takes 190. seconds for a sample of carbon dioxide to effuse through a porous plug and 295 seconds for the same volume of an unknown gas to effuse under the same conditions. What is the molar mass of the unknown gas? Enter your answer to 3 significant figures.

rate1/rate2 = sqrt(M2/M1)

I would assume some convenient volume such as 190 mL so
rate 1 = 190mL/190 sec = 1 mL/1 sec)
rate 2 = 190 mL/295 sec = ?
Substitute and solve for M2.
M1 = 44

If u take the porous membrane of porous of 190 sec and 295 sec I got stuck

To find the molar mass of the unknown gas, we can use Graham's law of effusion, which states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass.

First, let's define the variables:
r1 = rate of effusion of carbon dioxide
r2 = rate of effusion of the unknown gas
M1 = molar mass of carbon dioxide
M2 = molar mass of the unknown gas

According to Graham's law, we have the following relationship:

(r1/r2) = sqrt(M2/M1)

We are given that it takes 190 seconds for carbon dioxide to effuse and 295 seconds for the unknown gas to effuse.

So, r1 = 1/190 and r2 = 1/295.

Plugging these values into the equation, we have:

(1/190)/(1/295) = sqrt(M2/M1)

Simplifying,

295/190 = sqrt(M2/M1)

Squaring both sides of the equation,

(295/190)^2 = M2/M1

We know that the molar mass of carbon dioxide (M1) is approximately 44 g/mol.

Substituting this value,

(295/190)^2 = M2/44

Now, to solve for M2:

M2 = (295/190)^2 * 44

Calculating this expression:

M2 ≈ 108.68 g/mol

Therefore, the molar mass of the unknown gas is approximately 108.68 g/mol.