If the rate of effusion of oxygen (O_2) to an unknown gas is 0.935, what is the molar mass of the unknown gas? (Assume both gases are at the same temperature and pressure.)

• A. 28 g/mol
• B. 36.6 g/mol
• C. 44 g/mol
• D. 32 g/mol

Answer: (B)

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 \(r_1\) and \(r_2\) are the rates of effusion of gases 1 and 2, and \(M_1\) and \(M_2\) are their molar masses.
2. Identify the known values: the rate of effusion of oxygen (\(O_2\)) relative to the unknown gas is given as \(\frac{r_{O_2}}{r_{unknown}} = 0.935\), and the molar mass of oxygen is \(M_{O_2} = 32\) g/mol.
3. Set up the equation using Graham's law with oxygen as gas 1 and the unknown gas as gas 2: \$0.935 = \sqrt{\frac{M_{unknown}}{32}}$.
4. Square both sides of the equation to eliminate the square root: \((0.935)^2 = \frac{M_{unknown}}{32}\).
5. Solve for the molar mass of the unknown gas: \(M_{unknown} = 32 \times (0.935)^2\).