In-text question solution

1. Calculate the mass percentage of benzene () and carbon tetrachloride () if 22 g of benzene is dissolved in 122 g of carbon tetrachloride. Ans. Mass percentage of   = 15.28% Mass percentage of =  =84.72% Alternatively, Mass percentage of = (100 - 15.28) % = 84.72% 2. 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 () = (612 + 6)  = 78  ∴Number of moles of  = 0.3846 mol Molar mass of carbon tetrachloride ( ) = 112 + 4355 = 154  ∴Number of moles of  = 0.4545 mol Thus, the mole fraction of is given as:  = 0.458 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 diluted to 500 mL. Ans. Molarity is given by:  (a) Molar mass of Co(NO3)2 = 59 + 2 (14 + 3 16) + 618 = 291  Therefore, Moles of Co(NO3)2 = 30/291 mol = 0.103 mol Therefore, molarity  = 0.023 M (b) Number of moles present in 1000 mL of 0.5 M  ∴Number of moles present in 30 mL of 0.5 M  = 0.015 mol Therefore, molarity = 0.03 M 4. Calculate the mass of urea () required in making 2.5 kg of 0.25 molal aqueous solution. Ans. Molar mass of urea  = 60  0.25 molar aqueous solution of urea means: 1000 g of water contains 0.25 mol =  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 5. Calculate (a)molality (b)molarity and (c)mole fraction of KI if the density of 20% (mass/mass) aqueous KI is. Ans. (a) Molar mass of KI = 39 + 127 = 166  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 =  Therefore, Volume of 100 g solution =  = = 83.19 mL =  Therefore, molarity of the solution =  = 1.45 M (c) Moles of KI =  Moles of water =  Therefore, mole fraction of KI =  =  = 0.0263 6. , a toxic gas with rotten egg like smell, is used for the qualitative analysis. If the solubility of  in water at STP is 0.195 m, calculate Henry's law constant. Ans. It is given that the solubility of in water at STP is 0.195 m, i.e., 0.195 mol of is dissolved in 1000 g of water. Moles of water =  = 55.56 mol ∴Mole fraction of , x =  =  = 0.0035 At STP, pressure (p) = 0.987 bar According to Henry's law: p= KHx  = 282 bar 7. Henry's law constant for  in water is 1.67 x 108 Pa at 298 K. Calculate the quantity of in 500 mL of soda water when packed under 2.5 atm pressure at 298 K. Ans. It is given that: KH = 1.67 x 108 Pa = 2.5 atm = 2.5 x 1.01325 x 105 = 2.533 x 105 Pa According to Henry's law: PCO2 = KHx x=  = 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 = mol of water = 27.78 mol of water Now,  Hence, quantity of in 500 mL of soda water =  = 1.848 g 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. Ans. It is given that:  = 450 mm of Hg = 700 mm of Hg P total= 600 mm of Hg From Raoult's law, we have:  Therefore, total pressure,   600= (450-700)xA + 700 -100 = -250 xA xA = 100/250 = 0.4 Therefore,  = 1 - 0.4 = 0.6 Now,  =  = 180 mm of Hg  =  = 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 9. Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea () 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, of Hg Weight of water taken,  Weight of urea taken,  Molecular weight of water,  Molecular weight of urea,  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. 10. Boiling point of water at 750 mm Hg is . How much sucrose is to be added to 500 g of water such that it boils at . Molal elevation constant for water is. Ans. Here, elevation of boiling point = (100 + 273) – (99.63 + 273) = 0.37 K Mass of water,  Molar mass of sucrose ( ),  =  Molal elevation constant, Kb =  We know that:  = 121.67 g (approximately) Hence, 121.67 g of sucrose is to be added. 11. Calculate the mass of ascorbic acid (Vitamin C,) to be dissolved in 75 g of acetic acid to lower its melting point by.. Ans. Mass of acetic acid,  Molar mass of ascorbic acid  =  Lowering of melting point,  We know that:   = 5.08 g (approx) Hence, 5.08 g of ascorbic acid is needed to be dissolved. 12. 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. 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,  We know that: Osmotic pressure,   = 30.98 Pa = 31 Pa (approximately)

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