Argon is found to effuse through the hole at a rate of 1.6 × 10–3 mol in 215 s. How much O2 would effuse through the hole in the same amount of time?
O2 would effuse through the hole at a rate of 3.2 × 10–3 mol in 215 s.
To determine the amount of O2 that would effuse through the hole in the same amount of time, 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.
The rate of effusion of Argon (Ar) is given as 1.6 × 10–3 mol in 215 s. We can use this information to calculate the rate of effusion of O2.
Step 1: Determine the molar mass of Argon (Ar) and Oxygen (O2).
The molar mass of Argon (Ar) is approximately 39.95 g/mol.
The molar mass of Oxygen (O2) is approximately 32.00 g/mol (16.00 g/mol for each oxygen atom).
Step 2: Calculate the ratio of the molar masses.
The ratio of the molar masses can be found by dividing the molar mass of Argon by the molar mass of Oxygen: 39.95 g/mol / 32.00 g/mol ≈ 1.248.
Step 3: Calculate the square root of the ratio of molar masses.
The square root of the ratio of molar masses is approximately 1.118.
Step 4: Use Graham's Law to determine the rate of effusion of O2.
Since the rate of effusion of Argon is given as 1.6 × 10–3 mol in 215 s, we can use the square root of the ratio of molar masses to calculate the rate of effusion of Oxygen:
Rate of effusion of O2 = Rate of effusion of Ar × Square root of (Molar mass of Ar / Molar mass of O2)
Rate of effusion of O2 = 1.6 × 10–3 mol/s × 1.118 ≈ 1.790 × 10–3 mol/s.
Therefore, approximately 1.790 × 10–3 mol of O2 would effuse through the hole in the same amount of time as the Argon.
To find the amount of O2 that would effuse through the hole in the same amount of time, we need to use Graham's Law of Effusion.
Graham's Law of Effusion states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.
First, we need to find the molar mass of Argon (Ar) and Oxygen (O2).
The molar mass of Ar is approximately 39.95 g/mol.
The molar mass of O2 is approximately 32.00 g/mol (16.00 g/mol per O atom).
Now, let's calculate the ratio of effusion rates using Graham's Law.
Effusion rate of Argon (Ar):
1.6 × 10–3 mol / 215 s = 7.44 × 10–6 mol/s
Effusion rate of Oxygen (O2):
Ratio = √(molar mass of Ar / molar mass of O2)
= √(39.95 g/mol / 32.00 g/mol)
= 1.256
Effusion rate of O2 = Ratio * Effusion rate of Ar
= 1.256 * 7.44 × 10–6 mol/s
≈ 9.34 × 10–6 mol/s
Therefore, approximately 9.34 × 10–6 mol of O2 would effuse through the hole in the same amount of time.