A gas X takes 28.3 s to diffuse through a porous partition. Oxygen gas takes 20 s to diffuse through the same partition. Calculate the molar mass of gas X, given that the molar mass of oxygen is 32 g/

Graham's Law of Diffusion states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. For a fixed distance or amount of gas, the time taken for diffusion is directly proportional to the square root of its molar mass.
The formula is:

$$\frac{t_1}{t_2} = \sqrt{\frac{M_1}{M_2}}$$

Where:
$t_1$ = time for gas 1
$t_2$ = time for gas 2
$M_1$ = molar mass of gas 1
$M_2$ = molar mass of gas 2

Given:
Time for gas X ($t_X$) = $28.3 \, \text{s}$
Time for oxygen ($t_{O_2}$) = $20 \, \text{s}$
Atomic mass of Oxygen (O) = 16
Atomic mass of Sulphur (S) = 32

Oxygen gas is diatomic, so its formula is $O_2$.

$$M_{O_2} = 2 \times \text{atomic mass of O} = 2 \times 16 = 32 \, \text{g/mol}$$

Let gas 1 be gas X and gas 2 be oxygen.

$$\frac{t_X}{t_{O_2}} = \sqrt{\frac{M_X}{M_{O_2}}}$$ $$\frac{28.3 \, \text{s}}{20 \, \text{s}} = \sqrt{\frac{M_X}{32 \, \text{g/mol}}}$$

First, divide the times:

$$1.415 = \sqrt{\frac{M_X}{32}}$$

Square both sides of the equation to eliminate the square root:

$$(1.415)^2 = \frac{M_X}{32}$$ $$2.002225 = \frac{M_X}{32}$$

Now, multiply by 32 to find $M_X$:

$$M_X = 2.002225 \times 32$$ $$M_X = 64.0712 \, \text{g/mol}$$

Rounding to three significant figures (consistent with $28.3 \, \text{s}$), the molecular mass of gas X is $64.1 \, \text{g/mol}$.