60 cm3 of SO2 gas diffused through a porous membrane in x minutes. Under similar conditions, 360 cm3 of another gas (molar mass 4 g mol−1) diffused in y minutes. The ratio of x and y is:
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Graham’s law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.
To solve this problem, we will use 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. Mathematically, Graham's Law is expressed as:
r2r1=M1M2,
where:
r1 and r2 are the rates of diffusion of gases 1 and 2, respectively,
M1 and M2 are the molar masses of gases 1 and 2, respectively.
Step 1: Identify the gases and their molar masses
Gas 1: SO2 (sulfur dioxide), with molar mass M1=64g mol−1.
Gas 2: Another gas, with molar mass M2=4g mol−1.
Step 2: Express the rates of diffusion
The rate of diffusion r is given by the volume of gas diffused per unit time: