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18 July, 05:31

A gas of unknown molecular mass was allowed to effuse through a small opening under constant-pressure conditions. It required 125 s for 1.0 L of the gas to effuse. Under identical experimental conditions it required 28 s for 1.0 L of O2 gas to effuse.

Part A

Calculate the molar mass of the unknown gas. (Remember that the faster the rate of effusion, the shorter the time required for effusion of 1.0 L; that is, rate and time are inversely proportional.) Express your answer using two significant figures.

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  1. 18 July, 05:59
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    648g/mol

    Explanation:

    Let the rate of effusion of the unknown gas be R1 and that of O2 be R2

    For the unknown gas:

    Volume = 1L

    Time = 125sec

    R1 = volume / time

    R1 = 1/125

    R1 = 0.008L/s

    For O2:

    Volume = 1L

    Time = 28sec

    R2 = volume / time

    R2 = 1/28

    R2 = 0.036L/s

    Now, we can easily find the molar mass of the unknown gas using the Graham's law equation as shown below:

    R1/R2 = √ (M2/M1)

    R1 = 0.008L/s

    R2 = 0.036L/s

    M1 (molar Mass of unknown gas) = ?

    M2 (molar Mass of O2) = 16x2 = 32g/mol

    R1/R2 = √ (M2/M1)

    0.008/0.036 = √ (32/M1)

    Take the square of both sides

    (0.008/0.036) ^2 = 32/M1

    Cross multiply to express in linear form

    (0.008/0.036) ^2 x M1 = 32

    Divide both side by (0.008/0.036) ^2

    M1 = 32 / (0.008/0.036) ^2

    M1 = 648g/mol

    Therefore, the molar mass of the gas is 648g/mol
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