Under standard initial conditions, the rate of effusion of a gas having a molecular weight just doub — States of Matter and Gaseous State Chemistry Question
Question
Under standard initial conditions, the rate of effusion of a gas having a molecular weight just double that of hydrogen gas ($H_2$) is $30\text{ mL s}^{-1}$. Calculate the approximate rate of effusion of hydrogen gas (in $\text{mL s}^{-1}$) under identical conditions. (Use $\sqrt{2} = 1.414$ and round to the nearest integer).
Answer: 42
💡 Solution & Explanation
Let the unknown gas be X. $M_X = 2 \times M_{H_2} = 4\text{ g/mol}$. $r_X = 30\text{ mL/s}$. $r_{H_2} / r_X = \sqrt{M_X / M_{H_2}} = \sqrt{4 / 2} = \sqrt{2}$. Thus, $r_{H_2} = 30 \times \sqrt{2} = 30 \times 1.414 = 42.42\text{ mL/s}$. Rounded to nearest integer is 42.
💬Ask on WhatsApp →
Still have doubts about this question?
Send it to our AI chemistry tutor on WhatsApp — gets answered in minutes