QJEEM21Ch94e0: JEE Main 2021 Chemistry - Class 12 JEEM21Ch94e0 Solution: 2 molal solution of a weak acid HA has a freezing point of…

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

^\circ

C. The degree of dissociation of this acid is ___________

\times

10^{$-$ 3}. (Round off to the Nearest Integer).

[Given : Molal depression constant of water = 1.85 K kg mol^{$-$ 1} Freezing point of pure water = 0

^\circ

C]

Accepted Answer

\Delta

T_f = K_f (im)

\Rightarrow

3.885 = i

\times

1.85

\times

2

\Rightarrow

i = 1.05

Also, we know,

i = 1 + (n

-

1)

\alpha

here n = number of particle obtained upon the dissociation of one particle.

HA\rightleftharpoons H^{+}+A^{-}

here from one particle HA we get two particle H⁺ and A^$-$.

\therefore

n = 2

So, i = 1 + (2

-

1)

\alpha

\Rightarrow

1.05 = 1 +

\alpha

\Rightarrow

\alpha

= 0.05 = 50

\times

10^{$-$ 3}

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