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Calculate the mass of oxygen (in mg) dissolved in a 5.00 L bucket of water exposed to a pressure of 1.13 atm of air. Assume the mole fraction of oxygen in air to be 0.21 and the Henry's law constant for oxygen in water at this temperature to be 1.3 x 10-3 M/atm.

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4 answers

  1. This answer worked better for me

    P = P°*X

    => C = P°*X*k = 1.13atm * 0.21 * 1.3*10^-3M/atm = 3.085*10^-4M

    3.085*10^-4M * 5L = 1.54*10^-3mol = 1.54mmol

    1.54*MM(O2) = 1.54mmol * 32mg/mmol = 49.4mg

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  2. ^
    / \
    <(0o0)>
    { | | }
    / ( ) \
    | |
    ^. ^

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  3. 0O0

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  4. massO2=Hconstant*pressure*volume*molmass

    = 1.3E-3*1.13*5L*32 grams=235mg\\check it.

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