# The vapour pressures of A and B

Question:

The vapour pressures of $\mathrm{A}$ and $\mathrm{B}$ at $25^{\circ} \mathrm{C}$ are $90 \mathrm{~mm} \mathrm{Hg}$ and $15 \mathrm{~mm} \mathrm{Hg}$ respectively. If $\mathrm{A}$ and $\mathrm{B}$ are mixed such that the mole fraction of $\mathrm{A}$ in the mixture is $0.6$, then the mole fraction of B in the vapour phase is $x \times 10^{-1}$. The value of $x$ is (Nearest integer)

Solution:

Given $\mathrm{P}_{\mathrm{A}}{ }^{\circ}=90 \mathrm{~mm} \mathrm{Hg}$, at $25^{\circ} \mathrm{C}$

$\mathrm{P}_{\mathrm{B}}{ }^{\circ}=15 \mathrm{~mm} \mathrm{Hg}$

and $\left.\begin{array}{l}\mathrm{X}_{\mathrm{A}}=0.6 \\ \mathrm{X}_{\mathrm{B}}=0.4\end{array}\right\} \mathrm{P}_{\mathrm{T}}=\mathrm{X}_{\mathrm{A}} \mathrm{P}_{\mathrm{A}}^{\mathrm{o}}+\mathrm{X}_{\mathrm{B}} \mathrm{P}_{\mathrm{B}}^{\mathrm{o}}$

$=(0.6 \times 90)+(0.4 \times 15)$

$=54+6=60 \mathrm{~mm}$

Now mol fraction of B in the vapour phase

i.e. $\quad Y_{B}=\frac{P_{B}}{P_{T}}=\frac{X_{B} P_{B}^{0}}{60}=0.1=1 \times 10^{-1}$

therefore: x = 1