This problem involves Graham's Law of Diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Since the rate is inversely proportion

Here's the solution to the diffusion problem.

This problem involves Graham's Law of Diffusion, which states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass. Since the rate is inversely proportional to the time taken for diffusion, we can write: $\frac{t_{gas 2}}{t_{gas 1}} = \sqrt{\frac{M_{gas 1}}{M_{gas 2}}}$ where $\text{t}$ is the time taken for diffusion and $\text{M}$ is the molar mass.

Step 1: Identify the given values. • Time for gas X ($\text{t}_X$) = 25.3 s • Time for Oxygen ($\text{t}_{O_2}$) = 20 s • Relative atomic mass of Oxygen (O) = 16

Step 2: Calculate the molar mass of oxygen gas ($\text{O}_2$). Oxygen exists as a diatomic molecule, $\text{O}_2$. $M_{O_2} = 2 \times Relative atomic mass of O = 2 \times 16 = 32 g/mol$

Step 3: Apply Graham's Law of Diffusion. Let gas 1 be Oxygen ($\text{O}_2$) and gas 2 be the unknown gas X. $\frac{t_X}{t_{O_2}} = \sqrt{\frac{M_X}{M_{O_2}}}$ Substitute the known values into the equation: $\frac{25.3 s}{20 s} = \sqrt{\frac{M_X}{32 g/mol}}$

Step 4: Solve for the molecular mass of gas X ($\text{M}_X$). First, calculate the ratio of the times: $1.265 = \sqrt{\frac{M_X}{32}}$ Square both sides of the equation to remove the square root: $(1.265)^2 = \frac{M_X}{32}$ $1.600225 = \frac{M_X}{32}$ Now, multiply by 32 to find $\text{M}_X$: $M_X = 1.600225 \times 32$ $M_X = 51.2072 g/mol$ Rounding to three significant figures, the molecular mass of gas X is 51.2 g/mol.

The molecular mass of gas X is $\boxed{51.2 g/mol}$.

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