Anonymous
Anonymous asked in Science & MathematicsChemistry · 2 months ago

Find solubility in oxygen?

At 298 K, the Henry's law constant for oxygen is 0.00130 M/atm. Air is 21.0% oxygen.

1. At 298 K, what is the solubility of oxygen in water exposed to air at 1.00 atm?

2. At 298 K, what is the solubility of oxygen in water exposed to air at 0.894 atm?

3. If atmospheric pressure suddenly changes from 1.00 atm to 0.894 atm at 298 K, how much oxygen will be released from 4.80 L of water in an unsealed container?

1 Answer

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  • Dr W
    Lv 7
    2 months ago
    Favorite Answer

    henry's law states that

    .. c(O2) = kH * P(O2)

    where

    .. c(O2) = concentration of O2 in the liquid phase

    .. P(O2) = partial pressure of O2 in the vapor phase

    and we know from Daltons law of partial pressure

    .. P(O2) = mole fraction O2 * Ptotal

    then.. 

    .. (1).. c(O2) = 0.00130 M/atm * 0.210 * 1.00atm = ___ M

    .. (2).. c(O2) = 0.00130 M/atm * 0.210 * 0.894atm = ___ M

    .. (3).. 4.80L * (___ - ___) = ___ mol O2

    .. . .. ..plug in the answers from (1) and (2) and calc

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