Question

The vapour pressure of benzene at a certain temperature is $ 220\text{ }mm\text{ }hg $ .At the same temperature, the vapour pressure of a solution containing $ 2\text{ }g $ of non-volatile non-electrolyte solid in $ 78\text{ }g $ of benzene is $ 195\text{ }mm\text{ }hg $ . The molecular weight of solid is:
(A) 20
(B) 40
(C) 60
(D) 80

Answer 1

Hint: Benzene is an organic volatile compound. The relative lowering of vapour pressure is a colligative property.
Relative lowering in vapor pressure $ \begin {align}
  & \\
 & =\dfrac{\Delta P}{P{}^\circ }=\dfrac{P{}^\circ -P}{P{}^\circ } \\
 & P {} ^ \circ -{{P}_{s}}=\text{ lowering vapor pressure} \\
\end{align} $ .

Formula used $\dfrac{{{P}_{i}}-Ps}{{{P}_{i}}}=\dfrac{{{w}_{B}}.{{m}_{A}}}{{{m}_{A}}.{{w}_{A}}} $

Complete step by step solution:
Vapour pressure of pure solvent $ P {} ^\circ =200\text { } mm\text { } Hg $
Vapor pressure of solution $ = {{P}_{s}}=198\text{ }m $
${{w}_{A}} =2g \text{weight of solid} $
${{w}_{B}} =78\text{ } g\text{ } 10%\text {of benzene} $
${{m}_{A}} =\text{molecular weight of Benzene} =78m $
${{m}_{B}} =\text{molecular weight of solid=?} $
Using formula
$\dfrac{P{}^\circ -{{P}_{s}}}{P{}^\circ }=\dfrac{{{w}_{B}}\cdot {{m}_{A}}}{{{m}_{B}}.{{w}_{A}}} $
$ \dfrac{200-195}{200}=\dfrac{2\times 78}{{{m}_{B}}\times 78} $
$ {{m} {B}} =80 $
Option (D) correct answer.

Note
To solve this kind of problem firstly we have to recognize which colligative property is involved. The knowledge of Raoult’s law as well as lowering in vapor pressure is needed. Molecular weight is the weight of one mole of a given solid. Vapor pressure increases with temperature. When vapor pressure at the surface of liquid equals the pressure exerted by surroundings then it is defined as the boiling point of the liquid.
When a non volatile solute dissolves in the liquid, it tends to lower the vapour pressure of solution then the pure solvent. All the colligative properties only applicable when the dilute solution is used.