Hydrogen gas diffuses 4 times faster than a m | vedclass

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(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \f

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$3 : 1$

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(A) According to Graham's Law of Diffusion,the rate of diffusion $r$ is inversely proportional to the square root of the molar mass $M$: $r \propto \frac{1}{\sqrt{M}}$.Given that the rate of diffusion of $H_2$ is $4$ times that of the mixture,we have $\frac{r_{H_2}}{r_{mix}} = 4$.Therefore,$\frac{\sqrt{M_{mix}}}{\sqrt{M_{H_2}}} = 4$,which implies $\frac{M_{mix}}{M_{H_2}} = 16$.Since $M_{H_2} = 2 \ g/mol$,the average molar mass of the mixture $M_{mix} = 16 	imes 2 = 32 \ g/mol$.Let the mole fraction of $C_2H_4$ be $x$ and that of $CO_2$ be $(1-x)$.The molar mass of $C_2H_4$ is $28 \ g/mol$ and $CO_2$ is $44 \ g/mol$.$28x + 44(1-x) = 32$.$28x + 44 - 44x = 32$.$-16x = -12$,so $x = \frac{12}{16} = 0.75$.The mole fraction of $C_2H_4$ is $0.75$ and $CO_2$ is $0.25$.The molar ratio $C_2H_4 : CO_2 = 0.75 : 0.25 = 3 : 1$.

Hydrogen gas diffuses $4$ times faster than a mixture of $C_2H_4$ and $CO_2$. Find the molar ratio of $C_2H_4$ and $CO_2$ in the mixture.

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