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 28 s for 1.0 L of O2 gas to effuse.

Part A

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.) Express your answer using two significant figures.

Question

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Eduard22sly Eduard22sly · Answer 1 · 2020-03-21T20:39:14+01:00

Answer:

648g/mol

Explanation:

Let the rate of effusion of the unknown gas be R1 and that of O2 be R2

For the unknown gas:

Volume = 1L

Time = 125sec

R1 = volume / time

R1 = 1/125

R1 = 0.008L/s

For O2:

Volume = 1L

Time = 28sec

R2 = volume / time

R2 = 1/28

R2 = 0.036L/s

Now, we can easily find the molar mass of the unknown gas using the Graham's law equation as shown below:

R1/R2 = √(M2/M1)

R1 = 0.008L/s

R2 = 0.036L/s

M1 (molar Mass of unknown gas) =?

M2 (molar Mass of O2) = 16x2 = 32g/mol

R1/R2 = √(M2/M1)

0.008/0.036 = √(32/M1)

Take the square of both sides

(0.008/0.036)^2 = 32/M1

Cross multiply to express in linear form

(0.008/0.036)^2 x M1 = 32

Divide both side by (0.008/0.036)^2

M1 = 32/ (0.008/0.036)^2

M1 = 648g/mol

Therefore, the molar mass of the gas is 648g/mol