Two grams of hydrogen diffuse from a containe | 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

Vedclass · Accepted Answer

$0.5$

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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 $M$: $r \propto \frac{1}{\sqrt{M}}$.Since the time $t$ is the same for both gases,the rate of diffusion is proportional to the mass $m$ diffused: $r = \frac{m}{t}$.Therefore,$\frac{m_{H_2}}{m_{O_2}} = \sqrt{\frac{M_{O_2}}{M_{H_2}}}$.Given $m_{H_2} = 2 \ g$,$M_{H_2} = 2 \ g/mol$,and $M_{O_2} = 32 \ g/mol$.Substituting the values: $\frac{2}{m_{O_2}} = \sqrt{\frac{32}{2}} = \sqrt{16} = 4$.Thus,$m_{O_2} = \frac{2}{4} = 0.5 \ g$.

Two grams of hydrogen diffuse from a container in $10 \ min$. How many grams of oxygen would diffuse through the same container in the same time under similar conditions?

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