When $25 \mathrm{~g}$ of a non-volatile solute is dissolved in $100 \mathrm{~g}$ of water, the vapour…
When $25 \mathrm{~g}$ of a non-volatile solute is dissolved in $100 \mathrm{~g}$ of water, the vapour pressure is lowered by $2.25 \times 10^{-1} \mathrm{~mm}$. If the vapour pressure of water at $20^{\circ} \mathrm{C}$ is $17.5 \mathrm{~mm}$, what is the molecular weight of the solute?
$206$
$302$
$350$
$276$
Solution
Given,
weight of non-volatile solute,
$$
w=25 \mathrm{~g}
$$
Weight of solvent, $W=100 \mathrm{~g}$
Lowering of vapour pressure,
$$
p^{\circ}-p_s=0.225 \mathrm{~mm}
$$
Vapour pressure of pure solvent,
$$
p^{\circ}=17.5 \mathrm{~mm}
$$
Molecular weight of solvent $\left(\mathrm{H}_2 \mathrm{O}ight), M=18 \mathrm{~g}$
Molecular weight of solute, $m=$ ?
According to Raoult's law
$$
\begin{aligned}
\frac{p^{\circ}-p_s}{p^{\circ}} & =\frac{w \times M}{m \times W} \\
\frac{0.225}{17.5} & =\frac{25 \times 18}{m \times 100} \\
m & =\frac{25 \times 18 \times 17.5}{22.5} \\
& =350 \mathrm{~g}
\end{aligned}
$$