Using Graham's Law of Effusion, calculate the molar mass of gas A, given that it takes 146 s to effuse, while CO2 takes 115 s to effuse under the same conditions.

Graham's Law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. For two gases, A and B, effusing under identical conditions:

$$\frac{\text{Rate}_A}{\text{Rate}_B} = \sqrt{\frac{M_B}{M_A}}$$

Since the volume effused is the same for both gases, the rate is inversely proportional to the time taken for effusion ($\text{Rate} = \frac{\text{Volume}}{\text{Time}}$). Therefore, we can write:

$$\frac{t_B}{t_A} = \sqrt{\frac{M_B}{M_A}}$$

Where $t_A$ and $t_B$ are the times for effusion, and $M_A$ and $M_B$ are the molar masses.

The atomic mass of Carbon (C) is approximately $12.01 \text{ g/mol}$.
The atomic mass of Oxygen (O) is approximately $16.00 \text{ g/mol}$.

$$M_{\text{CO}_2} = 12.01 + (2 \times 16.00) = 12.01 + 32.00 = 44.01 \text{ g/mol}$$

Given:
$t_A = 146 \text{ s}$
$t_{\text{CO}_2} = 115 \text{ s}$
$M_{\text{CO}_2} = 44.01 \text{ g/mol}$

$$\frac{115 \text{ s}}{146 \text{ s}} = \sqrt{\frac{44.01 \text{ g/mol}}{M_A}}$$

Square both sides of the equation:

$$\left(\frac{115}{146}\right)^2 = \frac{44.01 \text{ g/mol}}{M_A}$$ $$(0.78767)^2 = \frac{44.01 \text{ g/mol}}{M_A}$$ $$0.62043 = \frac{44.01 \text{ g/mol}}{M_A}$$

Now, solve for $M_A$:

$$M_A = \frac{44.01 \text{ g/mol}}{0.62043}$$ $$M_A \approx 70.93 \text{ g/mol}$$

The molecular mass of gas A is $\text{70.9 g/mol}$.