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 using Graham's Law of Effusion:** Graham's Law states that the rate of effusion is inversely proportional to the square root of molar mass: $$\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}$$ **Given information:** - Rate of unknown gas = $\frac{1}{5}$ × Rate of helium - Therefore: $\frac{r_{gas}}{r_{He}} = \frac{1}{5}$ **Applying Graham's Law:** $$\frac{1}{5} = \sqrt{\frac{M_{He}}{M_{gas}}}$$ **Squaring both sides:** $$\frac{1}{25} = \frac{M_{He}}{M_{gas}}$$ **Solving for molecular weight of gas:** $$M_{gas} = 25 \times M_{He} = 25 \times 4 = 100 \text{ g/mol}$$ **Why this is correct:** The inverse relationship between effusion rate and molar mass means a slower-effusing gas must be much heavier. Since the gas effuses at only 1/5 the rate, squaring this ratio gives us 1/25, making the molecular weight 25 times that of helium (4 u), yielding **100 g/mol** — consistent with gases like $F_2$ or $Cl_2$ derivatives.