It takes 200. seconds for a sample of carbon dioxide to effuse through a porous plug and 280. seconds for the same volume of an unknown gas to effuse under the same conditions. What is the molar mass of the unknown gas? Enter your answer to 3 significant figures.
Choose a convenient volume such as 1 L. Then rate CO2 = 1L/200 s = ?
rate unknown = 1L/280 s =
(rate CO2/rate unk) = sqrt(MMunk/MMCO2)
MM = molar mass. Solve for MM unk.l
39.7
To find the molar mass of the unknown gas, we can use Graham's law of effusion. According to Graham's law, the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
The rate of effusion can be determined by taking the reciprocal of the time taken to effuse a certain volume:
Rate of effusion = 1 / time taken to effuse
Let's denote the rate of effusion of carbon dioxide as R1, and the rate of effusion of the unknown gas as R2.
R1 = 1 / 200.
R2 = 1 / 280.
Now, we can set up the equation using Graham's law:
R1 / R2 = √(M2 / M1)
Where M1 is the molar mass of carbon dioxide and M2 is the molar mass of the unknown gas.
Rearranging the equation:
M2 = (R2 / R1)² * M1
Now, let's substitute the known values into the equation:
M2 = (1 / 280.)² * M1
The molar mass of carbon dioxide (M1) is approximately 44.01 g/mol.
M2 = (1 / 280.)² * 44.01
Calculating the value:
M2 = 0.0284 * 44.01
M2 ≈ 1.25 g/mol
Therefore, the molar mass of the unknown gas is approximately 1.25 g/mol.