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青州's Question
Posted 2 months ago

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It takes 157 s157 \mathrm{~s} for an unknown gas to effuse through a porous wall and 96 s96 \mathrm{~s} for the same volume of N2\mathrm{N}_{2} gas to effuse at the same temperature and pressure. What is the molar mass of the unknown gas?
69 \times \mathrm{xol}
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Answer from Sia
Posted 2 months ago
Use Graham's Law of Effusion: Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass
Set up the equation: According to Graham's Law, Rate of effusion of gas 1Rate of effusion of gas 2=Molar mass of gas 2Molar mass of gas 1\frac{\text{Rate of effusion of gas 1}}{\text{Rate of effusion of gas 2}} = \sqrt{\frac{\text{Molar mass of gas 2}}{\text{Molar mass of gas 1}}}. Here, gas 1 is the unknown gas and gas 2 is N2\mathrm{N_2}
Calculate the rates of effusion: The rate of effusion is inversely proportional to the time taken, so Rate of effusion of unknown gas=1157 s\text{Rate of effusion of unknown gas} = \frac{1}{157 \mathrm{~s}} and \text{Rate of effusion of N_2} = \frac{1}{96 \mathrm{~s}}
Insert the values into the equation: 1/1571/96=28 g/molM\frac{1/157}{1/96} = \sqrt{\frac{28 \mathrm{~g/mol}}{M}}, where MM is the molar mass of the unknown gas
Solve for the molar mass of the unknown gas: 96157=28M\frac{96}{157} = \sqrt{\frac{28}{M}} leads to M=28×(157)2(96)2M = \frac{28 \times (157)^2}{(96)^2}
Calculate and find the molar mass: After solving the equation, we find the molar mass MM of the unknown gas
M=28×(157)2(96)244.1 g/molM = \frac{28 \times (157)^2}{(96)^2} \approx 44.1 \mathrm{~g/mol}
Key Concept
Graham's Law of Effusion
Graham's Law relates the rates of effusion of two gases to their molar masses, allowing us to calculate the molar mass of an unknown gas when compared to a known gas.

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