An effusion container is filled with 7 L of an unknown gas. It takes 109 s for the gas to effuse into a vacuum. From the same container under the same conditions--same temperature and initial pressure, it takes 380 s for 7.00 L of O2 gas to effuse. Calculate the molar mass (in grams/mol) of the unknown gas

 To calculate the molar mass of the unknown gas using effusion times, we can use 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 (M). The equation is as follows:

rate1 / rate2 = sqrt(M2 / M1)

where rate1 is the rate of effusion of the first gas (unknown gas), rate2 is the rate of effusion of the second gas (O2 gas), M1 is the molar mass of the first gas (unknown gas), M2 is the molar mass of the second gas (O2 gas).

Given that the volume of gas effused for both gases is the same, we can associate the effusion rates with the effusion times indirectly (since rate = volume/time). Thus:

time2 / time1 = sqrt(M1 / M2)

Rearrange this to solve for M1 (molar mass of the unknown gas):

M1 = M2 * (time1 / time2)^2

Given O2 has a molar mass (M2) of approximately 32.00 g/mol (since the molar mass of one O atom is 16.00 g/mol and O2 is a diatomic molecule with two O atoms), we can plug in the times into the formula:

M1 = 32.00 * (109 s / 380 s)^2 M1 = 32.00 * (0.2868)^2 M1 = 32.00 * 0.0823 M1 = 2.634 g/mol (rounded to three significant digits).

Thus, the molar mass of the unknown gas is approximately 2.63 grams per mole.