JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Solutions

1.22 g of an organic acid is separately dissolved in 100 g of benzene (

K_{b} = 2.6 K kg mol^{- 1}

) and 100 g of acetone (

K_{b} = 1.7 K kg mol^{- 1}

). The acid is known to dimerize in benzene but remain as a monomer in acetone. The boiling point of the solution in acetone increases by

0.1 7^{\circ} C

. The increase in boiling point of solution in benzene in

^{\circ} C

x \times 1 0^{- 2}

. The value of x is ______. (Nearest integer)
[Atomic mass :

C = 12.0, H = 1.0, O = 16.0

]

Answer: 13

Solution

From the acetone solution, where the acid stays monomeric,

Δ T_{b} = K_{b} m

gives

0.17 = 1.7 \times \frac{1.22/ M}{0.100}

, so

M = 122

g/mol. In benzene the acid dimerises, so the effective molar mass is

244

g/mol and the molality is halved:

m = \frac{1.22/244}{0.100} = 0.050 mol k g^{- 1}

. The boiling-point elevation in benzene is

Δ T_{b} = 2.6 \times 0.050 = 0.13^{\circ} C

, so with

Δ T_{b} = x \times 1 0^{- 2}

, the value is

x = 13

JEE Main · 2021 · 31 Aug 2021 · Shift 2 · Integer answer