Solutions and Colligative Properties 2 Question 33
33. A solution of a non-volatile solute in water freezes at $-0.30^{\circ} \mathrm{C}$. The vapour pressure of pure water at $298 \mathrm{~K}$ is $23.51 \mathrm{~mm} \mathrm{Hg}$ and $K_{f}$ for water is $1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}$. Calculate the vapour pressure of this solution at $298 \mathrm{~K}$. $\quad(1998,4 \mathrm{M})$
Show Answer
Solution:
- $-\Delta T_{f}=K_{f} \cdot m_{2}$
$$ \Rightarrow \quad m_{2}=\frac{0.3}{1.86}=0.1613 $$
Also,
$$ m_{2}=\frac{n_{2}}{n_{1}} \times \frac{1000}{M_{1}}=0.1613 $$
$$ \begin{aligned} \Rightarrow & & \frac{n_{2}}{n_{1}} & =\frac{0.1613 \times 18}{1000}=2.9 \times 10^{-3} \ \Rightarrow & & \frac{n_{2}}{n_{1}}+1 & =\frac{n_{2}+n_{1}}{n_{1}}=2.9 \times 10^{-3}+1 \ \Rightarrow & & \frac{n_{1}}{n_{1}+n_{2}} & =\chi_{1}=\frac{1}{1+2.9 \times 10^{-3}}=0.997 \ \Rightarrow & & p & =p_{0} \chi_{1}=23.51 \times 0.997=23.44 \mathrm{~mm} \end{aligned} $$
