Multiple Choice
A sample of bromine gas requires 84.5 s to effuse from a container. Under identical conditions, how long (in seconds) will it take a sample of neon gas to effuse?
7. Gases
Effusion
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The rate of effusion of oxygen compared to an unknown gas is 0.935. What is the molar mass of the unknown gas?
A
44 g/mol
B
32 g/mol
C
28 g/mol
D
16 g/mol
Verified step by step guidance1
Understand the concept of effusion: Effusion is the process by which gas molecules escape through a tiny hole into a vacuum. The rate of effusion is inversely proportional to the square root of the molar mass of the gas, according to Graham's law.
Apply Graham's law of effusion: Graham's law states that \( \frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \sqrt{\frac{M_2}{M_1}} \), where \( M_1 \) and \( M_2 \) are the molar masses of gas 1 and gas 2, respectively.
Set up the equation using the given rate: Let oxygen be gas 1 with a molar mass of 32 g/mol, and the unknown gas be gas 2. The rate of effusion of oxygen compared to the unknown gas is given as 0.935, so \( 0.935 = \sqrt{\frac{M_{\text{unknown}}}{32}} \).
Solve for the molar mass of the unknown gas: Square both sides of the equation to eliminate the square root, resulting in \( 0.935^2 = \frac{M_{\text{unknown}}}{32} \).
Rearrange the equation to find \( M_{\text{unknown}} \): Multiply both sides by 32 to isolate \( M_{\text{unknown}} \), giving \( M_{\text{unknown}} = 32 \times 0.935^2 \).
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