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At room temperature $25^{\large\circ}C$,the expression for total pressure of mixture of methanol and ethanol solution is given by P(in torr)$=135+120X$ where $X$ is mol fraction of methanol,Hence

$\begin{array}{1 1}(a)\;\text{vapour pressure of pure ethanol in 135 torr}\\(b)\;\text{vapour pressure of pure ethanol in 255 torr}\\(c)\;\text{both (a) and (b) correct}\\(d)\;\text{mixture is completely immiscible}\end{array}$

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Given $P=135+120X$
If $X=1$,then vapour pressure of methanol $P=P^0$
$\therefore P=135+120(1)$
$\Rightarrow 135+120=255torr$
If $X=0$,then vapour pressure of methanol $P=P^0$
$\therefore P=135+120(0)$
$\Rightarrow P=135$
Hence (c) is the correct answer.
answered Feb 14, 2014 by sreemathi.v

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