Gas Y effuses at half the rate of O_2. What is the molar mass of Gas Y?

• A. 64 g/mol
• B. 32 g/mol
• C. 16 g/mol
• D. 128 g/mol

Answer: (D)

Verified step by step guidance

1. Recall Graham's law of effusion, which relates the rates of effusion of two gases to their molar masses: \(\frac{r_1}{r_2} = \sqrt{\frac{M_2}{M_1}}\), where \(r\) is the rate of effusion and \(M\) is the molar mass.
2. Identify the gases: let Gas Y be gas 1 and oxygen (\(O_2\)) be gas 2. Given that Gas Y effuses at half the rate of \(O_2\), write \(\frac{r_{Y}}{r_{O_2}} = \frac{1}{2}\).
3. Substitute the known values into Graham's law: \(\frac{1}{2} = \sqrt{\frac{M_{O_2}}{M_Y}}\).
4. Square both sides to eliminate the square root: \(\left(\frac{1}{2}\right)^2 = \frac{M_{O_2}}{M_Y}\), which simplifies to \(\frac{1}{4} = \frac{M_{O_2}}{M_Y}\).
5. Rearrange to solve for the molar mass of Gas Y: \(M_Y = 4 \times M_{O_2}\). Since the molar mass of \(O_2\) is 32 g/mol, multiply by 4 to find \(M_Y\).