Mole Concept Solved Problems

Page # 3 MOLE CONCEPT SOLVED PROBLEMS Ex.1 8 litre of H2 and 6 litre of Cl2 are allowed to react to maximum possible extent. Find out the final volume of reaction mixture. Suppose P and T remains constant throughout the course of reaction (A) 7 litre (B) 14 litre (C) 2 litre (D) None of these. Sol. (B) H2 + Cl2  2 HCl Volume before reaction 8 lit 6 lit 0 Volume after reaction 2 0 12  Volume after reaction = Volume of H2 left + Volume of HCl formed = 2 + 12 = 14 lit Ex.2 Naturally occurring chlorine is 75.53% which has an atomic mass of 34.969 amu and 24.47% Cl37 which has a mass of 36.966 amu. Calculate the average atomic mass of chlorine (A) 35.5 amu (B) 36.5 amu (C) 71 amu (D) 72 amu Sol. (A) Average atomic mass Cl 35 % of I isotope × its atomic mass + = = % of II isotope × its atomic mass 100 75.53x34.969 + 24.47x36.96 100 = 35.5 amu. Ex.3 Calculate the mass in gm of 2g atom of Mg(A) 12 gm (B) 24 gm (C) 6 gm (D) None of these. Sol. (D)  1 gm atom of Mg has mass = 24 gm  2 gm atom of Mg has mass = 24 x 2 = 48 gm. Ex.4 In 5 g atom of Ag (At. wt. of Ag = 108), calculate the weight of one atom of Ag (A) 17.93 × 10–23gm (B) 16.93 × 10–23 gm (C) 17.93 × 1023 gm (D) 36 × 10–23 gm Sol. (A)  N atoms of Ag weigh 108 gm  108 N 1 atom of Ag weigh = 1 = 108 6.023 x10 23 = 17.93 × 10–23 gm. Ex.5 In 5g atom of Ag (at. wt. = 108), calculate the no. of atoms of Ag (A) 1 N (B) 3N (C) 5 N (D) 7 N. Sol. (C)  1 gm atom of Ag has atoms = N  5 gm atom of Ag has atoms = 5N. Ex.6 Calculate the mass in gm of 2N molecules of CO2 (A) 22 gm (B) 44 gm (C) 88 gm (D) None of these. Sol. (C)  N molecules of CO2 has molecular mass = 44.  2N molecules of CO2 has molecular mass = 44 x 2 = 88 gm. Ex.7 How many carbon atoms are present in 0.35 mol of C6H12O6 (A) 6.023 × 1023 carbon atoms (B) 1.26 × 1023 carbon atoms (C) 1.26 × 1024 carbon atoms (D) 6.023 × 1024 carbon atoms Sol. (C)  1 mol of C6H12O6 has = 6 N atoms of C  0.35 mol of C6H12O6 has = 6 × 0.35 N atoms of C = 2.1 N atoms = 2.1 × 6.023 × 1023 = 1.26 × 1024 carbon atoms Ex.8 How many molecules are in 5.23 gm of glucose (C6H12O6) (A) 1.65 × 1022 (B) 1.75 × 1022 21 (C) 1.75 × 10 (D) None of these Sol. (B)  180 gm glucose has = N molecules 5.23  6.023  10 23  5.23 gm glucose has = 180 = 1.75 × 1022 molecules Ex.9 What is the weight of 3.01 × 1023 molecules of ammonia (A) 17 gm (B) 8.5 gm (C) 34 gm (D) None of these Sol. (B)  6.023 × 1023 molecules of NH3 has weight = 17 gm Page # 4 MOLE CONCEPT  3.01 × 1023 = molecules of NH3 has weight 17  3.01 10 23 6.023  10 23 = 6.644 × 10–23 gm = 8.50 gm  weight of 'N' atoms of element = 6.644 × 10–23 × 6.023 × 1023 = 40 gm Ex.10 How many significant figures are in each of the following numbers (b) 6.023 × 1023 (a) 4.003 (c) 5000 (A) 3, 4, 1 (B) 4, 3, 2 (C) 4, 4, 4 (D) 3, 4, 3 Sol. (C)  40 gm of element has 1 gm atom. 40  10 3  40 x 103 gm of element has 40 = 103 gm atom. Ex.14 Calculate the number of Cl– and Ca+2 ions in 222 g anhydrous CaCl2 - Ex.11 How many molecules are present in one ml of water vapours at STP (A) 1.69 × 1019 (B) 2.69 × 10–19 (C) 1.69 × 10–19 (D) 2.69 × 1019 Sol.  weight of 1 atom of element (D)  22.4 litre water vapour at STP has = 6.023 × 1023 molecules  1 × 10–3 litre water vapours at STP has = 6.023  10 23 22.4 × 10–3 (A) 2N ions of Ca+2 4 N ions of Cl– (B) 2N ions of Cl– & 4N ions of Ca+2 (C) 1N ions of Ca+2 & 1N ions of Cl– (D) None of these. Sol.  mol. wt. of CaCl2 = 111 g  111 g CaCl2 has = N ions of Ca+2  222g of CaCl2 has Also Ex.12 How many years it would take to spend Calculate the standard molar volume of gas- (C) 19.098 × 109 years (D) None of these (C)  106 rupees are spent in 1sec.  6.023 × 1023 rupees are spent in = 10 6 10 6  60  60  24  365 (A) 22.4 lit. (B) 11.2 lit (C) 33.6 lit (D) 5.6 lit. Sol. (A)  1.429 gm of O2 gas occupies volume = 1 litre.  32 gm of O2 gas occupies = sec 1 6.023  10 23 2N  222 ions of Cl– 111 Ex.15 The density of O2 at NTP is 1.429g / litre. (B) 19.098 years 1 6.023  10 23 222 g CaCl2 has = = 4N ions of Cl – . million rupees in one second - =  111 g CaCl2 has = 2N ions of Cl –  Avogadro's number of rupees at the rate of 1 Sol. N  222 111 = 2N ions of Ca+2 = 2.69 × 10+19 (A) 19.098 × 1019 years (A) = 22.4 litre/mol. years = 19.098 × 109 year Ex.16 Which of the following will weigh maximum amount(A) 40 g iron Ex.13 An atom of an element weighs 6.644 × (B) 1.2 g atom of N 10–23 g. Calculate g atoms of element in 40 kg- (C) 1 × 1023 atoms of carbon (A) 10 gm atom (B) 100 gm atom (D) 1.12 litre of O2 at STP (C) 1000 gm atom (D) 104 gm atom Sol. Sol. (C) 32 1429 . (A) (A)Mass of iron = 40 g Page # 5 MOLE CONCEPT (B) Mass of 1.2 g atom of Ex.20 A silver coin weighing 11.34 g was N = 14 × 1.2 = 16.8 gm dissolved in nitric acid. When sodium chloride 1023 was added to the solution all the silver (present (C)Mass of 1 × = 12  1 10 23 6.023  10 23 atoms of C = 1.99 gm. (D)Mass of 1.12 litre of O2 at STP 32  1 .2 = = 1.6 g 22 .4 Ex.17 How many moles of potassium chlorate to be heated to produce 11.2 litre oxygen 1 1 (A) mol (B) mol 2 3 (C) 1 mol 4 (D) 2 mol. 3 Sol. (B) 2 KClO3   2KCl + 3O2 Mole for reaction  3 × 22.4 litre O2 is formed by 2 mol KClO3 2  11.2  11.2 litre O2 is formed by 3  22.4 1 = mol KClO3 3 Ex.18 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. (D)  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.  190 kg CaCO3 gives CaO = 56  190 = 106.4 kg. 100 Ex.19 The chloride of a metal has the formula MCl3. The formula of its phosphate will be(A) M2PO4 (B) MPO4 (C) M3PO4 (D) M(PO4)2 Sol. (B) AlCl 3 as it is AlPO4 as AgNO3) was precipitated as silver chloride. The weight of the precipitated silver chloride was 14.35 g. Calculate the percentage of silver in the coin (A) 4.8 % (B) 95.2% (C) 90 % (D) 80% Sol. (B) Ag + 2HNO3   AgNO3 + NO2 + H2O 108 AgNO3 + NaCl  AgCl + NaNO3 143.5  143.5 gm of silver chloride would be precipitated by 108 g of silver. or 14 .35 g of si l ver chl ori de would be precipitated 10.8 g of silver.  11.34 g of silver coin contain 10.8 g of pure silver.  100 g of silver coin contain 10.8 ×100=95.2 %. 1134 .