are 1, 2 and 3 gm respectively.) 29. A solution contain substances A and B in H2O (solvent). The mole fraction of A is 0.05 and molarity of B is 7 M. The solution has density 1.14 gm/ml. Calculate molarity of A''. [Molecular weight of A = 10 gm/mol; molecular weight of B = 30 gm/mol] 30. A fluorine disposal plant was constructed to carryout the following reactions: 1 O + 2NaF + H2O 2 2 2NaF + CaO + H2O CaF2 + 2NaOH F2 + 2NaOH Over a period of operation, 1900 kg of fluorine was fed into a plant and 10,000 kg of lime was required. What was the percentage utilisation of lime ? [Lime : CaO] PYQ'S (PAST YEAR QUESTIONS) UNCERTAINTY IN MEASUREMENT AND LAWS OF CHEMICAL COMBINATIONS 1. Which of the following have same number of significant figures? [8 April, 2023 (Shift-II)] A. 0.00253 B. 1.0003 C. 15.0 D. 163 Choose the correct answer from the options given below (a) A, B and C only (b) C and D only (c) A, C and D only (d) B and C only 2. Using the rules for significant figures, the correct answer 0.02858 0.112 for the expression 0.5702 [29 June, 2022 (Shift-II)] (a) 0.005613 (b) 0.00561 (c) 0.0056 (d) 0.006 3. The number of significant figures in 50000.020 10 3 is_______. [26 Feb, 2021 (Shift-I)] 4. To check the principle of multiple proportions, a series of pure binary compounds (PmQn) were analyzed and their composition is tabulated below. The correct option(s) is(are) [JEE Adv 2022] Compound Weight % of P Weight % of Q 1 50 50 2 44.4 55.6 3 40 60 Some Basic Concepts of Chemistry (a) If empirical formula of compound 3 is P3Q4. then the empirical formula of compound 2 is P3Q5. (b) If empirical formula of compound 3 is P3Q2 and atomic weight of clement P is 20, then the atomic weight of Q is 45. (c) If empirical formula of compound 2 is PQ, then the empirical formula of the compound 1 is P5Q4. (d) If atomic weight of P and Q are 70 and 35, respectively, then the empirical formula of compound 1 is P2Q. ATOMIC & MOLECULAR MASSES 5. The average molar mass of chlorine is 35.5 g mol 1. The ratio of 35Cl to 37Cl in naturally occurring chlorine is close to: [6 Sept, 2020 (Shift-II)] (a) 4 : 1 (b) 3 : 1 (c) 2 : 1 (d) 1 : 1 MOLE CONCEPT AND MOLAR MASSES 6. Match List-I with List-II. [10 April, 2023 (Shift-II)] A. B. Column-I 16g of CH4(g) 1g of H2(g) p. q. C. D. 1 mole of N2(g) 0.5 mol of SO2(g) r. s. Column-II Weighs 28 g 60.2 1023 electrons Weighs 32g Occupies 11.4 L volume at STP 75 Choose the correct answer from the options given below: (a) (A)-(p), (B)-(r), (C)-(q), (D)-(s) (b) (A)-(q), (B)-(r), (C)-(r), (D)-(p) (c) (A)-(q), (B)-(s), (C)-(r), (D)-(p) (d) (A)-(q), (B)-(s), (C)-(p), (D)-(r) 7. When 0.01 mol of an organic compound containing 60% carbon was burnt completely, 4.4 g of CO2 was produced. The molar mass of compound is _________ g mol 1 (Nearest integer) [29 Jan, 2023 (Shift-II)] 8. Production of iron in blast furnace follows the following equation Fe3O4(s) + 4CO(g) 3Fe(l) + 4CO2(g) When 4.640 kg of Fe3O4 and 2.520 kg of CO are allowed to react then the amount of iron (in g) produced is: [Given: Molar Atomic mass (g mol 1); Fe = 56 Molar Atomic mass (g mol 1); O = 16 Molar Atomic mass (g mol 1); C = 12] [29 June, 2022 (Shift-I)] (a) 1400 (b) 2200 (c) 3360 (d) 4200 9. Number of grams of bromine that will completely react with 5.0 g of pent-1-ene is _______________ 10 2 g. (Atomic mass of Br = 80 g/mol) (Nearest Integer) [25 June, 2022 (Shift-I)] 10. 4g equimolar mixture of NaOH and Na2CO3 contains x g of NaOH and y g of Na2CO3. The value of x is __________ g. (Nearest Integer) [20 July, 2021 (Shift-II)] 11. NaClO3 is used, even in spacecraft, to produce O2. The daily consumption of pure O2 by a person is 492L at 1 atm, 300 K. How much amount of NaClO3, in grams, is required to produce O2 for the daily consumption of a person at 1 atm, 300 K? NaClO3(s) + Fe(s) O2(g) + NaCl(s) + FeO(s) R = 0.082 L atm mol 1 K 1. [8 Jan, 2020 (Shift-II)] 12. 5 moles of AB2 weigh 125 10 3 kg and 10 moles of A2B2 weigh 300 10 3 kg. The molar mass of A (M A) and molar mass of B(M B ) in kg mol 1 are: [12 April, 2019 (Shift-I)] (a) MA = 50 10 3 and MB = 25 10 3 (b) MA = 25 10 3 and MB = 50 10 3 (c) MA = 5 10 3 and MB = 10 10 3 (d) MA = 10 10 3 and MB = 5 10 3 13. Aluminium reacts with sulfuric acid to form aluminium sulfate and hydrogen. What is the volume of hydrogen gas in liters (L) produced at 300 K and 1.0 atm pressure, when 5.4 g of aluminium and 50.0 mL of 5.0 M sulfuric acid are combined for the reaction? (Use molar mass of aluminium as 27.0 g mol 1, R = 0.082 atm L mol 1 K 1) [JEE Adv 2020] 14. The mole fraction of urea in an aqueous urea solution containing 900 g of water is 0.05. If the density of the solution is 1.2 g cm 3, the molarity of urea solution is ___. [JEE Adv 2019] (Given data: Molar masses of urea and water are 60 g mol 1 and 18 g mol, respectively) 76 PERCENTAGE COMPOSITION AND EMPIRICAL & MOLECULAR FORMULA 15. A metal chloride contains 55.0% of chlorine by weight. 100 mL vapours of the metal chloride at STP weigh 0.57 g. The molecular formula of the metal chloride is (Given: Atomic mass of chlorine is 35.5 u) [12 April, 2023 (Shift-I)] (a) MCl2 (b) MCl4 (c) MCl3 (d) MCl 16. An organic compound gives 0.220 g of CO2 and 0.126 g of H2O on complete combustion. If the % of carbon is 24, then the % hydrogen is ________ 10 1 . (Nearest integer) [13 April, 2023 (Shift-I)] 17. 116 g of a substance upon dissociation reaction yields 7.5 g of hydrogen, 60g of oxygen and 48.5 g of carbon. Given that the atomic masses of H, O and C are 1,16 and 12 g/mol respectively. The data agrees with how many formulae of the following? [27 June, 2022 (Shift-II)] (a) CH3COOH (b) HCHO (c) CH3OOCH3 (d) CH3CHO 18. A 2.0 g sample containing MnO 2 is treated with HCl liberating Cl2. The Cl2 gas is passed into a solution of KI and 60.0 mL of 0.1 M Na2S2O3 is required to titrate the liberated iodine. The percentage of MnO2 in the sample is____. (Nearest integer) [Atomic masses (in u ) Mn = 55; Cl = 35.5; O = 16, I = 127, Na = 23, K = 39, S = 32] [28 June, 2022 (Shift-I)] 19. Complete combustion of 750g of an organic compound provides 420 g of CO2 and 210 g of H2O. The percentage composition of carbon and hydrogen in organic compound is 15.3 and _________ respectively. (Round off to the nearest Integer). [16 March, 2021 (Shift-I)] 20. A 10 mg effervescent tablet containing sodium bicarbonate and oxalic acid releases 0.25 ml of CO2 at T = 298.15 K and p = 1 bar. If molar volume of CO2 is 25.0 L under such condition, what is the percentage of sodium bicarbonate in each tablet? [Molar mass of NaHCO3 = 84 g mol ] [11 Jan, 2019 (Shift-I)] (a) 0.84 (b) 33.6 (c) 16.8 (d) 8.4 STOICHIOMETRY & STOICHIOMETRIC CALCULATIONS 21. When a hydrocarbon A undergoes combustion in the presence of air, it requires 9.5 equivalents of oxygen and produces 3 equivalents of water. What is the molecular formula of A? [29 Jan, 2023 (Shift-II)] (a) C8H6 (b) C9H9 (c) C6H6 (d) C9H6 P W JEE (XI) Module-1 CHEMISTRY 22. 1 g of a carbonate (M2CO3) on treatment with excess HCl produces 0.01 mol of CO2. The molar mass of M2CO3 is ___ g mol 1. (Nearest integer) [13 April, 2023 (Shift-II)] 23. If a rocket runs on a fuel (C15H30) and liquid oxygen, the weight of oxygen required and CO2 released for every litre of fuel respectively are: [24 June, 2022 (Shift-I)] (Given: density of the fuel is 0.756 g / mL) (a) 1188 g and 1296 g (b) 2376 g and 2592 g (c) 2592 g and 2376 g (d) 3429 g and 3142 g 24. A 0.166 g sample of an organic compound was digested with conc. H2SO4 and then distilled with NaOH. The ammonia gas evolved was passed through 50.0 mL of 0.5 N H2SO4. The used acid required 30.0 mL of 0.25 N NaOH for complete neutralization. The mass percentage of nitrogen in the organic compound is__________. [24 June, 2022 (Shift-I)] Consider the above reaction where 6.1 g of Benzoic acid is used to get 7.8g of m- bromobenzoic acid. The percentage yield of the product is________. (Round off to the nearest integer). [Given: Atomic masses: C : 12.0 u, H : 1.0 u, O : 16.0 u, Br : 80.0 u] [18 March, 2021 (Shift-II)] 25. 28. A solution of phenol in chloroform when treated with aqueous NaOH gives compound P as a major product. The mass percentage of carbon in P is ___________. (to the nearest integer) (Atomic mass: C = 12 ; H = 1 ; O = 16) [6 September, 2020 (Shift-II)] 29. For a reaction, N2(g) + 3H2(g) 2NH3(g); Identify dihydrogen (H 2) as a limiting reagent in the following reaction mixtures. [9 April, 2019 (Shift-I)] (a) 14 g of N2 + 4 g of H2 (b) 28 g of N2 + 6 g of H2 (c) 56 g of N2 + 10 g of H2 (d) 35 g of N2 + 8 g of H2 O C O Cl + C6H5NHC6H5 0.140g C6H5 C N (C6H5)2 0.388g 0.210g Consider the above reaction. The percentage yield of amide product is _______. (Round off to the nearest integer). (Given : Atomic mass : C : 12.0 u, H : 1.0 u, N : 14.0 u, O : 16.0 u, Cl : 35.5 u) [17 March, 2021 (Shift-II)] COOH 26. COOH FeBr3 + Br2 + HBr Br 27. NO2 HNO3 H2SO4 In the above reaction, 3.9 g of benzene on nitration gives 4.92 g of nitrobenzene. The percentage yield of nitrobenzene in the above reaction is_____________ % (Round off to the nearest integer). (Given atomic mass : C : 12.0 u, H : 1.0 u, O : 16.0 u, N : 14.0 u) [17 March, 2021 (Shift-I)] 30. The stoichiometric reaction of 516 g of dimethyl dichlorosilane with water results in a tetrameric cyclic product X in 75% yield. The weight (in g) of X obtained is___. [Use, molar mass (g mol 1): H = 1, C = 12, O = 16, Si = 28, Cl = 35.5] [JEE Adv 2023] PW CHALLENGERS NUMERICAL TYPE QUESTIONS (ANSWER UPTO TWO DECIMAL PLACE) on the surface of the fabric or with traces of water to form the waterproofing film [(CH3)2SiO]n, by the reaction 1. Excess of calcium orthophosphate is reacted with magnesium to from Ca3P2 along with MgO. Ca3P2 on reaction with water liberated PH3 along with Ca(OH)2. PH3 is burnt in excess of oxygen to form P2O5 along with water. Oxides of magnesium and phosphorous react to give magnesium metaphosphate. Calculate grams of magnesium metaphosphate obtained if 1.92 gm of Mg is taken? n[(CH3)2SiCl2 + 2nOH 2nCl + nH2O + [(CH3)2SiO]n Where n stands for a large integer. The waterproofing film is deposited on the fabric layer upon layer. Each layer is 6.0 thick [the thickness of the (CH3)2SiO group]. How much (in g) (CH3)2 SiCl2 is needed to waterproof one side of a piece of fabric, 1.00 m by 3.70 m, with a film 300 layers thick? The density of the film is 1.0 g/cm3. (Si = 28) 2. In one process for waterproofing, a fabric is exposed to (CH3)2SiCl2 vapour. The vapour reacts with hydroxyl groups Molar mass of (CH3)2SiO = 74 g/mol Some Basic Concepts of Chemistry Molar mass of (CH3)2SiCl2 = 129 g/mol 77 INTEGER TYPE QUESTIONS 3. Cis-platin [Pt(NH 3) 2Cl 2], a compound used in cancer treatment is prepared by the reaction of ammonia with potassium tetrachloroplatinate. K2PtCl4 + 2NH3 2KCl + Pt(NH3)Cl2 (i) How many grams of cis-platin are formed from 41.7 gm K2PtCl4 amd 34 gm NH3, if the reaction takes place in 90% yield? (Ans. x) (ii) What is the maximum mass (in g) of KCl which can be produced if initially total 9 moles of reactant are taken. Assuming 100% reaction? (Ans. y) What is the value of (x + y)? [Pt = 195] 4. The number of Alkoxy groups in an organic compound A(OR)x may be determined by the sequential reaction. A(OR)x + xHI AI(OH)x + xRI RI + Ag+ + H2O ROH + AgI(s) + H+ When 4.8 gm of organic compound A(OR)x (molar mass = 240 gm mol 1) is treated as above, 9.4 gm AgI is precipitated. Then, calculate number of alkoxy groups in the compound A(OR)x. 5. A sample of ammonia contains only H1 and H2 isotopes of hydrogen in 4:1 ratio and N14 and N15 isotopes of nitrogen in 3 : 1 ratio. How many neutrons are present in 1.785 mg of ammonia? (Answer in the order of 1018) (NA = 6 1023) 6. In order to determine the composition of a mixture of halides containing MBr2 & NaI, 14 gm mixture was dissolved in water. To this solution, AgNO3 solution of certain molarity was added gradually. The mass of precipitate produced (in gm) were measured and it was plotted against volume of AgNO3 solution added (in ml). If it is known that AgI is precipitated first precipitation of Br does not start until the already precipitating I precipitates completely. Find out the value of AB CD where: AB = Atomic weight of metal M CD = Mole percent of NaI in original mixture. 24.44 Total Mass of ppt. formed 9.4 (in gm) 40 120 Vol. of AgNO3 added (in ml) (Molar mass of NaI = 149.89g/mol, Br = 79.9g/mol, AgI = 234.77g/mol) 7. A mixture of gases liberated upon decomposition of 33.12 gm Pb(NO3)2 is dissolved in 10ml of water. What is the mass (in g) of 0.1M KOH solution with density 1.05 g/ml required to neutralise this acid. The reactions involved are: (at. mass of Pb = 207) 2PbO+ 4NO2 + O2 2Pb(NO3)2 4NO2 + 2H2O + O2 4HNO3 KOH + HNO3 KNO3 + H2O 78 8. Water is the working Fluid in a thermal plant for producing electricity. Coal is Combusted for Generating heat as per reaction, C + O2 CO2. 0.01% of the released CO2 gas is absorbed in water and gets converted into weak acid, H2CO3 which dissociate to give H+ as H2CO3 2H+ + CO3 2 . The % dissociation of acid is 5%. Assume no ionisation of H2O. If in a certain application [H+] concentration can maximum be 10 5 M, then (a) Calculate maximum no. of moles of H+(x) and CO3 2 (y) in water if 109 litres of H2O is used. (b) Calculate maximum no. of mole of carbon (z) which can be burnt so that water remains fit to be used. What is the value of xy2/z? 9. A mother cell disintegrates into 60 identical cells and each daughter cell of mother disintegrates into 24 smaller cells, The smallest cell is uniform cylindrical in shape with diameter 120 and each cell is 6000 long. Determine molar mass of the mother cell, if density of the smallest cell is 1.12 gm/cm3. Using scientific notation if your answer is x 10y, their write the value of [x] + y, where [ ] is an integer function. Take Avogadro number as 6 1023. 10. Once Tom & Jerry entered a chemistry lab where a chemist was preparing 3L H2SO4 solution. He labelled the solution as 10 m (dsolution = 3.3 gm/ml). As the chemist left the lab, a mischief came in Tom's mind. He tried to throw the solution on Jerry but failed. In doing so some of the H2SO4 solution fell on the floor, so he added water to make it again to 3L. The chemist returned back & got astonished when he saw the result of analysis, that were d = 1.5 gm/ml and % w/w = 98. Find out the number of moles of H2SO4 which fell down on the floor. 11. A polymeric substance, tetrafluoroethylene, can be represented by the formula (C2F4)x, where x is a large number. The material was prepared by polymerizing C2F4 in the presence of a sulphur bearing catalyst that serves as a nucleus upon which the polymer grew. The final product was found to contain 0.012% S. What is the value of x, if each polymeric molecule contains one sulphur atom? Assume that the catalyst contributes a negligible amount to the total mass of the polymer. (F = 19, s = 32) SINGLE CORRECT TYPE QUESTIONS 12. Density of water varies with temperature as shown. d t = d 0 C (0.002 t C) ; where C d t C = density at t C and d 0 C is density at 0 C = 1gm/ml Calculate % change in molarity and molality respectively of pure water due to change in temperature from 0 C to 100 C. (a) 10%, no change (b) no change, 20% (c) 20%, no change (d) 20%, 20% P W JEE (XI) Module-1 CHEMISTRY Answer Key CONCEPT APPLICATION 1. (a) 2. (a) 3. (d) 4. (a) 5. (a) 6. (a) 7. (a) 8. (b) 9. (b) 10. (b) 11. (a) 12. (b) 13. (c) 14. (b) 15. (a) 16. (b) 17. (b) 18. (a) 19. (c) 20. (d) 21. (d) 22. (b) 23. (a) 24. (c) 25. (b) 26. (c) 27. (c) 28. (b) 29. (c) 30. (d) 31. (a) 32. (b) SCHOOL LEVEL PROBLEMS 1. (b) 2. (a) 3. (a) 11. (i)-(r), (ii)-(s), (iii)-(q), (iv)-(p) 4. (b) 5. (c) 6. (a) 7. (a) 12. (i)-(q), (ii)-(r), (iii)-(p), (iv)-(t), (v)-(s) 8. (b) 9. (c) 10. (a) 24. I-(b), II-(b), III-(a), IV-(a) PRARAMBH (TOPICWISE) 1. (b) 2. (b) 3. (c) 4. (c) 5. (c) 6. (c) 7. (a) 8. (b) 9. (a) 10. (d) 11. (a) 12. (a) 13. (a) 14. (b) 15. (b) 16. (a) 17. (a) 18. (c) 19. (c) 20. (d) 21. (b) 22. (a) 23. (a) 24. (d) 25. (c) 26. (b) 27. (b) 28. (c) 29. (c) 30. (c) 31. (c) 32. (a) 33. (a) 34. (a) 35. (d) 36. (a) 37. (c) 38. (d) 39. (b) 40. (c) 41. (a) 42. (d) 43. (d) 44. (d) 45. (a) PRABAL (JEE MAIN LEVEL) 1. (c) 2. (a) 3. (c) 4. (a) 5. (c) 6. (b) 7. (a) 8. (a) 9. (a) 10. (d) 11. (c) 12. (a) 13. (a) 14. (b) 15. (a) 16. (b) 17. (b) 18. (a) 19. (a) 20. (a) 21. (a) 22. (a) 23. (d) 24. (b) 25. (c) 26. (b) 27. (b) 28. (c) 29. (b) 30. (b) 31. (c) 32. (b) 33. (c) 34. (a) 35. (c) 36. (c) 37. (c) 38. (a) 39. (a) 40. (b) 41. (c) 42. (c) 43. (a) 44. (d) 45. (c) 46. (a) 47. [300] 48. [76] 49. [83] 50. [207] 51. [4] 52. [1167] 53. [30M, 75m] 54. [2] 55. [180] 4. (c) 5. (b) 6. (c) 11. (b,c,d) 12. (a,b,c) 13. (a,c) 14. (b,d) 15. (b,d) 16. (a,b) 21. (b) 22. (d) 24. (b) 25. (b) 26. [1811.5] 27. [87.6] 28. [80] 29. [3] PARIKSHIT (JEE ADVANCED LEVEL) 1. (c) 2. (d) 3. (a) 23. (c) 7. (a) 8. (d) 9. (b) 17. (a,b,d) 18. (a,b,d) 19. (a) 10. (a,b,c) 20. (b) 30. [28] PYQ's (PAST YEAR QUESTIONS) 1. (c) 2. (b) 3. [8] 4. (b, c) 5. (b) 11. [2130] 12. (c) 13. [6.15] 14. [2.98 or 2.99] 20. (d) 21. (a) 22. [100] 23. (c) 6. (d) 15. (a) 24. [63] 25. [77] 7. [200] 16. [56] 8. (c) 9. [1143] 10. [1] 17. [2] 18. [13] 19. [3] 26. [78] 27. [80] 28. [69] 29. (c) 30. [222] PW CHALLENGERS 1. [1.82] 2. [1.16] 3. [474] 4. [2] 5. [471] 6. [4994] 7. [2100] 8. [250] 9. [16] 10. [5] 11. [2667] 12. (c) Some Basic Concepts of Chemistry 79

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