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6 May, 23:52

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

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  1. 7 May, 00:50
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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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