Agas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. it required 125 s for 1.0 l of the gas to effuse. under identical experimental conditions it required 29 s for 1.0 l of o2 gas to effuse.

The molar mass of the other gas is 594.53 g/mol

Explanation:

The complete question is:  A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 125 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 29 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.)

Step 1: Data given

It required 125 s for 1.0 L of the gas to effuse.

Under identical experimental conditions it required 29 s for 1.0 L of O2 gas to effuse.

Step 2: Calculate the molar mass

Graham's law of effusion says:

r1/r2 = √(M2/M1)

⇒ with r1 = effusion rate 1 = L/125s

⇒ with r2 = effusion rate 2 = L/29s

⇒ with M1 = molar mass of the gas = x g/mol

⇒ with M2 = molar mass of oxygen = 32 g/mol

We should be careful if we plug in the effusion rate. Because the fractions could flip if we don't give the effusion rates as numeral values.

(L/125s)/(L/29s) =  29/125 = √(32/M1)

29/125 =√(32/X)

(29/125)² = 32/X

0.053824 =32/X

X = 32/0.053824

X = 594.53 g/mol

The molar mass of the other gas is 594.53 g/mol