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

Answer: 25

By raoult's law

$\frac{P_A^o-P_A}{P_A^o}=\frac{n_B}{n_A+n_B}$

Or $\frac{P_A^o-P_A}{P_A}=\frac{n_B}{n_A}$

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

$P_A^o$ = the vapour pressure of the pure component

$n_B$ : moles of solute

$n_A$ : moles of solvent

$\frac{P_A^o-P_A}{P_A}=\frac{n_B}{n_A}$

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

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

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

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

$P_A=25.0$