Question

The vapour pressure of pure water at $26^{o}C$ is $\text{25} \text{.5} \, \text{torr} \text{.}$ . The vapour pressure of a solution which contains $\text{20} \text{.0}$ glucose, $\left(\mathrm{C}_6 \mathrm{H}_{12} \mathrm{O}_6\right)$, water (in torr) is?

Answer 1

By raoult's law

$\frac{P_{A}^{ο} - P_{A}}{P_{A}^{ο}}=\frac{n_{B}}{n_{A} + n_{B}}$

Or $\frac{P_{A}^{ο} - P_{A}}{P_{A}}=\frac{n_{B}}{n_{A}}$

$P_{A}$ := the partial vapour pressure of the component {display style $\text{i}$ } in the gaseous mixture (above the solution),

$P_{A}^{ο}$ = the vapour pressure of the pure component

$n_{B}$ : moles of solute

$n_{A}$ : moles of solvent

$\frac{P_{A}^{ο} - P_{A}}{P_{A}}=\frac{n_{B}}{n_{A}}$

$\frac{25.5 - P_{A}}{P_{A}}=\frac{\frac{20}{180}}{\frac{100}{18}}$

$\frac{25.5 - P_{A}}{P_{A}}=\frac{20 \times 18}{100 \times 180}=0.02$

$25.5 - \text{P}_{\text{A}} = 0.02 \, \text{P}_{\text{A}}$

$1.02 \, \text{P}_{\text{A}} = 25.5$

$P_{A}=25.0$