According to Graham's law, what is the ratio of the rate of effusion of F_2 gas to Cl_2 gas?

• A. sqrt{M_{Cl_2} / M_{F_2}}
• B. M_{Cl_2} / M_{F_2}
• C. sqrt{M_{F_2} / M_{Cl_2}}
• D. M_{F_2} / M_{Cl_2}

Answer: (A)

Verified step by step guidance

1. Recall Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Mathematically, this is expressed as: \(\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\), where \(r\) is the rate of effusion and \(M\) is the molar mass.
2. Identify the gases involved: \(F_2\) and \(Cl_2\). Assign \(r_{F_2}\) and \(r_{Cl_2}\) as their respective rates of effusion, and \(M_{F_2}\) and \(M_{Cl_2}\) as their molar masses.
3. Set up the ratio of the rates of effusion according to Graham's law: \(\frac{r_{F_2}}{r_{Cl_2}} = \sqrt{\frac{M_{Cl_2}}{M_{F_2}}}\).
4. Understand that the heavier gas effuses more slowly, so the rate ratio depends on the square root of the inverse ratio of their molar masses.
5. To find the numerical ratio, you would substitute the molar masses of \(F_2\) and \(Cl_2\) into the formula, but since the problem only asks for the expression, the correct ratio is \(\sqrt{\frac{M_{Cl_2}}{M_{F_2}}}\).