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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, Question 2.1:, Calculate the mass percentage of benzene (C6H6) and carbon tetrachloride (CCl 4) 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%, 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, 1, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, Number of moles of, = 0.3846 mol, Molar mass of carbon tetrachloride (CCl 4) = 1 × 12 + 4 × 355, = 154 g mol−1, , Number of moles of CCl4, = 0.4545 mol, Thus, the mole fraction of C6H6 is given as:, , = 0.458, 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, , 2, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, (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, 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., 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, , 3, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, (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, , 4, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, = 0.0263, 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, Question 2.7:, A solution is obtained by mixing 300 g of 25% solution and 400 g of 40% solution by mass., Calculate the mass percentage of the resulting solution., Answer, , 5, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, Total amount of solute present in the mixture is given by,, , = 75 + 160, = 235 g, Total amount of solution = 300 + 400 = 700 g, Therefore, mass percentage (w/w) of the solute in the resulting solution,, = 33.57%, And, mass percentage (w/w) of the solvent in the resulting solution,, = (100 − 33.57)%, = 66.43%, 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:, , 6, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, 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, Question 2.9:, Vapour pressure of pure water at 298 K is 23.8 mm Hg. 50 g of urea (NH 2CONH2) is, dissolved in 850 g of water. Calculate the vapour pressure of water for this solution and, its relative lowering., , 7, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, 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 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., Question 2.10:, 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., Answer, , 8, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, Here, elevation of boiling point ∆Tb = (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:, , = 121.67 g (approximately), Hence, 121.67 g of sucrose is to be added., Note: There is a slight variation in this answer and the one given in the NCERT textbook., Question 2.11:, 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., Answer, 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, ∆Tf = 1.5 K We, know that:, , 9, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, , = 5.08 g (approx), Hence, 5.08 g of ascorbic acid is needed to be dissolved., Note: There is a slight variation in this answer and the one given in the NCERT textbook., Question 2.12:, 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 37°C., Answer, 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), , 10, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, Question 2.13:, The partial pressure of ethane over a solution containing 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?, Answer, Molar mass of ethane (C2H6) = 2 × 12 + 6 × 1, = 30 g mol−1, , Number of moles present in 6.56 × 10−2 g of ethane, = 2.187 × 10−3 mol, Let the number of moles of the solvent be x. According, to Henry’s law,, p = KHx, , (Since x >> 2.187 × 10−3), , Number of moles present in 5.00 × 10−2 g of ethane, = 1.67 × 10−3 mol According, to Henry’s law,, p = KHx, , (Since, x >> 1.67 × 10−3), , 11, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, = 0.764 bar, Hence, partial pressure of the gas shall be 0.764 bar., Question 2.14:, What is meant by positive and negative deviations from Raoult's law and how is the sign, of ∆solH related to positive and negative deviations from Raoult's law?, Answer, According to Raoult’s law, the partial vapour pressure of each volatile component in any, solution is directly proportional to its mole fraction. The solutions which obey Raoult’s law, over the entire range of concentration are known as ideal solutions. The solutions that do, not obey Raoult’s law (non-ideal solutions) have vapour pressures either higher or lower, than that predicted by Raoult’s law. If the vapour pressure is higher, then the solution is, said to exhibit positive deviation, and if it is lower, then the solution is said to exhibit, negative deviation from Raoult’s law., , Vapour pressure of a two-component solution showing positive deviation from, Raoult’s law, , 12, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, , Vapour pressure of a two-component solution showing negative deviation from, Raoult’s law, In the case of an ideal solution, the enthalpy of the mixing of the pure components for, forming the solution is zero., ∆solH = 0, In the case of solutions showing positive deviations, absorption of heat takes place., ∆solH = Positive, In the case of solutions showing negative deviations, evolution of heat takes place., ∆solH = Negative, Question 2.15:, An aqueous solution of 2% non-volatile solute exerts a pressure of 1.004 bar at the normal, boiling point of the solvent. What is the molar mass of the solute?, Answer, Here,, Vapour pressure of the solution at normal boiling point (p1) = 1.004 bar, Vapour pressure of pure water at normal boiling point, Mass of solute, (w2) = 2 g, Mass of solvent (water), (w1) = 98 g, Molar mass of solvent (water), (M1) = 18 g mol−1 According, to Raoult’s law,, , 13, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, , = 41.35 g mol−1, Hence, the molar mass of the solute is 41.35 g mol −1., , Question 2.16:, Heptane and octane form an ideal solution. At 373 K, the vapour pressures of the two, liquid components are 105.2 kPa and 46.8 kPa respectively. What will be the vapour, pressure of a mixture of 26.0 g of heptane and 35 g of octane?, Answer, Vapour pressure of, Vapour, octane, , pressure, , heptane, = 46.8 kPa, , of, , We know that,, Molar mass of heptane (C7H16) = 7 × 12 + 16 × 1, = 100 g mol−1, , Number of moles of heptane, = 0.26 mol, Molar mass of octane (C8H18) = 8 × 12 + 18 × 1, = 114 g mol−1, , Number of moles of octane, , 14, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, = 0.31 mol, Mole fraction of heptane,, = 0.456, And, mole fraction of octane, x2 = 1 − 0.456, = 0.544, Now, partial pressure of heptane,, = 0.456 × 105.2, = 47.97 kPa, Partial pressure of octane,, = 0.544 × 46.8, = 25.46 kPa, Hence, vapour pressure of solution, ptotal = p1 + p2, = 47.97 + 25.46, = 73.43 kPa, Question 2.17:, The vapour pressure of water is 12.3 kPa at 300 K. Calculate vapour pressure of 1 molal, solution of a non-volatile solute in it., Answer, 1 molal solution means 1 mol of the solute is present in 100 g of the solvent (water)., Molar mass of water = 18 g mol−1, , Number of moles present in 1000 g of water, = 55.56 mol, Therefore, mole fraction of the solute in the solution is, , ., It is given that,, , 15, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, = 12.3 kPa, , Vapour pressure of water,, , Applying the relation,, , ⇒ 12.3 − p1 = 0.2177, ⇒ p1 = 12.0823, = 12.08 kPa (approximately), Hence, the vapour pressure of the solution is 12.08 kPa., Question 2.18:, Calculate the mass of a non-volatile solute (molar mass 40 g mol −1) which should be, dissolved in 114 g octane to reduce its vapour pressure to 80%., Answer, Let the vapour pressure of pure octane be, Then, the vapour pressure of the octane after dissolving the non-volatile solute is, , Molar mass of solute, M2 = 40 g mol−1, Mass of octane, w1 = 114 g, Molar mass of octane, (C8H18), M1 = 8 × 12 + 18 × 1, = 114 g mol−1, Applying the relation,, , 16, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, , Hence, the required mass of the solute is 8 g., Question 2.19:, A solution containing 30 g of non-volatile solute exactly in 90 g of water has a, vapour pressure of 2.8 kPa at 298 K. Further, 18 g of water is then added to the, solution and the new vapour pressure becomes 2.9 kPa at 298 K. Calculate:, i., , molar mass of the solute, , ii., , vapour pressure of water at 298 K., , Answer, (i) Let, the molar mass of the solute be M g mol −1, , Now, the no. of moles of, , solvent (water),, , And, the no. of moles of solute,, , Applying the relation:, , 17, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, , After the addition of 18 g of water:, , 18, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, Again, applying the relation:, , Dividing equation (i) by (ii), we have:, , 19, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, , Therefore, the molar mass of the solute is 23 g mol −1., (ii) Putting the value of ‘M’ in equation (i), we have:, , Hence, the vapour pressure of water at 298 K is 3.53 kPa., Question 2.20:, A 5% solution (by mass) of cane sugar in water has freezing point of 271 K. Calculate the, freezing point of 5% glucose in water if freezing point of pure water is 273.15 K., Answer, Here, ∆Tf = (273.15 − 271) K, = 2.15 K, Molar mass of sugar (C12H22O11) = 12 × 12 + 22 × 1 + 11 × 16, = 342 g mol−1, , 20, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, 5% solution (by mass) of cane sugar in water means 5 g of cane sugar is present in (100, − 5)g = 95 g of water., , Now, number of moles of cane sugar, = 0.0146 mol, , Therefore, molality of the solution,, = 0.1537 mol kg−1, Applying the relation,, ∆Tf = Kf × m, , = 13.99 K kg mol−1, Molar of glucose (C6H12O6) = 6 × 12 + 12 × 1 + 6 × 16 =, 180 g mol−1, 5% glucose in water means 5 g of glucose is present in (100 − 5) g = 95 g of water., Number of moles of glucose, = 0.0278 mol, , Therefore, molality of the solution,, = 0.2926 mol kg−1, Applying the relation,, ∆Tf = Kf × m, = 13.99 K kg mol−1 × 0.2926 mol kg−1, = 4.09 K (approximately), Hence, the freezing point of 5% glucose solution is (273.15 − 4.09) K= 269.06 K., , 21, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, Question 2.21:, Two elements A and B form compounds having formula AB2 and AB4. When dissolved in, 20 g of benzene (C6H6), 1 g of AB2 lowers the freezing point by 2.3 Kwhereas 1.0 g of AB4, lowers it by 1.3 K. The molar depression constant for benzene is 5.1 Kkg mol −1., Calculate atomic masses of A and B., Answer, We know that,, , Then,, = 110.87 g mol−1, , = 196.15 g mol−1, Now, we have the molar masses of AB 2 and AB4 as 110.87 g mol−1 and 196.15 g mol−1, respectively., Let the atomic masses of A and B be x and y respectively., Now, we can write:, , Subtracting equation (i) from (ii), we have, 2y = 85.28, ⇒ y = 42.64, Putting the value of ‘y’ in equation (1), we have x, + 2 × 42.64 = 110.87, ⇒ x = 25.59, Hence, the atomic masses of A and B are 25.59 u and 42.64 u respectively., , 22, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, Question 2.22:, 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 the solution is 1.52 bars at the same temperature, what, would be its concentration?, Answer, Here,, T = 300 K π, = 1.52 bar, R = 0.083 bar L K−1 mol−1, Applying the relation, π =, CRT, , = 0.061 mol, Since the volume of the solution is 1 L, the concentration of the solution would be 0.061, M., Question 2.23:, Suggest the most important type of intermolecular attractive interaction in the following, pairs., (i), , n-hexane and n-octane, , (ii) I2 and CCl4, (iii) NaClO4 and water, (iv) methanol and acetone, (v) acetonitrile (CH3CN) and acetone (C3H6O). Answer, (i), , Van der Wall’s forces of attraction., , (ii) Van der Wall’s forces of attraction., (iii) Ion-diople interaction., 23, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, (iv) Dipole-dipole interaction., (v) Dipole-dipole interaction., Question 2.24:, Based on solute-solvent interactions, arrange the following in order of increasing solubility, in n-octane and explain. Cyclohexane, KCl, CH3OH, CH3CN., Answer, n-octane is a non-polar solvent. Therefore, the solubility of a non-polar solute is more than, that of a polar solute in the n-octane., The order of increasing polarity is:, Cyclohexane < CH3CN < CH3OH < KCl, Therefore, the order of increasing solubility is:, KCl < CH3OH < CH3CN < Cyclohexane, Question 2.25:, Amongst the following compounds, identify which are insoluble, partially soluble and highly, soluble in water?, (i) phenol (ii) toluene (iii) formic acid, (iv) ethylene glycol (v) chloroform (vi) pentanol., Answer, (i) Phenol (C6H5OH) has the polar group −OH and non-polar group −C6H5. Thus, phenol, is partially soluble in water., (ii) Toluene (C6H5−CH3) has no polar groups. Thus, toluene is insoluble in water., (iii), , Formic acid (HCOOH) has the polar group −OH and can form H-bond with water., Thus, formic acid is highly soluble in water., , (iv) Ethylene glycol, , has polar −OH group and can form H−bond. Thus, it is, , highly soluble in water., (v) Chloroform is insoluble in water., , 24, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, (vi) Pentanol (C5H11OH) has polar −OH group, but it also contains a very bulky nonpolar, −C5H11 group. Thus, pentanol is partially soluble in water., , Question 2.26:, If the density of some lake water is 1.25 g mL −1 and contains 92 g of Na+ ions per kg of, water, calculate the molality of Na+ ions in the lake., Answer, , Number of moles present in 92 g of Na+ ions =, = 4 mol, , Therefore, molality of Na+ ions in the lake, =4m, Question 2.27:, If the solubility product of CuS is 6 × 10 −16, calculate the maximum molarity of CuS in, aqueous solution., Answer, Solubility product of CuS, Ksp = 6 × 10−16 Let, s be the solubility of CuS in mol L−1., , Now,, =s×s, = s2, Then, we have, Ksp =, , 25, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, = 2.45 × 10−8 mol L−1, Hence, the maximum molarity of CuS in an aqueous solution is 2.45 × 10 −8 mol L−1., , Question 2.28:, Calculate the mass percentage of aspirin (C9H8O4) in acetonitrile (CH3CN) when 6.5 g of, C9H8O4 is dissolved in 450 g of CH3CN., Answer, 6.5 g of C9H8O4 is dissolved in 450 g of CH3CN., Then, total mass of the solution = (6.5 + 450) g, = 456.5 g, , Therefore, mass percentage ofC9H8O4, = 1.424%, Question 2.29:, 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 × 10−3m aqueous solution required for the above dose. Answer, The molar mass of nalorphene, , is given as:, , In 1.5 × 10−3m aqueous solution of nalorphene,, 1 kg (1000 g) of water contains 1.5 × 10−3 mol, , Therefore, total mass of the solution, This implies that the mass of the solution containing 0.4665 g of nalorphene is 1000.4665, g., Therefore, mass of the solution containing 1.5 mg of nalorphene is:, , 26, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, , Hence, the mass of aqueous solution required is 3.22 g., Note: There is a slight variation in this answer and the one given in the NCERT textbook., Question 2.30:, Calculate the amount of benzoic acid (C6H5COOH) required for preparing 250 mL of 0.15, M solution in methanol., Answer, 0.15 M solution of benzoic acid in methanol means,, 1000 mL of solution contains 0.15 mol of benzoic acid, Therefore, 250 mL of solution contains =, , mol of benzoic acid, , = 0.0375 mol of benzoic acid, Molar mass of benzoic acid (C6H5COOH) = 7 × 12 + 6 × 1 + 2 × 16, = 122 g mol−1, Hence, required benzoic acid = 0.0375 mol × 122 g mol −1, = 4.575 g, Question 2.31:, The depression in freezing point of water observed for the same amount of acetic acid,, trichloroacetic acid and trifluoroacetic acid increases in the order given above. Explain, briefly., Answer, , 27, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, Among H, Cl, and F, H is least electronegative while F is most electronegative. Then, F can, withdraw electrons towards itself more than Cl and H. Thus, trifluoroacetic acid can easily, lose H+ ions i.e., trifluoroacetic acid ionizes to the largest extent. Now, the more ions, produced, the greater is the depression of the freezing point. Hence, the depression in the, freezing point increases in the order:, Acetic acid < trichloroacetic acid < trifluoroacetic acid, Question 2.32:, Calculate the depression in the freezing point of water when 10 g of CH 3CH2CHClCOOH is, added to 250 g of water. Ka = 1.4 × 10−3, Kf = 1.86 K kg mol−1., Answer, Molar mass of, , No. of moles present in 10 g of, It is given that 10 g of is added to 250 g of water., , Molality of the solution,, , Let α be the degree of dissociation of, , 28, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, undergoes dissociation according to the following equation:, , Since α is very small with respect to 1, 1 − α ≈ 1, , Now,, , Again,, , Total moles of equilibrium = 1 − α + α + α, =1+α, , 29, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, , Hence, the depression in the freezing point of water is given as:, , Question 2.33:, 19.5 g of CH2FCOOH is dissolved in 500 g of water. The depression in the freezing point of, water observed is 1.0°C. Calculate the van’t Hoff factor and dissociation constant of, fluoroacetic acid., Answer, It is given that:, , We know that:, , Therefore, observed molar mass of, , 30, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, The calculated molar mass of, , Therefore, van’t Hoff factor,, , is:, , Let α be the degree of dissociation of, , Now, the value of Ka is given as:, , Taking the volume of the solution as 500 mL, we have the concentration:, , 31, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, , Therefore,, , Question 2.34:, Vapour pressure of water at 293 Kis 17.535 mm Hg. Calculate the vapour pressure of, water at 293 Kwhen 25 g of glucose is dissolved in 450 g of water., Answer, Vapour pressure of water,, , = 17.535 mm of Hg, , Mass of glucose, w2 = 25 g, Mass of water, w1 = 450 g, We know that,, Molar mass of glucose (C6H12O6), M2 = 6 × 12 + 12 × 1 + 6 × 16 =, 180 g mol−1, Molar mass of water, M1 = 18 g mol−1, , Then, number of moles of glucose,, , 32, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, = 0.139 mol, , And, number of moles of water,, = 25 mol, We know that,, , ⇒ 17.535 − p1 = 0.097, ⇒ p1 = 17.44 mm of Hg, Hence, the vapour pressure of water is 17.44 mm of Hg., Question 2.35:, Henry’s law constant for the molality of methane in benzene at 298 Kis 4.27 × 10 5 mm, Hg. Calculate the solubility of methane in benzene at 298 Kunder 760 mm Hg., Answer Here, p = 760, mm Hg kH = 4.27 × 105, mm Hg According to, Henry’s law,, p = kHx, , = 177.99 × 10−5, = 178 × 10−5 (approximately), Hence, the mole fraction of methane in benzene is 178 × 10 −5., , 33, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, Question 2.36:, 100 g of liquid A (molar mass 140 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., Answer, Number of moles of liquid A,, = 0.714 mol, Number of moles of liquid B,, = 5.556 mol, , Then, mole fraction of A,, , = 0.114, And, mole fraction of B, xB = 1 − 0.114, = 0.886, Vapour pressure of pure liquid B,, , = 500 torr, , Therefore, vapour pressure of liquid B in the solution,, , = 500 × 0.886, = 443 torr, Total vapour pressure of the solution, ptotal = 475 torr, Vapour pressure of liquid A in the solution,, pA = ptotal − pB = 475 − 443, = 32 torr, Now,, , 34, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, , = 280.7 torr, Hence, the vapour pressure of pure liquid A is 280.7 torr., Question 2.37:, Vapour pressure of pure acetone and chloroform at 328 K are 741.8 mm Hg and 632.8, mm Hg respectively. Assuming that they form ideal solution over the entire range of, composition, plot ptotal’ pchloroform’ and pacetone as a function of xacetone. The experimental data, observed for different compositions of mixture is., 100 ×xacetone, , 0, , 11.8, , 23.4, , 36.0, , 50.8, , 58.2, , 64.5, , 72.1, , pacetone /mm Hg, , 0, , 54.9, , 110.1, , 202.4, , 322.7, , 405.9, , 454.1, , 521.1, , pchloroform/mm Hg, , 632.8, , 548.1, , 469.4, , 359.7, , 257.7, , 193.6, , 161.2, , 120.7, , Answer, From the question, we have the following data, 100 ×xacetone, , 0, , 11.8, , 23.4, , 36.0, , 50.8, , 58.2, , 64.5, , 72.1, , pacetone /mm Hg, , 0, , 54.9, , 110.1, , 202.4, , 322.7, , 405.9, , 454.1, , 521.1, , pchloroform/mm Hg, , 632.8, , 548.1, , 469.4, , 359.7, , 257.7, , 193.6, , 161.2, , 120.7, , ptota(mm Hg), , 632.8, , 603.0, , 579.5, , 562.1, , 580.4, , 599.5, , 615.3, , 641.8, , 35, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, , It can be observed from the graph that the plot for the ptotal of the solution curves, downwards. Therefore, the solution shows negative deviation from the ideal behaviour., Question 2.38:, Benzene and toluene form ideal solution over the entire range of composition., The vapour pressure of pure benzene and naphthalene at 300 Kare 50.71 mm, Hg and 32.06 mm Hg respectively. Calculate the mole fraction of benzene in, vapour phase if 80 g of benzene is mixed with 100 g of toluene., Answer, Molar mass of benzene, , Molar mass of toluene, , 36, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, , Now, no. of moles present in 80 g of benzene, , And, no. of moles present in 100 g of toluene, , Mole fraction of benzene, xb, And, mole fraction of toluene,, It is given that vapour pressure of pure benzene,, And, vapour pressure of pure toluene,, Therefore, partial vapour pressure of benzene,, , And, partial vapour pressure of toluene,, , Hence, mole fraction of benzene in vapour phase is given by:, , Question 2.39:, 37, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, The air is a mixture of a number 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 Kif the, Henry’s law constants for oxygen and nitrogen are 3.30 × 107 mm and 6.51 × 107 mm, respectively, calculate the composition of these gases in water., Answer, Percentage of oxygen (O2) in air = 20 %, Percentage of nitrogen (N2) in air = 79%, Also, it is given that water is in equilibrium with air at a total pressure of 10 atm, that is,, (10 × 760) mm Hg = 7600 mm Hg, Therefore,, Partial pressure of oxygen,, = 1520 mm Hg, Partial pressure of nitrogen,, = 6004 mmHg, Now, according to Henry’s law:, p = KH.x, For oxygen:, , For nitrogen:, , 38, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, , Hence, the mole fractions of oxygen and nitrogen in water are 4.61 ×10 −5and 9.22 × 10−5, respectively., , Question 2.40:, 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 27°C., Answer, We know that,, , Here,, R = 0.0821 L atm K-1mol-1, M = 1 × 40 + 2 × 35.5, = 111g mol-1, , Therefore, w, , 39, Free web support in Education

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Chemistry, (www.tiwariacademy.com), , (Chapter 2)(Solutions), XII, = 3.42 g, Hence, the required amount of CaCl 2 is 3.42 g., Question 2.41:, Determine the osmotic pressure of a solution prepared by dissolving 25 mg of K2SO4 in 2, liter of water at 25° C, assuming that it is completely dissociated., Answer, When K2SO4 is dissolved in water,, , ions are produced., , Total number of ions produced = 3, i =3 Given, w = 25 mg = 0.025, g, V=2L, T = 250C = (25 + 273) K = 298 K Also,, we know that:, R = 0.0821 L atm K-1mol-1, M = (2 × 39) + (1 × 32) + (4 × 16) = 174 g mol -1, Appling the following relation,, , 40, Free web support in Education