Question

Equal weights of $C{H}_4$ and $H_2$ are mixed in an empty container at ${25}^{\circ }C$. The fraction of the total pressure exerted by $H_2$ is:

• A. $1/9$
• B. $1/2$
• C. $8/9$
• D. $16/17$

Answer 1

The correct option is C $8/9$

Ratio of number of moles of $C{H}_4$ and $H_2$ are $C{H}_4:H_2=\frac{x}{16}:\frac{x}{2}$
$=1:8$

Hence, partial pressure exerted by

$H_2=\frac{8}{1+8}\times p_{total}=\frac{8}{9}\times p_{total}$