Question

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