Question

When $25\, g$ of a non-volatile solute is dissolved in $100\,g$ of water, the vapour pressure is lowered by $2.25 \times 10^{-1}\, mm$. If the vapour pressure of water at $20^{\circ} C$ is $17.5\, mm$, what is the molecular weight of the solute?

• A. 206
• B. 302
• C. 350
• D. 276

Answer 1

Answer: (C)

Given,
Weight of non-volatile solute, $w=25\, g$
Weight of solvent, $W=100\, g$
Lowering of vapour pressure,
$p^{\circ}-p_{s}=0.225\, mm$
Vapour pressure of pure solvent,
$p^{\circ}=17.5\, mm$
Molecular weight of solvent
$\left(H_{2} O\right), M=18\, g$
Molecular weight of solute, $m=?$
According to Raoult's law
$p^{\circ}-p_{s}$
${p^{\circ}}=\frac{w \times M}{m \times W}$
$\frac{0.225}{17.5}=\frac{25 \times 18}{m \times 100}$
$m=\frac{25 \times 18 \times 17.5}{22.5}$
$=350\, g$