If H₂ has an effusion rate that is 3.72 times faster than a gas, what is the identity of the unknown gas?

• A. Cl₂
• B. CO₂
• C. N₂O₄
• D. N₂
• E. O₂

Answer: (D)

Verified step by step guidance

1. Understand that the problem involves effusion rates, which can be analyzed using Graham's Law of Effusion. This law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
2. Express Graham's Law mathematically: \( \frac{\text{Rate of effusion of } H_2}{\text{Rate of effusion of unknown gas}} = \sqrt{\frac{M_{\text{unknown gas}}}{M_{H_2}}} \), where \( M \) represents molar mass.
3. Given that the effusion rate of \( H_2 \) is 3.72 times faster than the unknown gas, set up the equation: \( 3.72 = \sqrt{\frac{M_{\text{unknown gas}}}{M_{H_2}}} \).
4. Square both sides of the equation to eliminate the square root: \( (3.72)^2 = \frac{M_{\text{unknown gas}}}{M_{H_2}} \).
5. Solve for \( M_{\text{unknown gas}} \) by multiplying both sides by \( M_{H_2} \). Use the known molar mass of \( H_2 \) (approximately 2.02 g/mol) to find the molar mass of the unknown gas, and compare it to the given options to identify the gas.