Question

Same mass of $C{H}_4$ and $H_2$ is taken in container. The partial pressure caused by $H_2$ is:

• A. $\frac{8}{9}$
• B. $\frac{1}{9}$
• C. $\frac{1}{2}$
• D. $1$

Answer 1

The correct option is A $\frac{8}{9}$

Equal mass to be taken $=16g$

No. of moles of $C{H}_4=\frac{16}{16}=1\text{mol}$

number of moles of $H_2=\frac{16}{2}=8$

Then, mole fraction of $H_2=\frac{8}{8+1}=\frac{8}{9}$

Partial pressure caused by $H_2=\frac{8}{9}\times \text{Total pressure}$

Total pressure $=1atm$ (as it is done in a open container)

Partial pressure of $H_2=\frac{8}{9}\times 1=\frac{8}{9}$