Question

The vapour pressure of benzene at a certain temperature is $640$ mm 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 $600$ mm of Hg, what is the molecular weight of the solid substance?

• A. $49.50$ g/mol
• B. $59.60$ g/mol
• C. $69.60$ g/mol
• D. $79.82$ g/mol

Answer 1

The correct option is C $69.60$ g/mol

Given, vapour pressure of benzene,

$p^o=640mm$ $Hg$

Vapour pressure of solution,

$p=600mm$ $Hg$

Weight of solute, $w=2-175g$

Weight of benzene, $W=39.08g$

The molecular weight of benzene,

$M=78g$

The molecular weight of solute, $m=?$

According to Raoult's law

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

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

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

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

$m=69.60$

Hence, the correct option is $C$