Calculating Molar Mass of an Unknown Gas from Effusion Rate

[Lilly]

effusion rates...please help!

Could someone please explain to me how to do this question...
The rate of effusion of a particular gas was measured to be 24.0 mL/min. Under the same conditions the rate of effusion of pure methane gas, (CH4) is 47.8 mL/min. What is the molar mass of the unknown gas?

 

[chem_tr]

Effusion rates are indirectly proportional with the molar mass; if the molecule is bigger, it can't move fast. So the particular gas has a greater molar mass.

Calculate methane's molar mass by taking C:12, and H:1 grams/mol, and find the unknown gas' molar mass.

Regards,
chem_tr

 

[ravenprp]

Sure, I'd be happy to help with this question! To calculate the molar mass of the unknown gas, we can use Graham's Law of Effusion. This law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. In other words, the lighter the gas, the faster it will effuse (or escape) through a small opening.

To apply this law to the given problem, we can set up a ratio of the unknown gas's effusion rate (24.0 mL/min) to the known methane gas's effusion rate (47.8 mL/min). This can be written as:

24.0 mL/min ÷ 47.8 mL/min = √(molar mass of unknown gas) ÷ √(molar mass of methane gas)

Simplifying this ratio, we get:

0.5021 = √(molar mass of unknown gas) ÷ √(16.04 g/mol)

To solve for the molar mass of the unknown gas, we can cross-multiply and square both sides of the equation:

0.5021 x √(16.04 g/mol) = √(molar mass of unknown gas)

Squaring both sides, we get:

0.5021^2 x 16.04 g/mol = molar mass of unknown gas

Solving for the molar mass, we get:

molar mass of unknown gas = 4.04 g/mol

Therefore, the molar mass of the unknown gas is approximately 4.04 g/mol. Keep in mind that this is just an estimation since we are using the effusion rate instead of the actual mass of the gas. I hope this helps! Let me know if you have any further questions.