An atmospheric concentration of 380 ppm CO2 corresponds to a partial pressure of 0.00038 atm. What percentage of the CO2 originally dissolve

Question

An atmospheric concentration of 380 ppm CO2 corresponds to a partial pressure of 0.00038 atm. What percentage of the CO2 originally dissolved in the solution in Part A remains in solution after the soft drink reaches equilibrium with the ambient atmosphere

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Tài Đức 5 years 2021-07-31T05:43:37+00:00 1 Answers 120 views 0

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  1. binhan
    0
    2021-07-31T05:45:31+00:00
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    Answer:

    8.44 * 10^-3 %

    Explanation:

    The solubility of CO2 gas in water is 0.15g/100 ml at a CO2 pressure of 760 mmHg.

    Determine the percentage of the CO2 that dissolved originally in the solution that remains

    Amount remaining ( Sg = kpg )

    Sg = solubility of gas = 0.15 / 100 g/ml

    pg. = partial pressure of gas = 760 mmHg = 1 atm

    therefore ; K = 1.5 * 10^-3 gm^-1 atm^-1

    solubility of gas inside container

    Sg = 0.675 * 10^-2 g/ml

    solubility of gas during/after dissolution ( amount that remains )

    sg = 1.5 * 10^-3 * 0.00038

        =  5.7 * 10^-7  g/ml

    therefore the percentage remaining

    = (5.7 * 10^-7  / 0.675 * 10^-2 ) * 100

    = 8.44 * 10^-3 %

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