logo

Ask Questions, Get Answers

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

A soda water bottle was opened and the soft drink is allowed to come at $27^{\large\circ}$C with the air containing carbon dioxide at $3.8\times 10^{-4}$atm.What would be the concentration of carbon dioxide in the soda water after it had stood open and come to equilibrium.Give value of Henry's constant for aqueous solution of carbon dioxide at $26^{\large\circ}$C is $3.1\times 10^{-2}$ mol lit $^{-1}$ atm$^{-1}$.

$\begin{array}{1 1}1.18\times 10^{-5}mollit^{-1}\\2.18\times 10^{-5}mollit^{-1}\\3.18\times 10^{-5}mollit^{-1}\\4.18\times 10^{-5}mollit^{-1}\end{array} $

Can you answer this question?
 
 

1 Answer

0 votes
According to Henry's law
$C=K_H.P$
Where given $K_H=3.1\times 10^{-2}$ mol lit $^{-1} atm^{-1}$
$P=3.8\times 10^{-4}$ atm
Hence,
$C=3.1\times 10^{-2}\times 3.8\times 10^{-4}$
$\;\;\;=1.18\times 10^{-5}mol lit ^{-1}$
answered Aug 7, 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
...