$(a)\;\large\frac{1}{2}\qquad(b)\;\large\frac{1}{4}\qquad(c)\;\large\frac{1}{3}\qquad(d)\;\large\frac{1}{6}$

Given

Weight of $CH_4$ = weight of $O_2$ = w g

Mole fraction of $CH_4 = \large\frac{w/16}{w/16 + w/32}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\; = \large\frac{2}{3}$

Mole fraction of $O_2 = \large\frac{w/32}{w/16 + w/32}$

$\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;= \large\frac{1}{3}$

$P'_{O_2} = P_M \times Mole\; fraction\; of\; O_2$

(Palton's law of partial pressure)

$\large\frac{P'_{O_2}}{P_M}$ = mole fraction of $O_2$

$\;\;\;\;\;\; = \large\frac{1}{3}$

Hence answer is (C)

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