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2.13    The partial pressure of ethane over a solution containing 6.56\times 10^{-3}\; g of ethane is 1 bar. If the solution contains 5.00\times 10^{-2}g of ethane, then what shall be the partial pressure of the gas?

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Using Henry's Law we can write,  

                                                           m = k.P

Putting value in this equation, we get :

                                              6.56\times10^{-3} = k\times 1

So, the magnitude of k is   6.56\times10^{-3}.

Now, we will again use the above equation for  m = 5.0\times10^{-2}\ g.

So the required partial pressure is :- 

                                           p = \frac{m}{k} = \frac{5.0\times10^{-2}}{6.56\times10^{-3}}

or                                                    p = 7.62\ bar

                                       

Posted by

Devendra Khairwa

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