Question

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 120 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 32 s for 1.0 L of O2 gas to effuse. You may want to reference (Pages 416 - 419) Section 10.8 while completing this problem. Part A Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.)

Answer 1

Answer:

450 g/mol

Explanation:

To solve the problem the problem, it is necessary to use Graham's Law of effusion

$\sqrt{\frac{M_{1} }{M_{2} } } = \frac{t_{1} }{t_{2}} \\\\{\frac{M_{1} }{32g/mol} = (\frac{120s}{32s} )^{2} = 450g/mol$

This answer is consistent with what we know. A heavier gas will require more time to effuse than a lighter gas