Question

At $363\mathrm{K}$, the vapour pressure of $\mathrm{A}$ is $21\mathrm{kPa}$ and that $\mathrm{B}$ is $18\mathrm{kPa}$ . One mole of $\mathrm{A}$ and $2$ moles of $\mathrm{B}$ are mixed. Assuming that solution is ideal, the vapour pressure of the mixture is ___________ $\mathrm{kPa}$. (Round off to the nearest integer).

Answer 1

Step 1:

Given;

${\mathrm{p}}_{\mathrm{A}}^{\circ }=21\mathrm{kPa},{\mathrm{p}}_{\mathrm{B}}^{\circ }=18\mathrm{kPa}$

Step 2:

Mole fraction $=$(Moles of single component/total moles of all components)

An ideal solution is prepared by mixing $1\mathrm{mol}\mathrm{A}$and $2\mathrm{mol}\mathrm{B}$.

$$\begin{gathered} {\mathrm{X}}_{\mathrm{A}}=\frac{1}{1+2} \\ {\mathrm{X}}_{\mathrm{A}}=\frac{1}{3} \end{gathered}$$

Similarly:

${\mathrm{X}}_{\mathrm{B}}=\frac{2}{3}$

According to Roult's law:

$$\begin{gathered} {\mathrm{P}}_{\mathrm{T}}={\mathrm{X}}_{\mathrm{A}}{\mathrm{P}}_{\mathrm{A}}^{\circ }+{\mathrm{X}}_{\mathrm{B}}{\mathrm{P}}_{\mathrm{B}}^{\circ } \\ {\mathrm{P}}_{\mathrm{T}}=\left(\frac{1}{3}\times 21\right)+\left(\frac{2}{3}\times 18\right) \\ =7+12 \\ {\mathrm{P}}_{\mathrm{T}}=\mathbf{19}\mathbf{kPa} \end{gathered}$$