Solutions and Colligative Properties - Result Question 62
####33. A solution of a non-volatile solute in water freezes at $-0.30^{\circ} C$. The vapour pressure of pure water at $298 K$ is $23.51 mm Hg$ and $K _f$ for water is $1.86 K kg mol^{-1}$. Calculate the vapour pressure of this solution at $298 K$. $\quad(1998,4 M)$
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Solution:
- $-\Delta T _f=K _f \cdot m _2$
$$ \Rightarrow \quad m _2=\frac{0.3}{1.86}=0.1613 $$
Also,
$$ m _2=\frac{n _2}{n _1} \times \frac{1000}{M _1}=0.1613 $$
$$ \begin{aligned} \Rightarrow & & \frac{n _2}{n _1} & =\frac{0.1613 \times 18}{1000}=2.9 \times 10^{-3} \ \Rightarrow & & \frac{n _2}{n _1}+1 & =\frac{n _2+n _1}{n _1}=2.9 \times 10^{-3}+1 \ \Rightarrow & & \frac{n _1}{n _1+n _2} & =\chi _1=\frac{1}{1+2.9 \times 10^{-3}}=0.997 \ \Rightarrow & & p & =p _0 \chi _1=23.51 \times 0.997=23.44 mm \end{aligned} $$
