Question

At ${20}^{o}C$, the vapour pressure of $0.1M$ solution of urea is $0.0311mm$ less than that of water and the vapour pressure of $0.1M$ solution of $KCl$ is $0.0574mm$ less than that of water. The apparent degree of dissociation of $KCl$ at this dilution is :

• A. $92.1\%$
• B. $84.6\%$
• C. $68.4\%$
• D. $54.1\%$

Answer 1

Answer: (B)

$84.6\%$

The Vant Hoff factor is the property of a solute. So it is directly proportional to any colligative property of solute (like vapour pressure).

So, vapour pressure $\propto$ Vant Hoff factor (i)

$⟹\frac{vapourpressur{e}_{\left(urea\right)}}{vapourpressur{e}_{\left(KCl\right)}}=\frac{i_{urea}}{i_{KCl}}$

But $i_{\left(urea\right)}=1$

$⟹\frac{vapourpressur{e}_{\left(KCl\right)}}{vapourpressur{e}_{\left(urea\right)}}=i_{KCl}=\frac{0.0574}{0.0311}$ ...(i)

$KCl\leftrightharpoons K^{+}+C{l}^{-}$

$1$ $0$ $0$

$1-\alpha$ $\alpha$ $\alpha$

For dissociation, we have:

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

Here, $n=2$

$⟹\alpha =\frac{i-1}{2-1}$

$⟹i=1+\alpha$

From (i), we have:

$\frac{0.0574}{0.0311}=1+\alpha$

$⟹\alpha =0.846$

So, $\%\alpha =\alpha \times 100=84.6\%$

Hence, the correct option is (B).