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).
(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
$\mathrm{P}_{\mathrm{T}}=\mathrm{X}_{\Lambda} \mathrm{P}_{\mathrm{A}}^{0}+\mathrm{X}_{\mathrm{B}} \mathrm{P}_{\mathrm{B}}^{0}$
$\Rightarrow \quad \mathrm{P}_{\mathrm{T}}=\left(\frac{1}{3} \times 21\right)+\left(\frac{2}{3} \times 18\right)$
$\Rightarrow \quad \mathrm{P}_{\mathrm{T}}=7+12=19 \mathrm{KPa}$
Click here to get exam-ready with eSaral
For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.