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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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The vapour pressure of a solution of 5g of a nonelectrolyte in 100g water at a particular temperature is 2950 Pa and that of pure water at the same temperature is 3000 Pa.The molar mass of the solute is

$\begin{array}{1 1}54gmol^{-1}\\119gmol^{-1}\\179gmol^{-1}\\229gmol^{-1}\end{array} $

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1 Answer

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Answer : $54gmol^{-1}$
$x_2=-\large\frac{\Delta p}{p_1^*}=\frac{(3000-2950) Torr}{3000 Torr}=\frac{1}{60}$.
Now $x_2=\large\frac{(m_2/M_2)}{(m_1/M_1)}$
Hence $\large\frac{1}{60}=\frac{(5g/M)}{(100g/18gmol^{-1})}$
This gives $M=54gmol^{-1}$
answered Jul 22, 2014 by sreemathi.v
 

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