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Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
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The vapour pressure of a pure liquid A is 10.0 Torr at $27^{\large\circ}C$,One gram of B is dissolved in 20g of A,the vapour pressure is lowered to 90 Torr.If the molar mass of A is $200gmol^{-1}$,the molar mass of B is

$\begin{array}{1 1}75gmol^{-1}\\85gmol^{-1}\\100gmol^{-1}\\115gmol^{-1}\end{array} $

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Answer : $100gmol^{-1}$
$x_2=-\large\frac{\Delta p}{p_1^*}=\frac{1 Torr}{10 Torr}=\frac{1}{10}$
Now $x_2=\large\frac{(m_2/M_2)}{(m_1/M_1)}$
Hence $\large\frac{1}{10}=\frac{(1g/M_2)}{(20g/200gmol^{-1})}$
This gives $M_2=100gmol^{-1}$
answered Jul 22, 2014 by sreemathi.v
 

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