Question

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

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

Answer 1

Answer: (B)

Assuming mass of $C{H}_4$ and $H_2$ are $1gram$ of each

$Mol{e}_{C{H}_4}=\frac{1}{16}$ ; $Mol{e}_{H_2}=\frac{1}{2}$

$P_{H_2}=$ Mole fraction $\times$ P_total

total moles $=\frac{1}{16}+\frac{1}{12}=\frac{1+8}{16}=\frac{9}{16}$

mole fraction of = $H_2=\frac{moleof{H}_2}{totalmoles}=\frac{1/2}{9/16}=\frac{8}{9}$ $\frac{1/2}{9/16}=\frac{8}{9}{\mathrm{P}}_{{\mathrm{H}}_2}=\frac{8}{9}\times {\mathrm{P}}_{\text{total }}$