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11 July, 16:20

701.28 grams of SO2 is held at 3606.96 torr and 603.2oC. What is the volume of the container

can some one answer this?

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  1. 11 July, 16:48
    0
    165.92L

    Explanation:

    First, let us calculate the number of mole SO2 in the container. This is illustrated below:

    Molar Mass of SO2 = 32 + (16x2) = 32 + 32 = 64g/mol

    Mass of SO2 = 701.28g

    Number of mole = Mass / Molar Mass

    Number of mole of SO2 = 701.28/64 = 10.96moles

    Now we can easily calculate the volume of the container by calculating the volume of SO2 as illustrated below:

    Data obtained from the question include:

    P = 3606.96 torr

    Recall: 760torr = 1atm

    Therefore, 3606.96 torr = 3606.96/760 = 4.746atm

    T = 603.2°C = 603.2 + 273 = 876.2K

    R = 0.082atm. L/Kmol

    n = 10.96moles

    V = ?

    Using the ideal gas equation: PV = nRT, the volume can be obtain as follow:

    PV = nRT

    V = nRT/P

    V = (10.96 x 0.082 x 876.2) / 4.746

    V = 165.92L

    Therefore, the volume of the container is 165.92L
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