February 14, 2024

Explore the comprehensive NCERT Solutions for Class 12 Science Chemistry Chapter 2 – Solutions, featuring easy-to-follow step-by-step explanations. These solutions have gained immense popularity among Class 12 Science students, serving as a valuable resource for completing homework efficiently and preparing for exams. All the questions and answers from Chapter 2 of the NCERT Book for Class 12 Science Chemistry are available here at no cost, ensuring convenient access for your academic needs. Page No 37: Question 2.1: Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride. ANSWER: Mass percentage of C6H6  Mass percentage of CCl4 Alternatively, Mass percentage of CCl4 = (100 − 15.28)% = 84.72% Page No 37: Question 2.2: Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. ANSWER: Let the total mass of the solution be 100 g and the mass of benzene be 30 g. ∴Mass of carbon tetrachloride = (100 − 30)g = 70 g Molar mass of benzene (C6H6) = (6 × 12 + 6 × 1) g mol−1 = 78 g mol−1 ∴Number of moles of  = 0.3846 mol Molar mass of carbon tetrachloride (CCl4) = 1 × 12 + 4 × 35.5 = 154 g mol−1 ∴Number of moles of CCl4 = 0.4545 mol Thus, the mole fraction of C6H6 is given as: = 0.458 Page No 37: Question 2.3: Calculate the molarity of each of the following solutions: (a) 30 g of Co(NO3)2. 6H2O in 4.3 L of solution (b) 30 mL of 0.5 M H2SO4 diluted to 500 mL. ANSWER: Molarity is given by: (a) Molar mass of Co (NO3)2.6H2O = 59 + 2 (14 + 3 × 16) + 6 × 18 = 291 g mol−1 ∴Moles of Co (NO3)2.6H2O = 0.103 mol Therefore, molarity  = 0.023 M (b) Number of moles present in 1000 mL of 0.5 M H2SO4 = 0.5 mol ∴Number of moles present in 30 mL of 0.5 M H2SO4 = 0.015 mol Therefore, molarity = 0.03 M Page No 37: Question 2.4: Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution. ANSWER: Molar mass of urea (NH2CONH2) = 2(1 × 14 + 2 × 1) + 1 × 12 + 1 × 16 = 60 g mol−1 0.25 molar aqueous solution of urea means: 1000 g of water contains 0.25 mol = (0.25 × 60)g of urea = 15 g of urea That is, (1000 + 15) g of solution contains 15 g of urea Therefore, 2.5 kg (2500 g) of solution contains  = 36.95 g = 37 g of urea (approximately) Hence, mass of urea required = 37 g Note: There is a slight variation in this answer and the one given in the NCERT textbook. Page No 37: Question 2.5: Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20% (mass/mass) aqueous KI is 1.202 g mL-1. ANSWER: (a) Molar mass of KI = 39 + 127 = 166 g mol−1 20% (mass/mass) aqueous solution of KI means 20 g of KI is present in 100 g of solution. That is, 20 g of KI is present in (100 − 20) g of water = 80 g of water Therefore, molality of the solution  = 1.506 m = 1.51 m (approximately) (b) It is given that the density of the solution = 1.202 g mL−1 ∴Volume of 100 g solution  = 83.19 mL = 83.19 × 10−3 L Therefore, molarity of the solution  = 1.45 M (c) Moles of KI  Moles of water  Therefore, mole fraction of KI  = 0.0263 Page No 41: Question 2.6: H2S, a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of H2S in water at STP is 0.195 m, calculate Henry’s law constant. ANSWER: It is given that the solubility of H2S in water at STP is 0.195 m, i.e., 0.195 mol of H2S is dissolved in 1000 g of water. Moles of water  = 55.56 mol ∴Mole fraction of H2S, x = 0.0035 At STP, pressure (p) = 0.987 bar According to Henry’s law: p = KHx = 282 bar Page No 41: Question 2.7: Henry’s law constant for CO2 in water is 1.67 × 108 Pa at 298 K. Calculate the quantity of CO2 in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K. ANSWER: It is given that: KH = 1.67 × 108 Pa = 2.5 atm = 2.5 × 1.01325 × 105 Pa = 2.533125 × 105 Pa According to Henry’s law: = 0.00152 We can write,  [Since, is negligible as compared to] In 500 mL of soda water, the volume of water = 500 mL [Neglecting the amount of soda present] We can write: 500 mL of water = 500 g of water = 27.78 mol of water Now,  Hence, quantity of CO2 in 500 mL of soda water = (0.042 × 44)g = 1.848 g Page No 47: Question 2.8: The vapour pressure of pure liquids A and B are 450 and 700 mm Hg respectively, at 350 K. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of the vapour phase. ANSWER: It is given that: = 450 mm of Hg = 700 mm of Hg ptotal = 600 mm of Hg From Raoult’s law, we have: Therefore, total pressure,  Therefore,  = 1 − 0.4 = 0.6 Now,  = 450 × 0.4 = 180 mm of Hg = 700 × 0.6 = 420 mm of Hg Now, in the vapour phase: Mole fraction of liquid A = 0.30 And, mole fraction of liquid B = 1 − 0.30 = 0.70 Page No 55: Question 2.9: Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH2CONH2) is dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and its relative lowering. ANSWER: It is given that vapour pressure of water, = 23.8 mm of Hg Weight of water taken, w1 = 850 g Weight of urea taken, w2 = 50 g Molecular weight of …

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