A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 125 s

Question

for 1.0 L
of the gas to effuse. Under identical experimental conditions it required 34 s
for 1.0 L
of O2
gas to effuse.
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.)

Answer 1

To find the molar mass of the unknown gas, we can use Graham's law of effusion:

Rate of effusion 1 / Rate of effusion 2 = √(Molar mass 2 / Molar mass 1)

Let the unknown gas be represented by gas 1, and O2 gas be represented by gas 2.

Rate of effusion 1 = 1/125
Rate of effusion 2 = 1/34

Plugging these values into the formula:

(1/125) / (1/34) = √(Molar mass O2 / Molar mass unknown gas)

34 / 125 = √(32 / Molar mass unknown gas)

Molar mass unknown gas = (32 * 125) / 34
Molar mass unknown gas = 118.53 g/mol

Therefore, the molar mass of the unknown gas is 118.53 g/mol.