Question

The vapour pressure of benzene at a certain temperature is $640mm$ of $Hg$ . A non-volatile and non-electrolyte solid weighing $2.175g$ is added to $39.08g$ of benzene. If the vapour pressure of the solution is $600mm$ of $Hg$ , what is the molecular weight of solid substance ?

• A. $49.50$
• B. $59.60$
• C. $69.60$
• D. $79.82$

Answer 1

Answer: (C)

Given, vapour pressure of benzene,

$p^{\circ }=640mmHg$

Vapour pressure of solution,

$p=600mmHg$

Weight of solute, $w=2.175g$

Weight of benzene, $W=39.08g$

Molecular weight of benzene,

$M=78g$

Molecular weight of solute, $m=$?

According to Raoult's law,

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

$\frac{640-600}{640}=\frac{2.175\times 78}{m\times 39.08}$

$\frac{40}{640}=\frac{2.175\times 78}{m\times 39.08}$

$m=\frac{16\times 2.175\times 78}{39.08}$

$m=69.60$