logo

Ask Questions, Get Answers

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

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} $

Can you answer this question?
 
 

1 Answer

0 votes
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
 

Related questions

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