An effusion container is filled with 7 L of an unknown gas. It takes 109 s for the gas to effuse into a vacuum. From the same container under the same conditions--same temperature and initial pressure, it takes 380 s for 7.00 L of O2 gas to effuse. Calculate the molar mass (in grams/mol) of the unknown gas

2.53g/mol

Explanation:

First, let us calculate the rate of effusion of the two gas. This is illustrated below:

For the unknown gas:

Volume = 7L

Time = 109secs

R1 = volume /time

R1 = 7/109

R1 = 0.064L/s

For O2:

Volume = 7L

Time = 380secs

R2 = volume /time

R2 = 7/380

R2 = 0.018L/s

Now, we can calculate the molar mass of the unknown gas using the Graham's law of diffusion equation:

R1/R2 = √(M2/M1)

R1 ( for the unknown gas) = 0.064L/s

M1( for the unknown gas) =?

R2 (for O2) = 0.018L/s

M2 (for O2) = 16x2 = 32g/mol

R1/R2 = √(M2/M1)

0.064/0.018 = √(32/M1)

Take the square of both side

(0.064/0.018)^2 = 32/M1

Cross multiply to express in linear form

(0.064/0.018)^2 x M1 = 32

Divide both side by (0.064/0.018)^2

M1 = 32/ (0.064/0.018)^2

M1 = 2.53g/mol

The molar mass of the unknown gas is 2.53g/mol

im not exactly sure but i think its b!