Determine the identity of an unknown gas that effuses at a rate 1.48 times as fast as chlorine gas.

• A. N₂
• B. O₂
• C. Ar
• D. CO₂
• E. CO

Answer: (B)

Verified step by step guidance

1. Understand the concept of effusion: Effusion is the process by which gas molecules escape through a tiny hole into a vacuum. According to Graham's law of effusion, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
2. Apply Graham's law of effusion: The formula for Graham's law is \( \frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \sqrt{\frac{M_2}{M_1}} \), where \( M_1 \) and \( M_2 \) are the molar masses of gas 1 and gas 2, respectively.
3. Set up the equation using the given information: Let the unknown gas be gas 1 and chlorine gas be gas 2. The rate of effusion of the unknown gas is 1.48 times that of chlorine gas. Therefore, \( 1.48 = \sqrt{\frac{M_{Cl_2}}{M_{unknown}}} \).
4. Solve for the molar mass of the unknown gas: Rearrange the equation to find \( M_{unknown} \). Square both sides to eliminate the square root: \( 1.48^2 = \frac{M_{Cl_2}}{M_{unknown}} \). Then, solve for \( M_{unknown} \) by multiplying both sides by \( M_{unknown} \) and dividing by \( 1.48^2 \).
5. Compare the calculated molar mass to the options provided: Calculate the molar mass of each option (\( N_2 \), \( O_2 \), \( Ar \), \( CO_2 \), \( CO \)) and identify which one matches the calculated molar mass of the unknown gas.