Problem: A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 115 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 31 s for 1.0 L of O2 gas to effuse.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.)

🤓 Based on our data, we think this question is relevant for Professor Nathanson's class at UW-MADISON.

FREE Expert Solution

We use Graham's Law of Effusion to solve this problem:

Where M is the Molar mass. 

We do not have the given rates so we would have to modify the equation to fit the given values. 

Now the rate is si given as below:

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Problem Details

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 115 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 31 s for 1.0 L of O2 gas to effuse.

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.)

Frequently Asked Questions

What scientific concept do you need to know in order to solve this problem?

Our tutors have indicated that to solve this problem you will need to apply the Effusion concept. You can view video lessons to learn Effusion. Or if you need more Effusion practice, you can also practice Effusion practice problems.

What professor is this problem relevant for?

Based on our data, we think this problem is relevant for Professor Nathanson's class at UW-MADISON.