Question

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 8.24 mL of O2 had passed through the membrane, but only 3.86 mL of of the unknown gas had passed through. What is the molar mass of the unknown gas?

Answer 1

Answer:

The molar mass of unknown gas is 145.82 g/mol.

Explanation:

Volume of oxygen gas effused under time t = 8.24 mL

Effusion rate of oxygen gas = $R=\frac{8.24 mL}{t}$

Molar mass of oxygen gas = 32 g/mol

Volume of unknown gas effused under time t = 3.86 mL

Effusion rate of unknown gas = $R'=\frac{3.86 mL}{t}$

Molar mass of unknown gas = M

Graham's Law states that the rate of effusion or diffusion of gas is inversely proportional to the square root of the molar mass of the gas. The equation given by this law follows the equation:

$\text{Rate of diffusion}\propto \frac{1}{\sqrt{\text{Molar mass of the gas}}}$

$\frac{R}{R'}=\sqrt{\frac{M}{32 g/mol}}$

$\frac{\frac{8.24 mL}{t}}{\frac{3.86 mL}{t}}=\sqrt{\frac{M}{32 g/mol}}$

$M=\frac{32 g/mol\times 8.24 \times 8.24}{3.86\times 3.86}=145.82 g/mol$