Equal moles of hydrogen and oxygen gases are | vedclass

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(C) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \fra

Vedclass · Accepted Answer

$1/8$

Vedclass · Answer

(C) According to Graham's Law of Effusion,the rate of effusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.The number of moles effused $n$ in time $t$ is given by $n = r 	imes t$.Therefore,the ratio of moles effused is $\frac{n_{H}}{n_{O}} = \frac{r_{H} 	imes t}{r_{O} 	imes t} = \frac{\sqrt{M_{O}}}{\sqrt{M_{H}}}$.Given $M_{H} = 2 \ g/mol$ and $M_{O} = 32 \ g/mol$.$\frac{n_{H}}{n_{O}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.So,$n_{H} = 4 	imes n_{O}$.We are given that $n_{H} = 1/2 \ mole$ (half of the initial $1 \ mole$).$1/2 = 4 	imes n_{O} \implies n_{O} = 1/8 \ mole$.Since we started with $1 \ mole$ of oxygen,the fraction of oxygen that escapes is $1/8$.

Equal moles of hydrogen and oxygen gases are placed in a container with a pin-hole through which both can escape. What fraction of the oxygen escapes in the time required for one-half of the hydrogen to escape?

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