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ISC Class XII Sample / Model Paper 2021 : Mathematics

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Devamanoharan Chitharanjan
The Vikasa School, Thoothukudi

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Some Basic Concept of Chemistry 1. 2.4 g of pure Mg (at. mass = 24) is dropped in 100 mL of 1M HCl. Which of the following statement is wrong? (A) 1.12 L of hydrogen is produced at S.T.P. (B) 0.01 mol of magnesium is left behind (C) HCl is the limiting reagent. (D) None of these Answer: [A] 800 g of solution has X = 320 g = Solvent in solution 800 = 230 480 g = X left after cooling 320 = 100 220 g Total weight of solution 2. In the reaction 3Cu + 8HNO3 3Cu(NO3 )2 + 2NO + 4H2O, equivalent weight of (A) M (B) = 480 + 220 = 700 g M 3 HNO3 ? what is the if molecular weight of (C) 3 M 4 (D) HNO3 is M 4 M 3 Answer: [D] 1 NO3 NO N+5 N+2 change = 3 N+5 2 N+2 total change = 6 n factor of per mole 3. A mixture of HNO3= 6 3 = 8 4 H2 and l2 (vapour) in molecular proportion 2 : 3 was heated at 440 C till the reaction H2 (g) l2 (g) 2HI(g) reached equilibrium state. Calculate the percentage of converted into HI. (K c at 440 C I2 is 0.02 and x is small compared to unity) (A) 10% (B) 5.77% (C) (D) 8.3% 20% Answer: [B] H2 (g) 2a x I2 (g) 2HI(g) 3a x 2x (2x)2 x = 0.02 = 1.73 10 1 (2a)(3a) a x % of I2 reacted = 100 =5.77% 3a 2 4. At 100 C and 1 atm, if the density of liquid water is and that of water vapour is 0.0006 0.0006 g cm 3 , 1g cm 3 then the volume occupied by water molecules in 1 litre of steam at that temperature is (A) 6 cm3 (B) 60 cm3 (C) 0.6 cm3 (D) 0.06 cm3 Answer: [C] Vol. of steam= 1It = 103 cm3 m= d.V = mass of 10 3 cm3 = steam density volume 0.0006 gm 3 0.6 gm 103 cm= cm3 Actual vol. occupied by H2O molecules is equal to vol. of water of same mass Actual vol. of H2O molecules in 6 g steam = Mass of steam/density of H2O = 0.6 = g / 1g / cm3 0.6 cm3 3 5. A carbon compound containing carbon and oxygen has approximate molar mass equal to 290. On analysis it is found to contain 50% by mass of each element. Therefore molecular formula of the compound is (A) C12O9 (B) C 4 O3 (C) C3 O 4 (D) C9 O12 Answer: [A] Element : Y by mass : C O 50 50 50 50 ratio by no. of atoms : 12 16 Simplest ratio by no of atoms : 4 3 emperical formula = C 4 O3 And let m.f is (C4 O3 )n 96n = 290 or n = 290 3 96 molecular formula is C12 O9 6. If 0.5 moles of maximum moles of (A) 0.2 (C) 0.3 is mixed with 0.2 moles of BaCl2 Ba3 (PO4 )2 Na3PO4 , the obtained is (B) 0.5 (D) 0.1 Answer: [D] 4 3BaCl2 + 2Na3PO4 Ba3 (PO4 )2 + 6NaCl The limiting reactant is Na PO , the no. of moles of Ba (PO 3 produced = 7. 3 ) 4 2 0.2 = 0.1mole 2 For the reaction: Fe2S3 FeSO4 + SO2 . is (M is the mol wt of Fe2S3 4 (A) M 4 (B) M 16 (C) M 22 (D) M 20 The equivalent mass of Fe2S3 ) Answer: [D] Fe2S3 + 5O2 Mgm 2FeSO4 + SO2 5 32 gm 5 32 gm O combines with 8 M gm 8 gm O combines with 8. The mass of with 34 g NH3 Mg3N2 Fe2S3 M M 8 = gm Fe2S3 5 32 20 produced if 48 g of Mg metal is reacted as is 3 Mg + 2 NH3 Mg3N2 + 3H2 (A) 200 3 (B) 100 3 5 (C) (D) 400 3 150 3 Answer: [A] Mg3N2 + 3H2 3Mg + 2NH3 48 34 = mole 2= 2 24 17 9. Mass of Mg3N2 = 1 200 2 (3 24 + 28) = 3 3 0.078 g of hydrocarbon occupy 22.4 ml of volume at 1 atm and 0 C. The empirical formula of the hydrocarbon is CH. The molecular formula is (A) C2H2 (B) C4H4 (C) C6H6 (D) C8H8 Answer: [C] w 1 22.4 10 3 = = M 78 g M 22.4 = n 78 = 6 13 MF = (CH)6 or C6H6 6 10. For SO3 118% labelled oleum if the no. of moles of H2SO4 be x & y respectively, the values of and free x+y x y is approximately (A) (B) 1.21 (C) 1.51 1.51 (D) 1.21 Answer: [B] 118% Oleum 18 g water = 1 mole water 1mole SO3 = 80 g SO3 y= 1 20 / 98 nH2SO4 in oleum( x ) = 20 1+ x+y 98 = 1.51 = x y 20 1 98 11. The average atomic mass of a mixture containing 79 mole % of Mg remaining 21 mole % of Mg and Mg is 24.31 %. 24 Mole of (A) 5 25 26 Mg 26 is (B) 20 7 (C) 10 (D) 15 Answer: [C] Let % mole of 26 Mg be X (21 X)25 + 26x + 79 24 = 24.31 100 X = 10% 12. 20 ml of a H PO solution needs 40 ml of 0.1 M NaOH to convert it into sodium dihydrogen phosphate. How much volume of 0.1 M Ca(OH)2 is needed to neutralise the same volume of same H PO completely (A) 120 ml (B) 20 ml (C) 40 ml (D) 60 ml Answer: [D] Equivalents of H PO = equivalents of NaOH 3 4 3 3 4 4 20 10 3 M 1= 40 0.1 10 3 M = 0.2 Equivalents of Ca(OH)2 = equivalents of H3PO4 V 2 0.1 10 3 = 3 0.2 10 3 20 V = 60 ml 13. NH3 is produced according to the following reaction : N2 (g) + 3H2 (g) 2NH3 (g). In an experiment 0.25 mol of NH3 is 8 formed when 0.5 mol of N2 is reacted with 0.5 mol of H2 . What is % yield? (A) 75% (B) 50% (C) 33% (D) 25% Answer: [A] % yield = Producedmolof NH3 100 0.25 = 100 = 75% max.possibleproducedmole of NH3 0.5 2 3 14. Which sample contains the largest number of atoms (A) 1 mg of C4H10 (C) 1 mg of Na (B) 1 mg of N2 (D) 1 mL of water Answer: [D] 1mg C4= H10 14N 10 3 atoms 58 1mg N2 = 2N 10 3 atoms 23 1mg Na = N 10 3 atoms 23 1m = H2O 1g = H2O 3N atoms 18 9 15. The pair of species having same percentage of carbon is (A) CH COOH and C H 3 (C) 6 12 O6 HCOOCH3 and C12H22O11 (B) CH3 COOH and C2H5 OH (D) C6H12O6 and C12H22O11 Answer: [A] Both have same empirical formula CH2O so same percentage. = V.D. Mol. wt = 120 2 16. Excess of aluminium is burn in the gaseous product (aluminium oxide) after thermal decomposition of potassium chlorate. If 2 moles of potassium chlorate is thermally decomposed then how many moles of aluminium oxide will form? (A) 2.5 (B) 2 (C) 3 (D) 1.5 Answer: [B] 2KClO3 2KCl + 3O2 3/2 moles of O2 mole of is formed by thermal decomposition of one KClO3 , thus 3 moles of decomposition of 2 moles of O2 is formed by thermal KClO3 . 10 4Al + 3 O2 2 Al2O3 . 2 moles of Aluminium oxide is formed with 3 moles of O . So, to form 2 moles of Aluminium oxide, 2 moles potassium chlorate is thermally decomposed. 2 17. Caffeine has a molecular mass 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 Answer: [A] Caffeine N 194 g 194 194 g NA = = 28.9 g 100 ? 28.9 194 g 100 NA 0.289 194 1 NA= atoms 4 NA 14 18. What will be the normality of a solution obtained by mixing 0.45 N and 0.60 N NaOH in the ratio 2: 1 by volume? 11 (A) 0.4 N (B) 0.5 N (C) 1.05 N (D) 0.15 N Answer: [B] N1V1 + N2 V2 0.45 2 + 0.6 1 1.5 = = = 0.5 N V1 + V2 2 +1 3 N = 19. The equivalent weight of K4 [Fe(CN)6] in the given reaction is [O] K 4 [Fe(CN)6 ] Fe3 + + NO3 + CO2 (A) M/20 (B) M/1 (C) M/60 (D) M/61 Answer: [D] +2 +1 +3 3+ +5 +4 +2 -4 -5 K4[Fe(CN)6] Fe + NO3 + CO2 5 +10 6=+60 +5 Net change in oxidn no. = + 60 + 1 = + 61 Eq. wt. = M 61 12 20. 105 ml of pure water at 4 C is saturated with NH (g) producing a solution of density 0.9 gm/ml. If this solution contain 30% of NH by mass, therefore the total volume of solution is 3 3 (A) 250 ml (B) 125 ml (C) 166.67 ml (D) 111.11 ml Answer: [C] Let. final vol. of solution is V ml wt of soln = V 0.9 gm and wt of NH3 = (V 0.9 105) gm V 0.9 105 =V 0.9 0.3 V 0.9 0.7 = 105 or = V 105 105 = = ml 166.67 ml 0.9 0.7 0.63 21. The solubility of substance X in pure ethanol is 0.1 gm/lit and in water is 0.01 gm/lit. To dissolve 11 gm of dry X we are adding 20 ml of fresh 50% (V/V) ethanol solution in each time on X . How many times we are to add this ethanol solution to dissolve X ? (A) 100 (B) 10 6 13 (C) 10 (D) 10 3 4 Answer: [D] In each time we are adding 20 ml ethanol solution. In 20 ml ethanol solution vol. C2H5 OH = 10 ml and vol of water = 10 ml In 10 ml ethanol mass of X dissolved = In 10 ml water mass of X dissolved = 0.1 10 gm 1000 In each time, total mass of X dissolved 0.1 10 gm 1000 = 1.1 gm 1000 No. of times of addition of ethanol solution 22. A mixture of CH4 and C2H2 11 = 1000 = 10 4 1.1 occupied a certain volume at a total pressure equal to 63 torr. The same gas mixture was burnt to CO and H O( ). The CO (g) alone was collected in the 2 2 2 same volume and at the same temperature, the pressure was found to be 69 torr. What was the mole fraction of CH4 in the original gas mixture? (A) 19 21 (B) 19 20 14 (C) 17 18 (D) 15 16 Answer: [A] Let no. of moles of CH4 Let no. of moles of C2H2 present present (n1 + n2 ) = 63 K = n1 mol. = n2 mol. ...(1) CH4 + 2O2 CO2 (g) + 2H2O( ) n1 mol n1 mol C2H2 + 3O2 2CO2 (g) + H2O( ) n2 mol 2n2 mol. After combustion total no. of moles =(n1 + 2n2 ) =69 K ...(2) = n2 6 K and = n1 57 K Mole fraction of CH4 in the original gas mix = 57K 19 = 63K 21 23. 12 g carbon combines with 64 g sulphur to form CS2 . 12 g carbon also combines with 32 g oxygen to form CO2 . 10 g sulphur combines with 10 g oxygen to form SO2 . These data illustrate the 15 (A) Law of multiple proportions (B) Law of definite proportions (C) Law of reciprocal proportions (D) Law of gaseous volumes Answer: [C] Ratio of the weights of S and O combining with fixed weight of C is 64: 32 = 2: 1. Ratio of the weights of S and O combining directly = 10: 10 = 1: 1. The two ratios are simple multiple of each other. This proves law of reciprocal proportions. (12 g) C 64 g S (10 g) O 32 g (10 g) 24. The weight of 305 mL of a diatomic gas at pressure is 1 g. The weight of one atom is 0 C and 2 atm (N is the Av. no.) : (A) 16/N (B) 32/N 16 (C) 16 N (D) 32 N Answer: [A] RTw PV Calculate M = = Wt. of one atom M / (N 2) 25. 10 mL of hydrogen contains 2 103 molecules of hydrogen at certain pressure and temperature. Calculate the number of molecules of oxygen whose volume is 200 mL at the same temperature and pressure (A) 2 104 molecules (B) (C) 4 102 molecules (D) None of these 4 104 molecules Answer: [B] At the same temperature and pressure equal volume of the gas contain equal number of molecules. Hence, 10 ml of 200 ml of 4X 104 H2 or H2 or O2 = 2X 103 number of molecules O2 = (2X 103 / 10)200 number of molecules 17 26. What is the empirical formula of a compound composed of O and Mn in equal weight ratio?[Mn = 55] (A) MnO (B) MnO2 (C) (D) Mn2O7 Mn2O3 Answer: [D] Mn2O7 mass of Mn = 55 2 = 110 Mass of O = 16 7 = 112 27. The number of moles of Cr O needed to oxidise 0.136 equivalents of N H by the reaction: N H + Cr O N Cr + H O is 2 2 2 7 + 5 2 (a) 0.136 (b) 0.272 (c) 0.816 (d) 0.0227 + 5 2 2 7 2 3+ 2 Answer: [d] 'n' factor of Cr2O72 Equivalents of Mole of is 6. Cr2O72 Cr2O72 = needed = equivalents of 0.136 6 N2H5+ = 0.136 = 0.0227 18 28. Ethanol C H OH, is the substance commonly called "alcohol". The density of liquid ethanol is 0.7893 g ml at 20 C. If 1.2 mol of ethanol are needed for a particular experiment, what volume of ethanol should be measured out? 2 5 1 (a) 55 ml (b) 58 ml (c) 70 ml (d) 79 ml Answer: [c] Mass of ethanol needed = 1.2 46 = 55.2 Volume ethanol = of measured out Mass of ethanol 55.2 = = 70 ml Density of ethanol 0.7893 29. 2 litre carbogen (at S.T.P.) is passed through a solution of lime water to get 0.5 g white precipitate. The percentage of carbon dioxide gas in carbogen by volume is (a) 8 (b) 11.2 (c) 5.6 (d) 22.4 Answer: [c] Ca(OH)2 + CO2 CaCO3 + H2O 0.5 g No. of moles of % of CO2 CO2 in 2 lit carbogen in carbogen by vol = = 5 1000 5 22.4 100 = 5.6 1000 2 19 30. 1000 ml of oxygen at STP were passed through an ozoniser and the resulting volume was 888 ml at STP. This quantity of ozonised oxygen is passed through excess of KI solution. The weight of l liberated is 2 (a) 25.4 g (c) 2.54 g Answer: [c] (b) 12.7 g (d) 1.27 g In ozoniser, never 100%. O2 is converted to O3 and transformation is 3O 2O3 2 1000 3x Where 2x 3x is the volume of O2 converted to O3 . 1000 3x + 2x = 888 x= 112 ml Volume of Moles of O = O3 at STP = 224 ml 224 = 0.01 22400 3 O3 + 2Kl + H2O 2KOH + l2 + O2 Moles of Weight of l2 = 0.01 l2 liberated = 0.01 254 = 2.54 g 31. The average molecular wt of a mixture of gas containing nitrogen and carbon dioxide is 36. The mixture contain 280 20 gm of nitrogen, therefore, the amount of CO2 present in the mixture is (A) 440 gm (B) 44 gm (C) 0.1 mole (D) 880 gm Answer: [A] The amount of CO2 present in the mixture is 440 gm. 32. In a reaction FeS is oxidised by O to Fe O and SO . If the equivalent of O O2 consumed is y, then the equivalents of Fe O and SO with respect to FeS are 2 2 2 3 2 2 2 2 3 2 and y (a) y and y (b) y 2 (c) (d) 10y y and 11 11 y 10y and 11 11 Answer: [c] +2 1 +3 +4 2Fe S2 + 11/ 2 O2 Fe2 O3 + 4 S O2 (n = 11) (n = 2) Equivalents of O2 = Moles of FeS2 = Moles of Fe2O3 = (n = 5) Equivalents of FeS2 = y y 11 y 2 11 Equivalents of Fe = 2 O3 y 2 y = 2 11 11 21 Moles of SO2 = 2y 11 Equivalents of = SO 2y 5 10y = 11 11 2 33. One mole of a mixture of CO and CO requires exactly 20 grams of NaOH to convert all the CO into Na CO . How many more grams of NaOH would it require for conversion into Na CO if the mixture (one mole) is completely oxidised to CO . 2 2 2 2 3 3 2 (a) 60 g (b) 80 g (c) 40 g (d) 20 g Answer: [a] 2NaOH + CO2 Na2CO3 + H2O Moles of NaOH= Moles of Moles of = CO CO + CO2 = 20 1 = 40 2 1 1 1 = 2 2 4 1 3 1 = 4 4 1 O2 CO2 2 Moles of CO2 produced Moles of NaOH extra = = 3 4 from CO 3 3 2 = 4 2 22 Mass of NaOH extra 3 = 40 =60 g 2 34. The equivalent weight of iron in Fe O would be (A) 18.6 (B) 28 (C) 56 (D) 112 Answer: [A] 112 parts of Fe combines with 48 parts of In Fe O , 2 56 = oxygen. Hence 8 parts of oxygen will combine with 2 2 = Fe 3 3 112 = 18.6 6 35. When 3.0 litre solution of normality N is mixed with 5.0 litre of 4M HCl, then the resultant solution has the normality 10. Find the value of N. (A) (D) Answer: [B] As we know, 10 25 (B) 20 (C) 15 N1V1 + N2 V2 = NV Normality of 4M HCl solution will be equal to 4N. Given, N , N = N , 4N and V , V = 3.0 and 5.0 L respectively. Normality of resultant solution = 10 and final volume = 3.0 L + 5.0 L= 8 L 1 2 1 2 N 3.0 + 4 5.0 = 10 8 3N + 20 = 80 3N = 60 60 N= = 20 3 23 36. A 0.1097 gm sample of As O required 26.10 mL of KMnO solution for its titration. The molarity of KMnO solution is (A) 0.02 (B) 0.04 (C) 0.018 (D) 0.3 Answer: [C] 2 3 4 4 n-factor = 5 As2O3 + MnO4 2AsO43 + Mn2+ n-factor = 4 Let molarity of KMnO solution be M Eq. of As O = Eq. of KMnO solution 4 2 3 4 0.1097 26.10 M 5 198 = 4 = (Equivalent weight As2O3 ) 198 1000 4 = Molarity 0.017M 0.018 37. 112 ml of a gas is produced at STP by the action of 0.412 gm of ROH alcohol with CH Mgl. Molecular mass of alcohol is: (A) 32g (B) 41.2g (C) 82.4g (D) 156g Answer: [C] 3 ROH + CH3MgI CH4 + MgI ( OR ) 1 mole 1 mole So gas produced is n= CH4 CH4 w ROH 112 = n= ROH 22400 MWKOH MW =0.412 22400 =82.4 g 112 38. 12 gm of Mg (At mass = 24) will react with an acid to give 24 (A) one mole of H (C) half moles of H Answer: [C] (B) one mole of O (D) half mole of O 2 2 2 2 Mg+ 2H+ Mg++ + H2 24g 1 mole 12 g of Mg gives half mole of H2 . 39. When pentane C H , is burned in excess oxygen, the products of the reaction are CO (g) and H O(l). In the balanced equation for this combustion C H (g) + _____ O (g) 5CO (g) + 6H O(l) the coefficient of oxygen should be (a) 16 (b) 12 (c) 11 (d) 8 Answer: [d] Using the POAC method, the balanced equation will be 5 12 2 5 12 2 2 2 2 C5H12 + 8O2 5CO2 + 6H2O Coefficient of oxygen is 8. 40. The volume of 0.25 M 0.03 M Ca(OH) is (A) 1.32 mL (C) 26.4 mL Answer: [B] H3PO3 required to neutralise 25 ml of 2 (B) 3 mL (D) 2.0 mL Meq. of H3PO4 = Meq. of Ca(OH)2 V 0.25 2 = 25 0.03 2(H3PO3 is diabasic acid) 25 = V 25 3 2 = 3mL 25 2 41. When 16.6 g of KI is treated with excess of KIO in presence of 6N HCl, ICl is produced. The amount of KIO consumed and the ICl formed are (a) 0.1 mol and 0.3 mol (b) 0.05 mol and 0.3 mol (c) 0.05 mol and 0.15 mol (d) 0.1 mol and 0.15 mol Answer: [c] 3 3 1 2K I + +5 +1 HCl 3 I Cl K I O3 = n 2= n 4 16.6 = 0 10.05 166 0.15 42. No. of moles in 11.2 L of (A) 0.5 (B) 0.3 (C) 0.2 (D) 0.1 Answer: [A] = n CO2 at NTP is 11.2 = 0.5 22.4 43. The number of ions present in 2.0 L of a solution of 0.8 M K [Fe(CN) ] is (B) 4.8 10 (A) 4.8 10 (C) 9.6 10 (D) 9.6 10 4 6 22 24 24 22 Answer: [B] 26 K4[Fe(CN)6] Molecules Ions 5 1 (0.8 2) 6 10 23 ? = 4.8 1024 44. Assuming full decomposition, the volume of CO released at STP on heating 9.85 g of BaCO (Atomic mass of Ba = 137) will be (A) 0.84 L (B) 2.24 L (C) 4.06 L (D) 1.12 L Answer: [D] 2 3 BaCO3 BaO + CO2 Molecular weight of BaCO = 137 + 12 + 3516 = 197 97 gm BaCO produces 22.4 L CO (g) at S.T.P. 3 2 3 9.85 gm produces 22.4 9.85 = 1.12L 197 at S.T.P. 45. If 30 mL of H and 20 mL of O react to form water, what is left at the end of the reaction? (A) 10 mL of H (B) 5 mL of H (C) 10 mL of O (D) 5 mL of O Answer: [D] 2 2 2 2 2 2 2H2 + O2 2H2O Combining volume ratio of H and O = 2 : 1 Here H is limiting reactant. 30 ml of H will react with 15 ml of O and 5 ml of O will remain unreacted at the end. 2 2 2 2 2 2 27

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