An unknown gas diffuses 0.25 times as fast as helium. what is it’s molar mass? what steps are to be done?

The molar mass of the gas which diffuses 0.25 times as fast as helium is 64.The steps to be taken are in line with the calculation techniques of the mathematical representation of Graham's law of diffusion.

According to Graham's law of diffusion;

At constant temperature and pressure, the rate of diffusion of a gas is inversely proportional to the square root of its molar mass.

The molar mass of helium, m2= 4.

Therefore, r1/r2 = 1/4 since the gas diffuses 0.25 times as fast as helium.

Therefore;

(0.25)² = 4/m10.0625 = 4/m1m1 = 4/0.0625m1 = 64.

Therefore, the molar mass of the gas is 64.

The steps to be taken are in line with the calculation techniques of the mathematical representation of Graham's law of diffusion.

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64.0 g/mol.

Explanation:

Thomas Graham found that, at a constant  temperature and pressure the rates of effusion  of various gases are inversely proportional to  the square root of their masses.

∨ ∝ 1/√M.

where, ∨ is the rate of diffusion of the gas.

M is the molar mass of the gas.

∨₁/∨₂ = √(M₂/M₁)

∨₁ is the rate of effusion of the unknown gas.

∨₂ is the rate of effusion of He gas.

M₁ is the molar mass of the unknown gas.

M₂ is the molar mass of He gas (M₂ = 4.0 g/mol).

∨₁/∨₂ = 0.25.

∵ ∨₁/∨₂ = √(M₂/M₁)

∴ (0.25) =√(4.0 g/mol)/(M₁)

By squaring the both sides:

∴ (0.25)² = (4.0 g/mol)/(M₁)

∴ M₁ = (4.0 g/mol)/(0.25)² = 64.0 g/mol.

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