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12 October, 00:41

Suppose that 0.00150 moles of CO2 (44.0 g/mol) effuse out of a pinhole in 1.00 hour. How many moles of N2 (28.0 g/mol) would effuse out of the same pinhole in 1.00 hour? Assume both gases are at the same temperature and pressure.

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  1. 12 October, 01:24
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    0.0019mol

    Explanation:

    To solve this problem, first we calculate for the rates of the effusion:

    For CO2,

    Number of mole = 0.00150 mol

    Time = 1h

    R1 = mol / time

    R1 = 0.0015/1

    R1 = 0.0015mol/h

    Molar Mass of CO2 (M1) = 44.0 g/mol

    For N2,

    R2 = ?

    Molar Mass of N2 (M2) = 28.0 g/mol

    Using Graham's law, we can obtain Rate (R2) of N2 as follows:

    R1/R2 = √ (M2/M1)

    0.0015/R2 = √ (28/44)

    0.0015/R2 = 0.7977

    R2 = 0.0015 / 0.7977

    R2 = 0.0019mol/h

    But rate = mole / time

    Time = 1h

    Rate od N2 = 0.0019mol/h

    0.0019 = mole / 1

    Mole = 0.0019mol

    Therefore, the number of mole of N2 is 0.0019mol
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