# Question #2438c

Jul 27, 2016

Trying to give an answer assuming that the normal atmospheric pressure is 760mm of Hg and the distillation process is brought about under this prssure.

Here it is given that in the vpour mixture at ${99.2}^{\circ} C$ the partial pressure of water vapour ${p}_{w} = 739 m m \text{ of Hg}$

So the partial pressure of the organic liquid will be
${p}_{A} = \left(760 - 739\right) m m = 21 m m \text{ of Hg}$ as guided by Dalton's law of partial pressure.Here total pressure is the normal atmospheric pressure 760mm of Hg.

The ratio of these partial pressures should be equal to the ratio of no.of moles of of respective substances present in the mixture of vapour forming distillate.
Let ${n}_{w} \mathmr{and} {n}_{A}$ represent the no.of moles of water and organic liquid respectively present in the mixture at vapour stage.

Then we can write

${n}_{w} / {n}_{A} = {p}_{w} / {p}_{A}$

The ratio ${n}_{w} / {n}_{A}$ will also be maintained in the distillate obtained by condensation of vapour. Now it is given that the mass of A condensed is 1 g and its$\text{ molar mass"=123g/"mol}$
$\therefore {n}_{A} = \frac{1}{123} m o l$

Now we have

${n}_{w} / {n}_{A} = {p}_{w} / {p}_{A}$

$\implies {n}_{w} = \frac{739}{21} \cdot {n}_{A} = \frac{739}{21} \cdot \frac{1}{123} m o l$

So the no.of gram of water to be condensed with 1 g of A will be

${n}_{w} \times 18 \frac{g}{\text{mol}} = \frac{739}{21} \cdot \frac{1}{123} \cdot 18 g \approx 5.15 g$