Question

The vapour pressure of a solution of a non-volatile electrolytes (A) in a solvent (B) is $95$% of the vapoure pressure of the solvent at the same temperature. If $M_B=0.3M_A$, where $M_B$ and $M_A$ are molar masses of $B$ and $A$ respectively, the mass ratio of the solvent and solute are:

Answer 1

Relative lowering of vapour pressure of a solvent $\left(B\right)$ upon addition of a non-volatile electrolyte$\left(A\right)$ is proportional to its mole fraction $x_A$ and is written as:

$\frac{p^{\circ }-p}{p^{\circ }}=x_A$

where $p^{\circ }$ =vapour pressure of pure solvent and $p$=vapour pressure of solution

Since $p=0.95\times p^{\circ }$

$\therefore p^{\circ }-p=0.05\times p^{\circ }$

and $x_A=\frac{n_A}{n_A+n_B}$=$\frac{m_A/M_A}{m_A/M_A+m_B/M_B}$

$x_A=\frac{m_A}{m_A+m_B\times \left(M_A/M_B\right)}$

Given: $M_B=0.3M_A$ or $M_A/M_B=1/0.3$

upon substitution we get:

$0.05=\frac{m_A}{m_A+m_B\times \left(1/0.3\right)}$=$\frac{m_A\times 0.3}{m_A\times 0.3+m_B}$

$\frac{1}{0.05}=\frac{m_A\times 0.3+m_B}{m_A\times 0.3}$=$1+\frac{m_B}{0.3m_A}$

Or$\frac{m_B}{m_A}=0.3\times (\frac{1}{0.05}-1)=5.7$

Therefore mass ratio of the solvent and solute is $5.7$.