Question

The weight of non-volatile solute having molecular weight 40, that should be dissolved in 43 g hexane to reduce its vapour pressure to 80% is:

• A. 5 g
• B. 47.2 g
• C. 95 g
• D. 80 g

Answer 1

The correct option is A 5 g

Solution:- (A) $5g$

Let the vapour pressure of pure octane be $p^{\circ }$.

Then, the vapour pressure of the octane after dissolving the non-volatile solute $\left(p_s\right)$ is-

$p_s=\frac{80}{100}{p}^{\circ }=0.8p^{\circ }$

Molar mass of solute $\left(M_2\right)=40g/mol$

Mass of hexane $\left(w_1\right)=43g$

Molar mass of hexane $\left(M_1\right)=86g$

From Rault's law,

$\frac{p^{\circ }-p_s}{p_s}=\frac{n_1}{n_2}$

$\frac{p^{\circ }-p_s}{p_s}=\frac{\left(\frac{w_2}{M_2}\right)}{\left(\frac{w_1}{M_1}\right)}$

$\frac{p^{\circ }-0.8p^{\circ }}{0.8p^{\circ }}=\frac{\left(\frac{w_2}{40}\right)}{\left(\frac{43}{86}\right)}$

$\Rightarrow w_2=\frac{0.2\times 20}{0.8}=5g$

Hence the weight of non-volatile solute reqired is $5gm$.