Methane gas,CH₄,effuses through a barrier at a rate of 0.568 mL/minute.If an unknown gas effuses through the same barrier at a rate of 0.343 mL/minute,what is the molar mass of the gas?

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Quizplus · Accepted Answer

Using Graham's law of effusion, $ \frac{rate_1}{rate_2} = \sqrt{\frac{M_2}{M_1}} $, where $rate_1$ and $rate_2$ are the effusion rates of two gases, and $M_1$ and $M_2$ are their molar masses. Given $rate_1 = 0.568$ mL/min for methane (CH4) with $M_1 = 16.04$ g/mol, and $rate_2 = 0.343$ mL/min for the unknown gas, we solve for $M_2$:$$ \frac{0.568}{0.343} = \sqrt{\frac{M_2}{16.04}} $$$$ 1.657 = \sqrt{\frac{M_2}{16.04}} $$$$ M_2 = 1.657^2 \times 16.04 \approx 43.9 \, \text{g/mol} $$

Methane Gas,CH4,effuses Through a Barrier at a Rate of 0

Methane Gas,CH₄,effuses Through a Barrier at a Rate of 0