450 g/mol
Explanation:
To solve the problem the problem, it is necessary to use Graham's Law of effusion
M1/32g(mol) = 120s/32s = 450/mol
A heavier gas will require more time to effuse than a lighter gas
Part A
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.)
Explanation:
To solve the problem the problem, it is necessary to use Graham's Law of effusion
M1/32g(mol) = 120s/32s = 450/mol
A heavier gas will require more time to effuse than a lighter gas
It's 125 sec and 34 sec. I get something like 432 but you should confirm that.
(1/125)/(1/34) = sqrt(32/M)
M = 432
Let's denote the rate of effusion of the unknown gas as r1, and the rate of effusion of O2 gas as r2. We are given the following information:
- Time for the unknown gas to effuse: t1 = 125 s
- Time for the O2 gas to effuse: t2 = 34 s
According to Graham's law, the rate of effusion for each gas can be calculated as:
r1 = 1 / t1
r2 = 1 / t2
Let's calculate the rates of effusion for each gas:
r1 = 1 / 125 = 0.008
r2 = 1 / 34 = 0.029
Now we can set up a ratio using these rates of effusion:
r1 / r2 = sqrt(M2 / M1)
where M1 and M2 are the molar masses of the unknown gas and O2 gas, respectively.
Let's plug in the known values:
0.008 / 0.029 = sqrt(M2 / M1)
To simplify the calculation, let's square both sides of the equation:
(0.008 / 0.029)^2 = (sqrt(M2 / M1))^2
0.2181 = M2 / M1
Since we want to find the molar mass of the unknown gas (M1), let's rearrange the equation:
M1 = M2 / 0.2181
Now, we need to know the molar mass of O2 gas (M2). The molar mass of O2 is 32 g/mol.
Plugging in the values:
M1 = 32 g/mol / 0.2181
M1 ≈ 146.6 g/mol
Therefore, the molar mass of the unknown gas is approximately 146.6 g/mol.
From the given information, we know that the rate of effusion for the unknown gas is inversely proportional to the time required for effusion. In other words, the faster the rate of effusion, the shorter the time required for effusion of 1.0 L.
Using this relationship, we can set up a proportion:
(rate of effusion of unknown gas) / (rate of effusion of O2) = (time for O2 to effuse) / (time for unknown gas to effuse)
Plugging in the given values:
(x) / (1.0) = (34) / (125)
Cross-multiplying and solving for x (rate of effusion of unknown gas):
x = (1.0) * (34) / (125)
x = 0.272 L/s
Now we can use the ideal gas law to relate the rate of effusion to the molar mass of the gas:
rate of effusion = (Molar mass)^(1/2) / (Square root of molar mass of O2)
Solving for the molar mass of the unknown gas:
(0.272) = (Molar mass of unknown gas)^(1/2) / (Square root of 32 g/mol)
Cross-multiplying and solving for Molar mass of unknown gas:
Molar mass of unknown gas = (0.272)^2 * (32 g/mol)
Molar mass of unknown gas = 0.238 g/mol
Therefore, the molar mass of the unknown gas is approximately 0.238 g/mol.