CH4 effuses through a pinhole at a rate of 97.0 mL/s at STP. An unknown gas effuses through the same pinhole at 46.2 mL/sec. What is the molar mass of the unknown gas?

To solve this problem, 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.

The formula for Graham's law is: (Rate1/Rate2) = sqrt(M2/M1)

Where:

• Rate1 and Rate2 are the rates of effusion of the two gases.
• M1 and M2 are the molar masses of the two gases.

Given in the problem:

• Rate1 (rate of effusion of CH4) = 97.0 mL/s
• Rate2 (rate of effusion of the unknown gas) = 46.2 mL/s
• M1 (molar mass of CH4) = 16.04 g/mol

We are asked to find M2, the molar mass of the unknown gas.

Substituting the given values into the formula, we get:

(97.0 mL/s / 46.2 mL/s) = sqrt(M2 / 16.04 g/mol)

Solving the left side of the equation gives us approximately 2.10.

So, we have:

2.10 = sqrt(M2 / 16.04 g/mol)

To isolate M2, we square both sides of the equation:

(2.10)^2 = M2 / 16.04 g/mol

4.41 = M2 / 16.04 g/mol

Finally, solving for M2 gives us:

M2 = 4.41 * 16.04 g/mol = 70.7 g/mol

So, the molar mass of the unknown gas is approximately 70.7 g/mol.