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Vapour pressure of an equimolar mixture of benzene and toluene at a given temperature was found to be 80 mm Hg. If vapour 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 $\frac{x + 2 y}{40}$

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Detailed Solution

$x + y = 160$

$y_{B} = \frac{x}{160} = x_{B}^{l}$

$y_{T} = \frac{y}{160} = x_{T}^{l}$

$100 = (x) (\frac{x}{160}) + y (\frac{y}{160})$

$(100) (160) = x^{2} + y^{2}$

$(100) (160) = x^{2} + {(160 - x)}^{2}$

$(160) (100) = x^{2} + {(160)}^{2} + x^{2} - 320 x$

$2 x^{2} - 320 x + 160 (60) = 0$

$x^{2} - 160 x + 80 (60) = 0$

$x = \frac{160 \pm \sqrt{{(160)}^{2} - 4 (80) (60)}}{2}$

$x = \frac{160 \pm \sqrt{(160) (160 - 120)}}{2}$

$x = \frac{160 \pm 4 \times 2 \times 10}{2} = \frac{160 \pm 80}{2}$

X=40, 120 (X=40 and Y=120 is rejected because benzene is more volatile then toluene) Y = 120, 40

$\frac{x + 2 y}{40} = \frac{120 + 80}{40} = \frac{200}{40} = 5$

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