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ICSE Class X Notes 2020 : Chemistry : Mole-Concept-and-Stoichiometry-A

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Sia Sen
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Important Question: ICSE 2010 Class 10th (Mole Concept and Stoichiometry A) Vapour Density and Molecular Weight. Num. 1. When heated, potassium permanganate decomposes according to the following equation : 2KMnO4 K 2 MnO4 + MnO2 + O2 Some potassium permanganate was heated in a test tube. After collecting one litre of oxygen at room temperature, it was found that the test-tube had undergone a loss in mass of 1.32 g. If one litre of hydrogen under the same condition of temperature and pressure has a mass of 0.0825 g, calculate the relative molecular mass of oxygen. Sol : - Mass of 1 lit. of oxygen = 1.32 g Mass of 1 lit. of hydrogen = 0.0825 g Vapour density of oxygen = mass of 1 lit. of oxygen / mass of 1 lit. of hydrogen = 1.32 /0.0825 = 16 g. Relative molecular mass of oxygen = 2 vapour density = 2 16 = 32. Q. 2. Name the term which defines the mass of a given volume of a gas compared to the mass of an equal volume of hydrogen. Ans.: - Vapour density. Num. 2. Find the relative molecular mass of a gas, 0.546 g of which occupies 360 cm3 at 87 C and 380 mm Hg pressure. [1 litre of hydrogen at s.t.p. weigh 0.09 g]. Sol : - P1 = 380 mm of Hg, V 1 = 360 cm3 = 360 ml, T1 = 87 + 273 = 360 K. P2 = 760 mm of Hg, V 2 = x ml, T2 = 273 K Using, P1 V1 /T1 = P2 V2 / T2 we get (380 360) / 360 = (760 x) / 273 => x = (273 380) / 760 = 136.5 l Mass of 136.5 ml = 0.546 g => mass of 1000 ml = (0.546/136.5) 1000 = 4 g Vapour density = mass of 1 lit. of the gas / mass of 1 lit. of hydrogen = 4/0.09 = 44.44 Molecular mass = 2 44.44 = 88.88 [Ans.] Percentage Composition. Num 1. Find the total percentage of oxygen in magnesium nitrate crystal : Mg(NO3)2.6H2 O [H = 1, N = 14, O = 16, Mg = 24]. Sol :- gram molecular mass of Mg(NO 3 )2.6H2O = 24 + (14 + 48) 2 + 6 18 = 256. Mass of oxygen in magnesium nitrate = 192 g % of oxygen = (192 / 256) 100 = 75 %. Num 2. What is the mass of nitrogen in 1000 kg of urea [CO(NH2)2]. [Answer to the nearest kg.] Sol.: - Molecular weight of urea [CO(NH2)2] = 12 + 16 + (14 + 2) 2 = 60. 60 kg of urea contains 28 kg of nitrogen 1000 kg of urea will contain (28/60) 1000 = 467kg of nitrogen . Num 3. Calculate the percentage of boron [B] in borax Na2B4O7.10H2O [H = 1, B = 11, O = 16, Na = 23]. Sol : - Gram molecular weight of Borax = 46 + 44 + 112 + 180 = 382 g Weight of Boron in Borax = 44 g. Percentage of Boron = (44/382) 100 = 11.5 %. Num 4. If a crop of wheat removes 20 kg of nitrogen per hectare of soil, what mass of the fertilizer calcium nitrate Ca(NO3 )2 would be required to replace the nitrogen in a 10 hectare field? [N = 14; O = 16; Ca = 40] (Give your answer to the nearest kg). Sol : - Amount of nitrogen removed from 10 hectare field = 10 20kg = 200 kg. Molar mass of Calcium nitrate Ca(NO 3 )2 = 40 + (14 + 16 3) 2 = 164 Mass of nitrogen in 164 kg of Ca(NO3 )2 is 28 kg For 28 kg of nitrogen, amount of calcium nitrate required is 164 kg For 200 kg of nitrogen, amount of calcium nitrate required = (164/28) 200 kg = 1171 kg. Empirical Formula and Molecular Formula. Num.1. Determine the empirical formula of the compound whose composition by mass is 42 % nitrogen, 48 % oxygen and 9 % hydrogen. (H = 1: N = 14; O = 16). Sol.: Element Percentage Atomic Composition weight Relative No. of Atoms Simplest ratio Nitrogen 42 14 42/14 = 3.0 3.0/3.0 = 1 Oxygen 48 16 48/16 = 3.0 3.0/3.0 = 1 Hydrogen 9 1 9/1 = 9.0 9.0/3.0 = 3 Thus empirical formula of the compound is NOH3. Num. 2. A metal M forms a volatile chloride containing 65.5% chlorine. If the density of the chloride relative to hydrogen is 162.5, find the molecular formula of the chloride. (M = 56, Cl = 35.5). Sol.: - Percentage of M in the chloride = 100 65.5 = 34.5 Element Percentage Atomic weight Relative No. of Atoms Simplest ratio M 34.5 56 34.5/56 = 0.61 0.61/0.61 = 1 Cl 65.5 35.5 65.5/35.5 = 1.84 1.84/0.61 = 3 Empirical formula of the chloride = MCl3 Empirical formula mass = 56 + 35.5 3 = 162.5 g Vapour density = 162.5 Molar mass of the chloride = 2 V.D. = 2 162.5 = 325.0 g Molecular formula = (Empirical formula)n Or, n = Molar mass /Empirical formula mass = 325.0 g / 165.5 g = 2 Molecular formula = (MCl3 )2 = M2Cl6 Num. 3. The percentage composition of sodium phosphate as determined by analysis is 42.1 % sodium, 18.9 % phosphorus and 39 % oxygen. Find the empirical formula of the compound. [Na = 23, P = 31, O = 16]. Sol : Element Percentage Atomic composition weight Relative No. of atoms Simplest ratio Sodium 42.1 23 42.1/23 = 1.83 1.83/0.61 = 3 Phosphorus 18.9 31 18.9/31 = 0.61 0.61/0.61 = 1 Oxygen 39.0 16 39.0/16 = 2.44 2.44/0.61 = 4 Thus empirical formula is Na3 PO 4 . Num. 4. An experiment showed that in a lead chloride solution, 6.21g of lead combined with 4.26 g of chlorine. What is the empirical formula of this chloride? [Pb = 207; Cl = 35.5]. Sol : Element Mass Atomic weight Relative No. of atoms Simplest ratio Lead 6.21 g 207 6.21/207 = 0.03 0.03/0.03 = 1 Chlorine 4.26 g 35.5 4.26/35.5 = 0.12 0.12/0.03 = 4 Hence, empirical formula is PbCl4. Num.5. Determine the empirical formula of a compound containing 47.9% potassium, 5.5% beryllium and 46.6% fluorine by mass.[Atomic weight of Be = 9; F = 19; K = 39]. Work to one decimal place. Sol : Element Percentage composition Atomic weight Relative No. of atoms Simplest ratio Potassium 47.9 39 47.9/39 = 1.2 1.2/0.6 = 2 Beryllium 5.5 9 5.5/9 = 0.6 0.6/0.6 = 1 Fluorine 46.6 19 46.6/19 = 2.4 2.4/0.6 = 4 Hence empirical formula is K2 BeF4 . Num.6. A compound has the following percentage composition by mass : Carbon 14.4%, hydrogen 1.2% and chlorine 84.5%. Determine the empirical formula of this compound. Work correct to 1 decimal place. The relative molecular mass of this compound is 168, so what is its molecular formula. [C = 12; H = 1; Cl = 35.5] Sol : Element Percentage composition Atomic weight Relative No. of atoms Simplest ratio Carbon 14.4 12 14.4/12 = 1.2 1.2/1.2 = 1 Hydrogen 1.2 1 1.2/1 = 1.2 1.2/1.2 = 1 Chlorine 35.5 84.5/35.5 = 2.38 2.38/1.2 = 2 84.5 Hence empirical formula is CHCl2 . n = Molecular mass/Empirical formula mass = 168/(12 + 1 + 70) = 168/83 = 2. Therefore, molecular formula = (CHCl2 )2 = C 2H2Cl4. Q.1.What is the empirical formula of octane ? Ans.: - C 8H18 . Problem based on Chemical Equations Num.1. Given that the relative molecular mass [molecular weight] of copper oxide is 80, what volume of ammonia [measured at s.t.p.]is required to completely reduce 120 g of copper oxide. The equation for the reaction is : 3CuO + 2NH3 3Cu + 3H2 O + N 2 . [Vol. occupied by 1 mole of gas at s.t.p. is 22.4 lit.] Sol : - 3CuO + 2 NH3 3Cu + 3H2 O + N2 3 mole 2 mole 3 80 g 2 22.4 l = 240 g = 44.8 l For 240 g of CuO, ammonia consumed is 44.8 l For 120 g of CuO, ammonia consumed = (44.8/240)/120 = 22.4 lit. Num. 2. The equations given below relate to the manufacture of sodium carbonate [Mol. Wt. of Na2CO3= 106] (i) NaCl + NH3 + CO2 + H2 O NaHCO3 + NH4Cl (ii) 2NaHCO4 Na2 CO3 + H2 O + CO22. Questions (a) and (b) are based on the production of 21.2 g of sodium carbonate. (a) What mass of sodium hydrogen carbonate must be heated to give 21.2 g of sodium carbonate. [Molecular weight of NaHCO2 = 84]. (b) To produce the mass of sodium hydrogen carbonate calculated in (a), what volume of carbon dioxide, measured at s.t.p., would be required. Sol : - (a) 2NaHCO3 Na2 CO3 + H 2O + CO 2 2 84 = 168 g 106 g 106 g of Na2CO 3 is obtained from 168 g of NaHCO3 21.2 g of Na2CO3 is obtained from (168/106) 21.2 = 33.6 g of NaHCO 3. (b) NaCl + NH3 + CO2 + H 2 O NaHCO 3 + NH 4Cl 1 mole 1 mole 22.4 l 84 g 84 g of NaHCO3 requires 22.4 l of NH4 Cl 33.6 g of NaHCO 3 requires (22.4/84) 33.6 = 8.96 l of CO2. Num. 3. The reaction of potassium permanganate with acidified iron [II] sulphate is given below : 2KMnO4 + 10FeSO4 + 8H2 SO4 K 2 SO4 + 2MnSO4 + 5Fe2(SO4)3 + 8H2O. If 15.8 g of potassium permanganate was used in the reaction, calculate the mass of iron [II] sulphate used in the above reaction. [K = 39, Mn = 55, Fe = 56, S = 32, O = 16]. Sol : - Molecular weight of KMnO 4 = 39 + 55 + 16 4 = 158 Molecular weight of K2 SO 4 = 2 39 + 32 + 16 4 = 174 Molecular weight of FeSO 4 = 56 + 32 + 16 4 = 152 2 158 g of KMnO4 yields 174 g of K2 SO 4 15.8 g of KMnO 4 will yield [(174)/(2 158)] 15.8 g = 8.7 g of K2SO 4. 174 g of K2SO4 yields 152 g of FeSO4 8.7 g of K2SO 4 will yield (152/174) 8.7 = 7.6 g of FeSO 4. Hence, 7.6 g of Iron (II) sulphate is used . Num. 4. 10 g of a mixture of sodium chloride and anhydrous sodium sulphate is dissolved in water. An excess of barium chloride solution is added and 6.99 g of barium sulphate is precipitated according to the equation : Na2SO4 + BaCl2 BaSO4 + 2NaCl Calculate the percentage of sodium sulphate in the original mixture. [O = 16, Na = 23, S = 32, Ba = 137]. Sol : - Na2SO 4 + BaCl2 BaSO4 + 2NaCl Molecular weight of BaSO 4 = 137 + 32 + 16 4 = 233 Molecular weight of Na2 SO 4 = 2 23 + 32 + 16 4 = 142 233 g of BaSO4 is obtained from 142 g of Na2SO4 6.99 g of BaSO 4 will be obtained from (142/233) 6.99 g = 4.26 g of Na2SO4 Percentage of Na2 SO 4 in 10 g of mixture = (4.26/10) 100 = 42.6%. Num. 5. From the equation : C + 2H2SO4 CO2 _ 2H2 O + 2SO2 . Calculate : The mass of carbon dioxide by 49 g of sulphuric acid [C = 12, rel. mol. Mass of H2SO4 = 98] ii. The volume of sulphur dioxide measured at s.t.p., liberated at the same time. [Volume occupied by 1 mole of a gas at s.t.p is 22.4 dm3]. i. Sol : C + 2H2 SO 4 CO2 + 2H 2 O + 2SO 2 Molecular mass of CO2 = 12 + 32 = 44. As, 98 2 g of sulphuric acid produces 44 g of carbon dioxide. Therefore, 49 g of sulphuric acid will provide (44/196) 49 = 11 g of CO 2. [Ans.] 2 moles of sulphur dioxide is produced. Volume = 222.4 dm3 = 44.8 dm3. As, 196 g of sulphuric acid provide 44.8 dm3 of SO2 Therefore, 49 g of sulphuric acid will provide (44/196) 44.8 dm3 = 11.2 dm3. [Ans.]

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