during an effusion experiment, oxygen gas passed through a tiny hole 2.5 times faster than the same number of moles in another gas under the same conditions. What is the molar mass of the unknown gas?
The easy way to do these is to assume some convenient number for the rate of effusion. The number 1 is an easy one to use.
unknown gas rate = 1L/min
oxygen gas rate = 2.5*1L/min = 2.5 L/min
(rate O2/rate unk) = swrt(MMunk/MMO2)
(2.5/1) = sqrt(MMunk/32)
Solve for molar mass (MM) unk.
To find the molar mass of the unknown gas, we can use Graham's law of effusion.
Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Let's denote the rate of effusion for oxygen gas as R1 and the rate of effusion for the unknown gas as R2.
According to the problem, oxygen gas flows 2.5 times faster than the unknown gas. So we can write the following relationship:
R1 = 2.5 * R2
The molar mass of oxygen (O2) is approximately 32 grams per mole. Now let's apply Graham's law:
√(Molar mass of oxygen / Molar mass of the unknown gas) = R2 / R1
√(32 / Molar mass of the unknown gas) = R2 / (2.5 * R2)
Simplifying the equation:
√(32 / Molar mass of the unknown gas) = 1 / 2.5
Cross-multiplying:
2.5 * √(32 / Molar mass of the unknown gas) = 1
Now let's isolate the molar mass of the unknown gas:
√(32 / Molar mass of the unknown gas) = 1 / 2.5
Squaring both sides:
32 / Molar mass of the unknown gas = (1/2.5)^2
32 / Molar mass of the unknown gas = 1 / 6.25
Cross-multiplying:
Molar mass of the unknown gas = 32 * 6.25
Molar mass of the unknown gas = 200 grams per mole
Therefore, the molar mass of the unknown gas is 200 grams per mole.
To determine the molar mass of the unknown gas, we can use Graham's Law of Effusion. Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
Let's assume the molar mass of oxygen gas (O2) is M1, and the molar mass of the unknown gas is M2.
According to the information given, the oxygen gas is passing through a tiny hole 2.5 times faster than the unknown gas, when they are at the same conditions. This can be expressed mathematically as:
(rate of effusion of oxygen gas) / (rate of effusion of unknown gas) = 2.5
Using Graham's Law, we can rewrite this equation as:
(sqrt(M2) / sqrt(M1)) = 2.5
Taking the square of both sides gives:
(M2 / M1) = (2.5)^2
Simplifying:
M2 = (2.5)^2 * M1
Since we know the molar mass of oxygen gas (M1 = 32 g/mol), we can substitute it into the equation:
M2 = (2.5)^2 * 32
Calculating this expression, we find:
M2 ≈ 200 g/mol
Therefore, the molar mass of the unknown gas is approximately 200 g/mol.