Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  JEEMAIN and AIPMT  >>  Chemistry  >>  Solutions
0 votes

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} $

Can you answer this question?

1 Answer

0 votes
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

Related questions

Ask Question
student study plans
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App