A sample of He gas (2.0 mmol)effused through a pinhole in 53 s.The same amount of an unknown gas,under the same conditions,effused through the pinhole in 248 s.The molecular mass of the unknown gas is ________ g/mol.
A)0.19
B)5.5
C)88
D)19
E)350

Question

Quizplus · Accepted Answer

The molecular mass of the unknown gas can be found using 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. The formula derived from Graham's law is:$$ \frac{\text{Rate}_1}{\text{Rate}_2} = \sqrt{\frac{M_2}{M_1}} $$where Rate_1 and Rate_2 are the effusion rates of gas 1 and gas 2, respectively, and M_1 and M_2 are their molar masses. Given that helium (He) has a molar mass of 4 g/mol and effused in 53 s, and the unknown gas effused in 248 s, we can set up the equation as follows, with the rate being inversely proportional to time:$$ \frac{53}{248} = \sqrt{\frac{M_{\text{unknown}}}{4}} $$Solving for M_unknown gives a molar mass of approximately 88 g/mol.

A Sample of He Gas (2