'x' cm\(^3\) of CH\(_4\) gas diffused through a porous membrane in 25 min, Under the same conditions 'y' cm\(^3\) of another gas of molar mass 64 g mol\(^{-1}\) diffused in 20 min. The ratio of x and y is:
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When comparing rates of diffusion, it's important to consider both the square root of the molar mass ratio and the conditions under which the diffusion occurred (same temperature and pressure).
Step 1: Apply Graham's Law of Effusion. Graham's law states that the rate of effusion of a gas is inversely proportional to the square root of its molar mass. Step 2: Calculate the relative rates of diffusion. Given that: Molar mass of CH\(_4\) = 16 g/mol. Molar mass of the other gas = 64 g/mol. \[ \frac{{Rate of CH}_4}{{Rate of other gas}} = \frac{\sqrt{64}}{\sqrt{16}} = \frac{8}{4} = 2 \] Step 3: Relate the rates to the volumes diffused over time. Using the given times for diffusion: \[ \frac{x}{25} = 2 \times \frac{y}{20} \] Step 4: Solve for the ratio \(\frac{x}{y}\). \[ \frac{x}{y} = 2 \times \frac{25}{20} = 2.5 \] Since the ratio \(2.5\) is equivalent to \(5:2\), the correct answer is confirmed to be option (4).
