Under identical conditions, separate samples of O2 and an unknown gas were allowed to effuse through identical membranes simultaneously. After a certain amount of time, it was found that 6.51 mL of O2 had passed through the membrane, but only 3.05 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas

145.78g/mol

Explanation:

Let the time taken for the two gas to effuse be x

For O2:

Volume = 6.51mL

Time = x

Rate (R1) = volume/time

R1 = 6.51/x

For the unknown gas:

Volume = 3.05mL

Time = x

Rate (R2) = Volume/time

R2 = 3.05/x

Now, we can obtain the molar mass of the unknown gas as follows:

R1 = 6.51/x

M1 (O2) = 16 x 2 = 32g/mol

R2 = 3.05/x

M2(unknown gas)

R1/R2 = √(M2/M1)

(6.51/x)/(3.05/x) = √(M2/32)

6.51/3.05 = √(M2/32)

Take the square of both side

(6.51/3.05)^2 = M2/32

Cross multiply to express in linear form:

M2 = (6.51/3.05)^2 x 32

M2 = 145.78

Therefore, the molar mass of the unknown gas is 145.78g/mol

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