States of Matter - Result Question 40
####40. Equal weights of methane and oxygen are mixed in an empty container at $25^{\circ} C$. The fraction of the total pressure exerted by oxygen is
$(1981,1 M)$
(a) $\frac{1}{3}$
(b) $\frac{1}{2}$
(c) $\frac{2}{3}$
(d) $\frac{1}{3} \times \frac{273}{298}$
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Solution:
- If $x g$ of both oxygen and methane are mixed then :
$$ \begin{aligned} \text { Mole of oxygen } & =\frac{x}{32} \ \text { Mole of methane } & =\frac{x}{16} \end{aligned} $$
$$ \Rightarrow \quad \text { Mole fraction of oxygen }=\frac{\frac{x}{32}}{\frac{x}{32}+\frac{x}{16}}=\frac{1}{3} $$
According to law of partial pressure
Partial pressure of oxygen $\left(p _{O _2}\right)=$ Mole fraction $\times$ Total pressure
$$ \Rightarrow \quad \frac{p _{O _2}}{p}=\frac{1}{3} $$
