Ask Questions, Get Answers

Want to ask us a question? Click here
Browse Questions
Home  >>  CBSE XII  >>  Chemistry  >>  Solutions
0 votes

Henry’s law constant for the molality of methane in benzene at 298 K is $4.27 \times 10^5$ mm Hg. Calculate the solubility of methane in benzene at 298 K under 760 mm Hg.

$\begin{array}{1 1}1.78\times 10^{-3}\\2.78\times 10^{-3}\\3.78\times 10^{-3}\\4.78\times 10^{-3}\end{array} $

Can you answer this question?

1 Answer

0 votes
Given :
$K_H=4.27\times 10^5$ mm
Applying Hentry's law
$\;\;\;=\large\frac{760m}{4.27\times 10^5mm}$
$\;\;\;=1.78\times 10^{-3}$
(ie) Mole fraction of methane in benzene =$1.78\times 10^{-3}$
answered Aug 6, 2014 by sreemathi.v
Ask Question
student study plans
JEE MAIN, CBSE, AIPMT Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App