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5 April, 14:15

An effusion container is filled with 7 L of an unknown gas. It takes 109 s for the gas to effuse into a vacuum. From the same container under the same conditions--same temperature and initial pressure, it takes 380 s for 7.00 L of O2 gas to effuse. Calculate the molar mass (in grams/mol) of the unknown gas

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  1. 5 April, 14:44
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    2.53g/mol

    Explanation:

    First, let us calculate the rate of effusion of the two gas. This is illustrated below:

    For the unknown gas:

    Volume = 7L

    Time = 109secs

    R1 = volume / time

    R1 = 7/109

    R1 = 0.064L/s

    For O2:

    Volume = 7L

    Time = 380secs

    R2 = volume / time

    R2 = 7/380

    R2 = 0.018L/s

    Now, we can calculate the molar mass of the unknown gas using the Graham's law of diffusion equation:

    R1/R2 = √ (M2/M1)

    R1 (for the unknown gas) = 0.064L/s

    M1 (for the unknown gas) = ?

    R2 (for O2) = 0.018L/s

    M2 (for O2) = 16x2 = 32g/mol

    R1/R2 = √ (M2/M1)

    0.064/0.018 = √ (32/M1)

    Take the square of both side

    (0.064/0.018) ^2 = 32/M1

    Cross multiply to express in linear form

    (0.064/0.018) ^2 x M1 = 32

    Divide both side by (0.064/0.018) ^2

    M1 = 32 / (0.064/0.018) ^2

    M1 = 2.53g/mol

    The molar mass of the unknown gas is 2.53g/mol
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