A sample of Br2(g) takes 38.0 min to effuse through a membrane. How long would it take the same number of moles of Ar(g) to effuse through the same membrane?

You worked the Xe/CO; why is this one a problem? I would assume some mass Br2, calculate mols and go from there.

The time taken for a gas to effuse through a membrane is inversely proportional to the square root of its molar mass. Therefore, we can use Graham's law of effusion to find the time it would take for the same number of moles of Ar(g) to effuse through the same membrane.

The ratio of the effusion rates of two gases is given by:

Rate1 / Rate2 = √(Molar mass2 / Molar mass1)

Let's denote the time taken for Ar(g) to effuse through the membrane as t. We know that the molar mass of bromine (Br2) is 2 * 79.90 g/mol = 159.80 g/mol, and the molar mass of argon (Ar) is 39.95 g/mol.

Using Graham's law, we can set up the following equation:

t / 38.0 min = √(159.80 g/mol / 39.95 g/mol)

Simplifying the equation, we get:

t = 38.0 min * √(39.95 g/mol / 159.80 g/mol)

Calculating the expression:

t = 38.0 min * √(0.2498)

t = 38.0 min * 0.4998

t ≈ 19.0 min

Therefore, it would take approximately 19.0 minutes for the same number of moles of Ar(g) to effuse through the same membrane.

To determine the time it would take for the same number of moles of Ar(g) to effuse through the same membrane, 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 formula for Graham's Law of Effusion is:
Rate1 / Rate2 = √(Molar mass2 / Molar mass1)

In this case, we have the rate of effusion of Br2(g) and we want to find the rate of effusion of Ar(g). Let's assign the subscripts 1 and 2 to Br2(g) and Ar(g) respectively.

Rate1 (Br2(g)) = 1
Rate2 (Ar(g)) = ?

Molar mass1 (Br2(g)) = 159.808 g/mol (from periodic table)
Molar mass2 (Ar(g)) = 39.948 g/mol (from periodic table)

Using Graham's Law of Effusion, we can set up the equation as follows:

1 / Rate2 (Ar) = √(39.948 g/mol / 159.808 g/mol)

To solve for Rate2, we need to isolate it in the equation.

Taking the reciprocal of both sides of the equation:

Rate2 (Ar) = 1 / √(39.948 g/mol / 159.808 g/mol)

Simplifying the equation:

Rate2 (Ar) = √(159.808 g/mol / 39.948 g/mol)

Rate2 (Ar) = √4

Rate2 (Ar) = 2

Therefore, the rate of effusion for Ar(g) is 2 times that of Br2(g).

Since it took 38.0 min for Br2(g) to effuse, it would take 38.0 min / 2 = 19.0 min for the same number of moles of Ar(g) to effuse through the same membrane.