Solution Solved Problems

Page # 28 SOLUTION SOLVED PROBLEMS Q.1 Calculate the mole fraction of benzene in solution containing 30% by mass in carbon tetrachloride. Sol. 30% by mass of benzene in CCl4 means 30 g benzene dissolved in 70 g of CCl4 Molecular mass of C6H6 = 78 Molecular mass of CCl4 = 154 Number of moles of benzene = Number of moles of CCl4 = xBenzene = Q.2 30 = 0.38 mol 78 70 = 0.45 mol 154 0.38 0.38 = = 0.485 0.38  0.45 0.83 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 Sol. (a) Molarity of Co(NO3)2.6H2O M= Mass of Co(NO3 )2 .6H2O  1000 Molecular mass  Volume in mL M= 30 1000  = 0.023 M 310.7 4300 (b) We know that M 1 V1 = M 2 V2 30 × 0.5 = M2 × 500 M2 = 30  0.5 = 0.03 M 500 Q.3 Calculate the mass of urea (NH2CONH2) required in making 2.5 kg of 0.25 molal aqueous solution. Sol. Urea has molar mass = 60 g mol–1 w2 = Mas sof solute w1 = Mass of the solvent M2 = Molecular mass of solute w 2 1000 m= M  w 2 1 0.25 = w2 = w 2 1000  60 2500 0.25  60  2500 = 37.50 g 1000 Page # 29 SOLUTION Q.4  Mass  Calculate (a) molality (b) molarity and (c) mole fraction of KI if the density of 20%    Mass  –1 aqueous solution KI is 1.202 gL . Sol.  Mass  20%   KI means 20 g of KI is dissolved in 80 g of water..  Mass  Molar mass of KI = 166 g w 2 1000 20 1000 (a) Molality of KI = M  w  166  80 = 1.5 m 2 1 w2 1000  Density 20 1000  1.202 (b) Molarity of KI = M  Solution in gram  166  = 1.448 m 100 1 80 = 04.4 mol 18 nKI 0.12  = n n 4 . 4  0.12 KI H2O (c) No. of moles of KI = xKI xKI = Q.5 Sol. 0.12 = 0.0265 4.52 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. Molarity = 0.195 m 0.195 mole of H2S dissolved in 1000 g of solvent water. 0.195 0.195  55.55  0.195 55.745 p = KHx Pressure at STP = 0.987 bar xH2S   0.987 = KH × 0.195 = KH × 0.0035 55.745 KH = 282 bar Q.6 Sol. 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. KH = 1.67 × 108 Pa PCO2 = 2.5 atm = 2.5 × 101325 Pa Applying Henry's law, PCO2 = 'KH × x CO2  x CO2 = pCO 2 KH  2.5  101325 Pa 1.67  10 8 Pa = 1.517 × 10–3 i.e., nCO 2 nH2O  nCO 2 For 500 mL of soda water, water present 500 mL = 500 g =  nCO2 27.78 = 1.517 × 10–3 500 = 2778 moles i.e., nH2O = 27.78 moles 18 nCO2 = 42.14 × 10–3 mole = 42.14 m mol = 42.14 × 10–3 × 44 g = 1.854 g ~ nCO2 nH2O = 1.517 × 10–3 Page # 30 SOLUTION Q.7 The vapour pressure of pure liquid A and B are 450 and 700 mm Hg at 350 K respectively. Find out the composition of the liquid mixture if total vapour pressure is 600 mm Hg. Also find the composition of vapour phase. Sol. For A and B liquids P = pºAxA × pºBxB So, 600 = 450 (xA) + 700 (1 – xA) (  x A + xB = 1 ) xA = 2 5 250 xA = 100  PA = pºAxA = 450 × xB = 3 5 2 = 180 mm Hg 5 PB = pºBxB = 700 × 3 = 420 mm Hg 5  Mole fraction of A component in the vapour phase =  Mole fraction of B component in the vapour phase = 180 = 0.3 600 420 = 0.7 600 Q.8 Boiling point of water at 750 mm Hg is 99.63º C. How much sucrose is to be added to 500 g of water such that it boils 100º C? Sol. Tºb = 99.63ºC, Tb = 100ºC, M2 = 342 g mol–1, Kb = 0.52 K kg mol–1 Tb = Tb – Tºb = 100 – 99.63 = 0.37ºC M2 = K b  1000  W2 Tb  W1 342 = W2 = 342  0.37  500 0.52  1000 W2 = 121.7 g 0.52  1000  W2 0.37  500 Q.9 A solution is obtained by mixing 300 g of a 25 % and 400 g of a 40% solution by mass. Calculate the mass percentage of the resulting solution. Sol. 25% solution means 25 g solute in 100 g solution. 40% solution means 40 g solute in 100 g solution  Mass of solute in 300 g solution = 25  300 = 75 g 100  Mass of solute in 400 g solution = 40  400 = 160 g 100 Total mass of solute = 75 + 160 = 235 g Total mass of the solution = 300 g + 400 g = 700 g So, mass percentage of solute in mixture = 235  100 = 33.57% 700 mass percentage of solvent in mixture = 100 - 33.57 = 66.43 % Page # 31 SOLUTION Q.10 An antifreeze solution is prepared from 222.6 g of ethylene glycol (C2H6O2) and 200 g of water. Calculate the molality of the solution. If the density of the solution is 1.072 g mL–1, then what shall be the molarity of the solution? Sol. Molecular mass of C2H6O2 = 2 × 12 + 6 × 1 + 2 × 16 = 62 According to question, WB = 222.6 g, MB = 62, WA = 200 g WB  1000 Molality (m) = M  W B A 222.6  1000 = 17.95 m 62  200 Molality of ethylene glycol = Mass of solution = Mass of glycol + Mass of water 222.6 + 200 = 422.6 g Volume of solution (V) = Mass of solution 422.6 = mL = 394.2 mL Density 1.072 WB  1000 Molarity (M) = M  V(mL ) B Molarity of ethylene glycol = 222.6  1000 = 9.11 M 62  394.21 Q.11 the partial pressure of ethane over a solution 6.56 × 10–3 g of ethane is 1 bar. If the solution contains 5.00 × 10–2 g of ethane, then what shall be the partial pressure of the gas? Sol. According to Henry's law m  P, thus, and m = KHP 6.56 × 10–3 = KH × 1 (P = 1 bar) 5.00 × 10–2 = KH × 1 .............(i) ...............(ii) Compare equation (i) and (ii), we have 6.56  10 3 5.0  10 2  KH  1 1 5.0  10 2 P= = 7.62 bar K HP 6.56  10 3 Q.12 An aqueous solution of a 2 percent non-volatile solution exerts a pressure of 1.004 bar at the normal boiling point of the solvent. What is the molar mass of the solute? Sol. Given ps = 1.004 bar, pº = 1.013 bar 9at boiling point p = 1.013 bar) WB = 2g, WA = 100 – 2 = 98 g, MA = 18 (H2O) According to Raoult's law : p0A  p p0A MB = WB  WA = M W B A 2  18  1.013 98(1.013  1.004 ) WB  WA p0A MB = × 0 WA pA  p MB = 2  18  1.013 98  0.009 = 41.35 g mol–1 Q.13 A 55 solution (by mass ) of cane sugar in water has freezing point of 271 K. Calculate the freezing point of a 5% glucose in water if freezing point of pure water is 273.15 K. Page # 32 SOLUTION Sol. 5% glucose 5 g glucose in 100 g of solution WB = 5 g, WA = 100 – 5 = 95 g, Tf = 273.15 – 271 = 2.15 K, MB = 342 (for sugar) Tf = 1000  K f  WB MB  WA For sugar, 2.15 = 1000  K f  WB 342  95 For Glucose , Tf =  2.15 = Kf × 0.154 1000  K f  5  Tf = Kf × 0.292 180  95 ..........(i) ..........(ii) Compare the equation (i) and (ii), we have 2.15 K f  0.154  Tf K f  0.292 Tf = 2.15  0.292 = 4.08 K 0.154 Q.14 At 300 K, 36 g of glucose present in a litre of its solution has an osmotic pressure of 4.98 bar. If the osmotic pressure of solution is 1.52 bar at the same temperature, what would be its concentration? Sol. given, Osmotic pressure = 4.98 bar, W = 36 g, V = 1 litre Osmotic pressure = 1.52 bar, 4.98 × 1 = For case II : (Case II ) WB B = W RT B For case I :  (Case I ) 36  R  T or 4.98 = 0.2 RT 180 1.52 = C × R × T  1.52 = CRT ................(i) ................(ii) WB [C = M  V ] B Comparing the equations (i) and (ii), we have 4.98 0.2 RT  1.52 CRT C= 0.2  1.52 = 0.061 M 4.98 Q.15 Calculate the mass percentage of aspirin (C9H8O4) is acetonitrile (CH3CN) when 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN. Sol. Mass % of solute = Q.16 Nalorphene (C19H21NO3) similar to morphine, is used to combat withdrawal symptoms in narcotic users. Dose of nalorphene generally given is 1.5 mg. Calculate the mass of 1.5 × 10–3 m aqueous solution required for the above dose. Mass of solute  100 Mass of solution  Mass % of solute aspirin = 6.5 × 100 = 1.424 % 450  6.5 Page # 33 SOLUTION Sol. WB 1000 As we know m = M  W B A For nalorphene MB = 311 g mol–1, m = 1.5 × 10–3 m, WB = 1.5 mg = 1.5 × 10–3 g 1.5 × 10–3 = 1.5  10 3 1000  311 WA WA = 3.215 g Mass of solution = 3.215 g + 1.5 × 10–3 g = 3.2165 g Q.17 Sol. Q.18 Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15 M solution in methanol. WB 1000 W 1000 Molarity (M) = M . V(in mL ) 0.15 M = B . 122 250 B Calcualte the depression in the freezing point of water when 10 g of CH3CH2CHClCOOH is added to 250 g of water. Ka = 1.4 × 10–3, Sol. WB = 4.575 g Kf = 1.86 K kg mol–1 W2 1000 Molarity = M  W 2 1 For CH3CH2CHCOOH ` | Cl M2 = 122.5 m= 10 1000  = 0.33 122 .5 250 For monobasic acid. HA = Ka = C H+ + A– 1.4  10 3 = 0.065 0.33 i = 1 +  = 1 + 0.065 = 1.065 Tf = i, Kf m = (1.065) × (1.86) × 0.33 = 0.65º Q.19 Sol. 19.5 g CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of water observed is 1.0ºC. Calculate the van't Hoff factor and dissociation constant of fluoroacetic acid. W2 1000 19.5 1000 1 Molality = M  W  78  500  2 2 1 Tf = Kf × m Tf = 1.86 × = Ka = 1 = 0.93 2 i 1 1.0753  1 = m 1 2 1 i= ( T )obs 1 = = 1.0753 ( T ) exp 0.93  = 0.0753 C 2 (0.5)2  (0.075 )2  = 3.04 × 10–3 1  (1  0.075 ) Page # 34 SOLUTION Q.20 Vapour pressure of water at 293 K is 17.535 mm Hg. Calculate the vapour pressure of water at 298 K when 25 of glucose is dissolved in 450 g of water. Sol. As we know for dilute solution n2 WB MA p10  p A = x2 = n  M  W 0 p1 1 B A 17.535  p A 25  18 = 17.535 180  450 17.535  p A 1 = 17.535 180 17.535 ×180 – 180 pA = 17.535 pA = 17.44 mm of Hg. Q.21 Henry's law constant for the molarity of methane in benzene at 298 K is 4.27 × 105 mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg. Sol. According to Henry's law p = KH × x x= 760 mm of Hg 5 4.27  10 mm of Hg x= 760  10 5 = 178 × 10–5 4.27 Q.22 100 g of liquid A (molar mas s140 g mol–1) was dissolved in 1000 g of liquid B (molar mass 180 g mol–1). The vapour pressure of pure liquid B was found to be 500 torr. Calculate the vapour pressure of pure liquid A and its vapour pressure in the solution if the total vapour pressure of the solution is 475 torr. Sol. nA = 100 = 0.714 140 nB = xA = 0.714 0.714  = 0.114 5.555  0.714 6.269 xB = 1 – 0.114 = 0.886 1000 = 5.555 180 ptotal = pºAxA + pºBxB 475 = pºA × 0.114 + 0.886 × 500 pºA = 475  0.886  500 = 280.7 torr 0.114 We also know that pA = pºAxA = 280 .7 × 0.114 = 32 torr. Q.23 Benzene and toluene form ideal solution over the entire range of composition. The vapour pressure of pure benzene and naphthalene at 300 K are 50.71 mm Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in vapour phase of 80 g benzene is mixed with 100 g of naphthalene. Sol. No. of moles of benzene = 80 = 1.026 mol 78 No. of moles of toluene = 100 = 1.087 mol 92 Page # 35 SOLUTION xB = 1.026 = 0.486 1.026  1.087 xB + xT =1  xT = 1 – 0.486 = 0.514 Total pressure P = pºB × xB + pºT × xT = 50.71 × 0.486 + 32.06 × 0.514 = 41.13 Mole fraction benzene in vapour phase = 50.71 0.486 = 0.60 mol 41.13 Q.24 The air is a mixture of no. of gases. The major components are oxygen and nitrogen with approximate proportion of 20% is to 79% by volume at 298 K. the water is in equilibrium with air at a pressure of 10 atm. At 298 K if the Henry's law constant for oxygen and nitrogen are 3.3 × 107 mm and 6.51 × 107 mm respectively then calculate the composition of these gases in water. Sol. Consider 1 mole of air and let its volume at 100 atm be V Volume O2 = V  20 V  79 = 0.2 V; Volume of N2 = 0.79 V 100 100 Partial pressure of O2(Po2) = 10  0.79 V = 7.9 atm V Partial pressure of N (pN2 ) = Solutbility of O2 ( xO2 ) = Solutbility of N2 ( xN2 ) = 10  0.2V = 2 atm V pO2 kH p N2 KH = = 2  750(mm ) 3.30  10 7 (mm ) 7.9  750(mm ) 6.51 10 7 (mm ) = 4.606 × 10–5 = 9.22 × 10–5 Q.25 Determine the amount of CaCl2(i = 2.47) dissolved in 2.5 litre of water such that its osmotic pressure is 0.75 atm at 27ºC. Sol. We know that WB RT  = iCRT = i M  V B WB =   V  MB i  RT Put the values according to question, WB =  No. of moles of CaCl2 = 0.75  2.5  111 = 3.42 g 2.47  0.0821  300 3.42 = 0.03 mol 111