JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Solutions

A solute A dimerizes in water. The boiling point of a 2 molal solution of A is

100.5 2^{\circ} C

. The percentage association of

A

is ____. (Round off to the nearest integer).
[Use:

K_{b}

for water

= 0.52 K kg mol^{- 1}

, Boiling point of water

= 10 0^{\circ} C

]

Answer: 100

Solution

The observed elevation is

Δ T_{b} = 100.52 - 100 = 0.52

K, whereas a solute staying as single molecules would give

K_{b} m = 0.52 \times 2 = 1.04

K. The van't Hoff factor is

i = \frac{0.52}{1.04} = 0.5

For the association

2 A ⇌ A_{2}

with degree of association

α

, one mole of A leaves

1 - α + \frac{α}{2}

moles of particles, so

i = 1 - \frac{α}{2}

. Then

0.5 = 1 - \frac{α}{2}

gives

α = 1

, and the percentage association is

100

JEE Main · 2021 · 18 Mar 2021 · Shift 2 · Numerical answer

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