JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Solutions

2 molal solution of a weak acid HA has a freezing point of

3.88 5^{\circ} C

. The degree of dissociation of this acid is ____

\times 1 0^{- 3}

. (Round off to the nearest integer).
[ Given : Molal depression constant of water

= 1.85 K kg mol^{- 1}

, Freezing point of pure water

= 0^{\circ} C

]

Answer: 50

Solution

The depression of freezing point is

Δ T_{f} = 0 - (- 3.885) = 3.885 K

. With

Δ T_{f} = i K_{f} m

i = \frac{3.885}{1.85 \times 2} = 1.05

One molecule of

HA

that dissociates gives two ions, so for a fraction

α

dissociated the van't Hoff factor is

i = 1 + α

. Hence

α = 0.05

, which is

50 \times 1 0^{- 3}

JEE Main · 2021 · 18 Mar 2021 · Shift 1 · Integer answer

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