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Equal weights of $CH_4$ and $O_2$ are mixed in an empty container of one litre at $27^{\large\circ}C$. Calculate the fraction of total pressure exerted by $O_2$.

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

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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)
answered Apr 3, 2014 by sharmaaparna1
 

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