14. A solution of a nonvolatile solute in water freezes at – 0.30°C. The vapour pressure of pure water at 298 K is 23.51 mm Hg and Kf for water is 1.86 K kg mol^{–1}. Calculate the vapour pressure of this solution at 298 K. (1998  4 Marks)
Ans : 23.44 mm Hg
Solution :
TIPS/Formulae :
Depression in freezing point, ΔT_{f} = 0  (0.30) = 0.30
Now we know that ΔT_{f} = K_{f}m
According to Raoult’s law
(∵ No. of moles of H_{2} O = 1000/18)
On usual calculations
p = 23.51– 23.51´0.0020898 = 23.51 – .068
p = 23.44 mm Hg
15. Nitrobenzene is formed as the major product along with a minor product in the reaction of benzene with a hot mixture of nitric acid and sulphuric acid. The minor product consists of carbon : 42.86%, hydrogen : 2.40% , nitrogen : 16.67%, and oxygen: 38.07% (i) Calculate the empirical formula of the minor product. (ii) When 5.5 g of the minor product is dissolved in 45 g of benzene, the boiling point of the solution is 1.84 °C higher than that of pure benzene. Calculate the molar mass of the minor product and determine its molecular and structural formula. (Molal boiling point elevation constant of benzene is 2.53 K kg mol–1.) (1999  10 Marks)
Ans : 168 g, C_{6}H_{4} (NO_{2} )_{2} , mdinitrobenzene
Solution :
TIPS/Formulae : ΔT_{b} = kb ´ m
∴Empirical formula of the minor product is C_{3}H_{2}NO_{2 }Molar empirical formula mass of the minor prodouct = 3 × 12 + 2 × 1 + 1 × 14 + 2 × 16 = 84 g mol^{–2}
Let M be the molar mass of the minor product. For 5.5 g of the minor product dissolved in 45 g benzene, the molality
(m) of the solution =
Substituting this in the expression of elevation of boiling point,
ΔT_{b} = K_{b}m ⇒ 1.84K = (2.53 Kmol^{1) }
or M = 168 g mol^{–1}
No. of unit of empirical formula in molecular formula
Hence the molecular formula of the minor product is
2 (C_{3}H_{2}NO_{2}), i.e., C_{6}H_{4}(NO_{2})_{2}
The product is m  dinitrobenzene .
16. To 500 cm^{3} of water, 3.0 × 10^{–3} kg of acetic acid is added. If 23% of acetic acid is dissociated, what will be the depression in freezing point? Kf and density of water are 1.86 K kg^{–1} mol^{–1} and 0.997 g cm^{–3}, respectively. (2000  3 Marks)
Ans : 0.228 K
Solution :
TIPS/Formulae :
ΔT_{f} = ixK_{f} x m
Weight of water = 500 × 0.997 = 498.5 g
( weight = volume × denisty)
No. of moles of acetic acid
Since 498.5 g of water has 0.05 moles of CH_{3}COOH
Therefore molality of the solution = 0.1
Determination of van’t Hoff factor, i
Now we know that
ΔT_{f} = ixK_{f} xm = 1.23 x 1.86 x 0.1= 0.228K
17. 1.22g of benzoic acid is dissolved in 100 g of acetone and 100 g of benzene separately. Boiling point of the solution in acetone increases by 0.17º C, while that in the benzene increases by 0.13º C; Kb for acetone and benzene is 1.7 K kg mol^{–1} and 2.6 K kg mol^{–1}. Find molecular weight of benzoic acid in two cases and justify your answer. (2004  2 Marks)
Ans : 122, 224
Solution :
TIPS/Formulae :
ΔT_{b} = K_{b} × M
In first case,
Thus the benzoic acid exists as a monomer in acetone
(ii) In second case,
NOTE : Double the expected molecular weight of benzoic acid (244) in acetone solution indicates that benzoic acid exists as a dimer in acetone.
18. 75.2 g of C_{6}H_{5}OH (phenol) is dissolved in a solvent of K_{f} = 14. If the depression in freezing point is 7 K then find the % of phenol that dimerises. (2006  6M)
Ans : 75%
Solution :
Total number of moles at equilibrium =
ΔT_{f} =iK_{f} ×(molality)
∴ a = 0.75So the percentage of phenol that dimerises = 75%.
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