JEE Advanced (Subjective Type Questions): Solutions- 3 | Chapter-wise Tests for JEE Main & Advanced PDF Download

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_fm

According to Raoult’s law 

(∵ No. of moles of H₂ 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, C₆H₄ (NO₂ )₂ , m-dinitrobenzene

Solution : 

 TIPS/Formulae : ΔT_b = kb ´ m

∴Empirical formula of the minor product is C₃H₂NO₂  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_bm ⇒ 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 (C₃H₂NO₂), i.e., C₆H₄(NO₂)₂

The product is m - dinitrobenzene . 

16. To 500 cm³ 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 CH₃COOH

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 C₆H₅OH (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%.