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11 April, 16:45

2. 20.83 g. of a gas occupies 4.167 L at 79.97 kPa at 30.0 °C. What is its molecular weight?

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  1. 11 April, 18:05
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    f. wt. = 158 g/mol

    Explanation:

    Given:

    mass = 20.83 g

    Vol = 4.167 L

    Pressure = 79.97 KPa x 0300987 Atm/KPa = 0.78924 Atm

    T = 30°C + 273 = 303K

    R = 0.08206 L·Atm/mol·K

    PV = nRT = (mass/f. wt.) RT = > f. wt. = mass x R x T / P x V

    f. wt. = 20.83g x 0.08206 L·Atm/mol·K x 303K / 0.78924 Atm x 4.167L

    = 158 g / mol
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