A certain gas effuses through a small opening of a vessel at a rate which is exactly one-fifth the r — States of Matter and Gaseous State Chemistry Question

💡 Solution & Explanation

# Solution: Effusion Rate and Molecular Weight **Apply Graham's Law of Effusion:** The rate of effusion is inversely proportional to the square root of molar mass: $$\frac{\text{Rate}_{\text{gas}}}{\text{Rate}_{He}} = \sqrt{\frac{M_{He}}{M_{\text{gas}}}}$$ **Given Information:** - Rate of gas = $\frac{1}{5}$ × Rate of He - $M_{He} = 4$ g/mol **Substitute into the equation:** $$\frac{1}{5} = \sqrt{\frac{4}{M_{\text{gas}}}}$$ **Square both sides:** $$\frac{1}{25} = \frac{4}{M_{\text{gas}}}$$ **Solve for $M_{\text{gas}}$:** $$M_{\text{gas}} = 4 \times 25 = 100 \text{ g/mol}$$ **Answer: The molecular weight of the gas is 100 g/mol.** This corresponds to a gas like $N_2O_4$ or $C_6H_6$ (benzene). The heavier the gas, the slower it effuses—here, a gas 25 times heavier than He effuses at one-fifth the rate.