# At 363K, the vapour pressure of A is 21kPa and that of B is 18 kPa.

Question:

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

Solution:

(19)

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

$\rightarrow$ An Ideal solution is prepared by mixing mol $\mathrm{A}$ and $2 \mathrm{~mol} \mathrm{~B}$

$\rightarrow X_{A}=\frac{1}{3}$ and $X_{B}=\frac{2}{3}$

$m$ Acc to Raoult's low

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

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

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