A gas A effuses through a porous partition in 115 s, while CO2 effuses through the same partition in 146 s. Calculate the molar mass of gas A.

We are given the effusion times for gas A and CO$_2$ for the same volume. Graham's Law of Effusion states that the ratio of the rates of effusion of two gases is inversely proportional to the square root of the ratio of their molar masses. Since rate is inversely proportional to time for a fixed volume, we can write:

$$\frac{t_{\text{CO}_2}}{t_A} = \sqrt{\frac{M_{\text{CO}_2}}{M_A}}$$

Given:
$t_A = 146 \text{ s}$
$t_{\text{CO}_2} = 115 \text{ s}$

The molar mass of carbon (C) is approximately $12.01 \text{ g/mol}$.
The molar mass of oxygen (O) is approximately $16.00 \text{ g/mol}$.

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

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

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

The molecular mass of A is approximately 70.93 g/mol.

$$70.93 \text{ g/mol}$$