Question

Derive the relationship between relative lowering of vapour pressure and mole fraction of the volatile liquid.

Answer 1

As we know that, for any solution, the partial pressure of each volatile component in the solution is directly proportional to its mole fraction.

$\therefore p_A=x_A$; where $p_A$ is vapour pressure of solvent having mole fraction $x_A$.

$p_A={p^0}_A\times x_A.....\left(1\right)$

But,

$x_A+x_B=1\Rightarrow x_A=1-x_B$, where $x_B$ is mole fraction of non-volatile solute $B$

Now from ${eq}^n\left(1\right)$, we have

$p_A={p^0}_A(1-x_B)$

$\Rightarrow p_A={p^0}_A-{p^0}_A{x}_B$

As we know that, total vapour pressure $\left(p\right)$ is-

$p=p_A+p_B$

As non-volatile does not have any vapour pressure.

$\therefore p_B=0$

$\therefore p=p_A.....\left(2\right)$

$\Rightarrow p={p^0}_A-{p^0}_A{x}_B$

$\Rightarrow {p^0}_A{x}_B={p^0}_A-p$

$\Rightarrow x_B=\frac{{p^0}_A-p}{{p^0}_A}$

As,

$p=p_A\left(\text{From}\left(2\right)\right)$

$\therefore x_B=\frac{{p^0}_A-p_A}{{p^0}_A}$

Hence the relationship between relative lowering of vapour pressure and mole fraction of the volatile liquid is-

$x_B=\frac{{p^0}_A-p_A}{{p^0}_A}$