At a fixed temperature, how many times faster does nitric oxide (NO) effuse compared to nitrogen dioxide (NO_2)?

• A. sqrt{30/46}
• B. sqrt{46/30}
• C. 30/46
• D. 46/30

Answer: (B)

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: Gas 1 is nitric oxide (NO) and Gas 2 is nitrogen dioxide (NO_2).
3. Calculate the molar masses of each gas: For NO, sum the atomic masses of nitrogen (approximately 14 g/mol) and oxygen (approximately 16 g/mol) to get \(M_{NO}\). For NO_2, sum the atomic masses of nitrogen and two oxygens to get \(M_{NO_2}\).
4. Set up the ratio of effusion rates using Graham's law: \(\frac{r_{NO}}{r_{NO_2}} = \sqrt{\frac{M_{NO_2}}{M_{NO}}}\).
5. Interpret the result: The ratio tells you how many times faster NO effuses compared to NO_2 at the same temperature.