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8 December, 21:22

Determine the molecular mass of a gas where 3.87 g occupies 0.896 l at standard conditions.

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  1. 8 December, 21:34
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    Let's assume that the gas has ideal gas behavior.

    Ideal gas law,

    PV = nRT (1)

    Where, P is the pressure of the gas (Pa), V is the volume of the gas (m³), n is the number of moles of gas (mol), R is the universal gas constant (8.314 J mol ⁻¹ K⁻ ¹) and T is temperature in Kelvin.

    n = m/M (2)

    Where, n is number of moles, m is mass and M is molar mass.

    From (1) and (2),

    PV = (m/M) RT

    By rearranging,

    M = (mRT) / PV (3)

    P = standard pressure = 1 atm = 101325 pa

    V = 0.896 L = 0.896 x 10⁻³ m³

    R = 8.314 J mol⁻¹ K⁻¹

    T = Standard temperature = 273 K

    m = 3.87 g = 3.87 x 10⁻³ kg

    M = ?

    By appying the formula,

    M = (3.87 x 10⁻³ kg x 8.314 J mol⁻¹ K⁻¹ x 273 K) / 101325 pa x 0.896 x 10⁻³m³

    M = 0.0967 kg

    M = 96.7 g.

    Hence, the molar mass of the gas is 96.7 g.
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