Question:
At $20^{\circ} \mathrm{C}$, the vapour pressure of benzene is 70 torr and that of methyl benzene is 20 torr. The mole fraction of benzene in the vapour phase at $20^{\circ} \mathrm{C}$ above an equimolar mixture of benzene and methyl benzene is ________$\times 10^{-2}$. (Nearest integer)
Solution:
$\mathrm{P}_{\mathrm{B}}^{\mathrm{o}}=40$ $\mathrm{P}_{\mathrm{T}}^{\mathrm{o}}=20$ $\mathrm{K}_{\mathrm{B}}=0.5=\mathrm{K}_{\mathrm{T}}$
Now $\quad \mathrm{y}_{\mathrm{B}}=\frac{\mathrm{K}_{\mathrm{B}} \mathrm{P}_{\mathrm{B}}^{\mathrm{o}}}{\mathrm{K}_{\mathrm{B}} \mathrm{P}_{\mathrm{B}}^{\mathrm{o}}+\mathrm{K}_{\mathrm{T}} \mathrm{P}_{\mathrm{T}}^{\mathrm{o}}}$
$=\frac{70 \times 0.5}{70 \times 0.5+20 \times 0.5}$