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5 November, 05:57

What is the pressure in mm of Hg, of a gas mixture that contains 1g of H2, and 8.0 g of Ar in a 3.0 L container at 27°C.

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  1. 5 November, 06:25
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    4362.4 mmHg

    Explanation:

    Step 1:

    Determination of the number of mole of H2 and the number of mole of Ar.

    Molar Mass of H2 = 2x1 = 2g/mol

    Mass of H2 = 1g

    Number of mole of H2 = ?

    Number of mole = Mass/Molar Mass

    Number of mole of H2 = 1/2 = 0.5 mole

    Molar Mass of Ar = 40g/mol

    Mass of Ar = 8g

    Number of mole of Ar = ?

    Number of mole = Mass/Molar Mass

    Number of mole of Ar = 8/40 = 0.2 mole

    Step 2:

    Data obtained from the question.

    Volume (V) of the mixture = 3L

    Temperature (T) = 27°C = 27°C + 273 = 300K

    Number of mole (n) of the mixture = 0.5 + 0.2 = 0.7 mole

    Gas constant (R) = 0.082atm. L/Kmol

    Pressure (P) of the mixture = ?

    Step 3:

    Determination of the pressure of the mixture.

    The pressure of the mixture can be obtained as follow:

    PV = nRT

    P x 3 = 0.7 x 0.082 x 300

    Divide both side by 3

    P = (0.7 x 0.082 x 300) / 3

    P = 5.74 atm

    Step 4:

    Conversion of the pressure obtained from atm to mmHg.

    1 atm = 760mmHg

    Therefore, 5.74 atm = 5.74 x 760 = 4362.4 mmHg
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