According to Graham's law of effusion, how many times faster do ammonia (NH_3) molecules effuse compared to carbon monoxide (CO) molecules?

• A. 1.37 times faster
• B. 2.00 times faster
• C. 1.00 times faster
• D. 0.73 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: \[\frac{\text{rate}_1}{\text{rate}_2} = \sqrt{\frac{M_2}{M_1}}\] 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: ammonia (NH\(_3\)) and carbon monoxide (CO). Assign ammonia as gas 1 and carbon monoxide as gas 2 for the formula.
3. Calculate the molar masses of each gas: - For NH\(_3\): Nitrogen (N) has an atomic mass of approximately 14 g/mol, and Hydrogen (H) has about 1 g/mol. So, \(M_{NH_3} = 14 + 3 \times 1 = 17\) g/mol. - For CO: Carbon (C) is about 12 g/mol, Oxygen (O) is about 16 g/mol. So, \(M_{CO} = 12 + 16 = 28\) g/mol.
4. Substitute the molar masses into Graham's law formula to find the relative rate of effusion of ammonia compared to carbon monoxide: \[\frac{\text{rate}_{NH_3}}{\text{rate}_{CO}} = \sqrt{\frac{28}{17}}\]
5. Evaluate the square root expression to determine how many times faster ammonia effuses compared to carbon monoxide. This value will tell you the relative speed of effusion.