🤓 Based on our data, we think this question is relevant for Professor Heider's class at UVU.
Recall: Graham's Law of Effusion allows us to compare the rate of effusion of two gases. Graham's Law states that the rate of effusion of a gas is inversely proportional to its molar mass.
This means that when comparing two gases:
Let's designate CO2 as gas 1 and the unknown as gas 2. We're given:
rateCO2 = 102 mL/s MCO2 = 12.01 g/mol C + 2(16.00 g/mol O) = 44.01 g/mol
rateunk = 83.3 mL/s Munk = ?
A gas of unknown identity diffuses at a rate of 83.3 mL/s in a diffusion apparatus in which carbon dioxide diffuses at the rate of 102 mL/s. Calculate the molecular mass of the unknown gas.
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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.
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Based on our data, we think this problem is relevant for Professor Heider's class at UVU.
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Our data indicates that this problem or a close variation was asked in Chemistry - OpenStax 2015th Edition. You can also practice Chemistry - OpenStax 2015th Edition practice problems.