객관식
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?
675
views
7. Gases
Effusion
객관식
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
검증된 단계별 안내1
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 \( M_{\text{unknown}} = 32 \times 0.935^2 \).
관련 영상
관련 실천
