Under identical conditions, N_2(g) effuses at a rate that is approximately how many times faster than Cl_2(g)?

• A. 1.41 times faster
• B. 2.00 times faster
• C. 1.00 times faster
• D. 0.71 times faster

Answer: (A)

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: \(\text{Rate}_1 / \text{Rate}_2 = \sqrt{M_2 / M_1}\), where \(\text{Rate}_1\) and \(\text{Rate}_2\) are the effusion rates of gases 1 and 2, and \(M_1\) and \(M_2\) are their molar masses.
2. Identify the gases involved: gas 1 is \(\mathrm{N_2}\) and gas 2 is \(\mathrm{Cl_2}\). Find their molar masses using the periodic table: - Molar mass of \(\mathrm{N_2}\) is approximately 28 g/mol (14 g/mol per nitrogen atom times 2). - Molar mass of \(\mathrm{Cl_2}\) is approximately 70.9 g/mol (35.45 g/mol per chlorine atom times 2).
3. Set up the ratio of effusion rates using Graham's law: \(\frac{\text{Rate}_{\mathrm{N_2}}}{\text{Rate}_{\mathrm{Cl_2}}} = \sqrt{\frac{M_{\mathrm{Cl_2}}}{M_{\mathrm{N_2}}}}\).
4. Substitute the molar masses into the equation: \(\frac{\text{Rate}_{\mathrm{N_2}}}{\text{Rate}_{\mathrm{Cl_2}}} = \sqrt{\frac{70.9}{28}}\).
5. Calculate the square root to find how many times faster \(\mathrm{N_2}\) effuses compared to \(\mathrm{Cl_2}\). This will give you the factor by which \(\mathrm{N_2}\) effuses faster under identical conditions.