At 363   K , the vapour pressure of A is 21   kPa and that of B is 18 kPa . One mole of A and 2…

Given ${\mathrm{P}}_{\mathrm{A}}^0=21 \mathrm{kPa} \Rightarrow {\mathrm{P}}_{\mathrm{B}}^0=18 \mathrm{kPa}$

$\to$ An Ideal solution is prepared by mixing $1$ mole $\mathrm{A}$ and $2$ mole $\mathrm{B}$.

$\to {\mathrm{X}}_{\mathrm{A}}=\frac{1}{3}$ and ${\mathrm{X}}_{\mathrm{B}}=\frac{2}{3}$

$\to$ According to Raoult's law

${\mathrm{P}}_{\mathrm{T}}={\mathrm{X}}_{\mathrm{A}}{\mathrm{P}}_{\mathrm{A}}^0+{\mathrm{X}}_{\mathrm{B}}{\mathrm{P}}_{\mathrm{B}}^0$

$\Rightarrow {\mathrm{P}}_{\mathrm{T}}=\left(\frac{1}{3}\times 21\right)+\left(\frac{2}{3}\times 18\right)$

$\Rightarrow {\mathrm{P}}_{\mathrm{T}}=7+12=19 \mathrm{kPa}$