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  • 2 molal solution of a weak acid HA has a freezing point of 3.885oC. The degree of dissociation of this acid is _________ (Round off to the Nearest Integer). [

2 molal solution of a weak acid HA has a freezing point of 3.885oC. The degree of dissociation of this acid is _________\times 10^{-3}. (Round off to the Nearest Integer). [Given : Molal depression constant of water \mathrm{=1.85\; K\; kg\; mol^{-1}} Freezing point of pure water = 0o C]
 

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best_answer

Given,

2 molal solution of a weak acid HA has a freezing point of 3.885ºC.

\mathrm{T}_{\mathrm{f} \text { sol. }}=3.885^{\circ} \mathrm{C}

m = 2

Molal depression constant of water (Kf) = 1.85 K kg mol-1

The freezing point of pure water  Ti = 0ºC

van't Hoff factor i is related to the degree of dissociation \alpha \alpha as, if n = number of ions,

\alpha=\frac{i-1}{n-1}

Here n = 2 (H+and A-) 

So, 

i = 1+\alpha

We know the formula of depression at the freezing point.

\mathrm{\Delta T = i\times K_f\times m}

\mathrm{\Delta T = T_f -T_i= i\times K_f\times m}

3.885 - 0= i\times 1.85\times 2

i = 1.05

Now,

i = 1+\alpha

1.05 = 1+\alpha

\alpha=0.05

\alpha=50\times10^{-3}

Ans = 50

Posted by

Kuldeep Maurya

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