The rate of diffusion of a gas having a molecular weight precisely double that of nitrogen gas () is — States of Matter and Gaseous State Chemistry Question
Question
The rate of diffusion of a gas having a molecular weight precisely double that of nitrogen gas ($N_2$) is $56\text{ mL/sec}$. What will be the rate of diffusion of nitrogen gas under identical conditions?
Answer: A
💡 Solution & Explanation
$r_{N2} / r_X = \sqrt{M_X / M_{N2}} = \sqrt{2M_{N2} / M_{N2}} = \sqrt{2} \approx 1.414$. $r_{N2} = r_X \times 1.414 = 56 \times 1.414 = 79.19\text{ mL/sec}$.
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