Vapour pressure of an equimolar mixture of benzene and toluene at a given temperature was found to be 80 mm Hg. If vapours above the liquid phase is condensed in a beaker, vapour pressure of this condensate at the same temperature was found to be 100 mm Hg. If the pure state vapour pressure of benzene and toluene is respectively x and y. then determine value of $\frac{x + 2 y}{50}$

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Detailed Solution

Let vapour pressure of pure benzene = $P_{B}^{\circ}$
Vapour pressure of pure toluene $= P_{T}^{\circ}$
$X_{B} = \frac{1}{2} X_{T} = \frac{1}{2}$
$80 = \frac{1}{2} P_{B}^{\circ} + \frac{1}{2} P_{T}^{\circ} - - - - (1)$
$Y_{B} = \frac{X_{B} P_{B}^{\circ}}{P} = \frac{P_{B}^{\circ}}{160} Y_{T} = \frac{X_{T} P_{T}^{\circ}}{P} = \frac{P_{T}^{\circ}}{160}$
For the vapours of condensate
$100 = Y_{B} P_{B}^{\circ} + Y_{T} P_{T}^{\circ} 100 = \frac{{(P_{B}^{\circ})}^{2}}{160} + \frac{{(P_{T}^{\circ})}^{2}}{160} 16000 = {(P_{B}^{\circ})}^{2} + {(P_{T}^{\circ})}^{2}$
On solving
$P_{B}^{\circ} = 120 P_{T}^{\circ} = 40 X = 120 Y = 40$