Question

The vapour pressure of benzene at ${30}^{\circ }C$ is $121.8mm$. By adding $15g$ of non-volatile solute in $250g$ of benzene, its vapour pressure is decreased to $120.2mm$. The molecular weight of solute is:

• A. $156.6g$
• B. $267.4g$
• C. $356.3g$
• D. $467.4g$

Answer 1

Answer: (C)

$356.3g$

According to Raoult's law,

$\frac{p^0-p_s}{p_s}=\frac{w\times M}{m\times W}$

$\frac{121.8-120.2}{121.8}=\frac{15}{m}\times \frac{78}{250}$

$\therefore m=\frac{15\times 78\times 121.8}{250\times 1.6}=356.265g$

Hence, the correct option is $C$