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# Solutions

In our day to day life we rarely come across pure substances. Most of these contain homogenous mixtures of two or more pure substances. During summer after returning home from work it feels refreshing to have a glass lemon juice. What is it actually? It is nothing but a mixture of salt, sugar, lemon juice in water. It is a solution. A solution is a homogenous mixture of two or more components. The component present in larger quantity is referred to as solvent and the components present in smaller quantity is referred to as solute. In this chapter we will study different types of solutions, liquid solution, properties of the solutions and various alternatives in which concentrations of a solute can be expressed in solution.

•  Q1 Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl4) if 22 g of benzene is dissolved in 122 g of carbon tetrachloride. Ans: We know that: Mass percentage of C6H6  and we also know that: Mass percentage of CCl4  Alternatively, Mass percentage of CCl4= (100 - 15.28)% = 84.72% Q2 Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. Ans: 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 C6H6  =30/78 mol = 0.3846 mol Molar mass of carbon tetrachloride (CCl4) = 1 × 12 + 4 × 35.5 = 154 g mol - 1 ∴ Number of moles of CCl4 = 70/154 mol = 0.4545 mol Thus, the mole fraction of C6H6 is given as: = 0.3846 / (0.3846 +0.4545) = 0.458 Q3 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. Ans: Molarity is given by: Molarity =  moles of solute / Volume of solution in litre (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  = 30 / 291 mol = 0.103 mol Therefore, molarity  = 0.103 mol / 4.3 L = 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   = (0.5 X 30 ) / 1000 mol = 0.015 mol Therefore, molarity = 0.015 mol / 0.5 L = 0.03 M Q4 Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution. Ans: 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 =  (15 X 2500) / (1000+15) g = 36.95 g = 37 g of urea (approximately) Hence, mass of urea required = 37 g Q5 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. Ans: (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 = Moles of KI / Mass of water in kg = 20/166 / 0.08 m = 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 = Mass /  Density = 100g / 1.202g mL-1 = 83.19 mL = 83.19 × 10 - 3 L Therefore, molarity of the solution = 20/166 mol / 83.19 × 10 - 3 L = 1.45 M   (c) Moles of KI = 20/166 = 0.12 mol Moles of water = 80/18 = 4.44 mol Therefore, mole fraction of KI = Moles of KI / (Moles of KI + Moles of water) = 0.12 / (0.12+4.44) = 0.0263 Q6 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. Ans: 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 = 1000g / 18g mol-1 = 55.56 mol   ∴Mole fraction of H2S, x =  Moles of H2S / Moles of H2S+Moles of water 0.195 / (0.195+55.56) = 0.0035   At STP, pressure (p) = 0.987 bar According to Henry's law: p= KHx ⇒ KH = p / x = 0.0987 / 0.0035 bar = 282 bar Q7 Henry's law constant for CO2 in water is 1.67 x 108Pa at 298 K. Calculate the quantity of CO2in 500 mL of soda water when packed under 2.5 atm CO2 pressure at 298 K. Ans: It is given that: KH= 1.67 × 108Pa PCO2 = 2.5 atm = 2.5 × 1.01325 × 105Pa = 2.533125 × 105Pa   According to Henry's law: PCO2 = KHX ⇒ x =  PCO2 / KH = 2.533125 × 105 / 1.67 × 108 = 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 =500 / 18 mole of water = 27.78 mol of water Now, nCO2 / nH2O = x nCO2 / 27.78 = 0.00152 nCO2 = 0.042 mol Hence, quantity of CO2 in 500 mL of soda water = (0.042 × 44)g = 1.848 g Q8 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. Ans: It is given that: PAo = 450 mm of Hg PBo = 700 mm of Hg ptotal = 600 mm of Hg From Raoult's law, we have: ptotal = PA + PB Therefore, xB = 1 - xA = 1 - 0.4 = 0.6   Now,  PA = PAo xA = 450 × 0.4 = 180 mm of Hg and PB = PBo xB = 700 × 0.6 = 420 mm of Hg Now, in the vapour phase: Mole fraction of liquid A =  PA / (PA + PB ) =180 / (180+420) = 180/600 = 0.30 And, mole fraction of liquid B = 1 - 0.30 = 0.70 Q9 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. Ans: It is given that vapour pressure of water,PIo = 23.8 mm of Hg Weight of water taken, w1= 850 g Weight of urea taken, w2= 50 g Molecular weight of water, M1= 18 g mol - 1 Molecular weight of urea, M2= 60 g mol - 1 Now, we have to calculate vapour pressure of water in the solution. We take vapour pressure as p1. Now, from Raoult's law, we have: Hence, the vapour pressure of water in the given solution is 23.4 mm of Hg and its relative lowering is 0.0173. Q10 Boiling point of water at 750 mm Hg is 99.63°C. How much sucrose is to be added to 500 g of water such that it boils at 100°C.Molal elevation constant for water is 0.52 K kg mol-1. Ans: Here, elevation of boiling point ΔTb= (100 + 273) - (99.63 + 273) = 0.37 K Mass of water, wl = 500 g Molar mass of sucrose (C12H22O11), M2= 11 × 12 + 22 × 1 + 11 × 16 = 342 g mol - 1 Molal elevation constant, Kb= 0.52 K kg mol - 1 We know that: = (0.37 x 342 x 500) /  (0.52 x 1000) = 121.67 g (approximately) Hence, 121.67 g of sucrose is to be added. Q11 Calculate the mass of ascorbic acid (Vitamin C, C6H8O6) to be dissolved in 75 g of acetic acid to lower its melting point by 1.5°C. Kf= 3.9 K kg mol-1. Ans: Mass of acetic acid, w1= 75 g Molar mass of ascorbic acid (C6H8O6), M2= 6 × 12 + 8 × 1 + 6 × 16 = 176 g mol - 1 Lowering of melting point, ΔTf = 1.5 K We know that: =  (1.5 x 176 x 75) / (3.9 x 1000)   = 5.08 g (approx) Hence, 5.08 g of ascorbic acid is needed to be dissolved. Q12 Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 g of polymer of molar mass 185,000 in 450 mL of water at 37°C. Ans: It is given that: Volume of water, V= 450 mL = 0.45 L Temperature, T = (37 + 273)K = 310 K Number of moles of the polymer, n = 1 / 185000 mol We know that: Osmotic pressure, = 1/185000 mol X 1/0.45L X 8.314 X 103 Pa L K-1mol-1 X 310K    = 30.98 Pa = 31 Pa (approximately)