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A solution contains 10g of a solute and 116g of acetone at $30^{\large\circ}$C.Its vapour pressure is 290 Torr.If the vapour pressure of pure acetone is 300 Torr at $30^{\large\circ}$C,the molar mass of solute is

$\begin{array}{1 1}75gmol^{-1}\\100gmol^{-1}\\125gmol^{-1}\\150gmol^{-1}\end{array} $

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Answer : $150gmol^{-1}$
$x_2=-\large\frac{\Delta p}{p_1^*}$
$\Rightarrow \large\frac{10\;Torr}{300\;Torr}=\frac{1}{30}$
Now $x_2=\large\frac{(m_2/M_2)}{(m_1/M_1)}$
Hence $\large\frac{1}{30}=\frac{(10g/M_2)}{(116g/58gmol^{-1})}$
This gives $M_2=150gmol^{-1}$
answered Jul 22, 2014 by sreemathi.v

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