2 molal solution of a weak acid HA has a freezing point of 3 . 885 ° C . The degree of dissociation of this acid is ______ × 10 ^ - 3 . (Round off to the Nearest Integer). [ Given : Molal depression constant of water =

Correct answer: 0 Δ T _ f = ( 1 + α ) K _ f × m3 . 885 = ( 1 + α ) × 1 . 85 × 2 α = 0 . 05 = 50 × 10 ^ - 3

JEE Main 202118 MarchMorning Shift

JEE Main 2021 · 18 March · Morning Shift · Chemistry · Solutions

2 molal solution of a weak acid HA has a freezing point of 3 . 885...

2

molal solution of a weak acid

HA

has a freezing point of

3 . 885 ° C .

The degree of dissociation of this acid is ______

\times {10}^{- 3} .

(Round off to the Nearest Integer).

[

Given : Molal depression constant of water

= 1 . 85Kkg{mol}^{- 1}

Freezing point of pure water

\left= 0 ° C\right]

$\Delta{T}_{f} = ( 1 + \alpha ) {K}_{f} \times m3 . 885 = ( 1 + \alpha ) \times 1 . 85 \times 2$

$\alpha = 0 . 05 = 50 \times {10}^{- 3}$