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28 December, 19:07

Air is a mixture of gases that is about 78.0% N2 by volume. When air is at standard pressure and 25.0 ∘C, the N2 component will dissolve in water with a solubility of 4.88*10-4 M. What is the value of Henry's law constant for N2 under these conditions?

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  1. 28 December, 19:33
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    The value of Henry's law constant for N2 = 6.26 * 10^-4 M / atm

    Explanation:

    Step 1: Data given

    Temperature = 25.0 °C

    The N2 component will dissolve in water with a solubility of 4.88*10^-4 M

    air is at standard pressure = 1.00 atm

    Step 2: Calculate Henry's law constant for N2

    C=k*Pgas

    ⇒ with C = the solubility of a gas at a fixed temperature in a particular solvent

    ⇒ with k = Henry's law constant

    ⇒ with Pgas = the partial pressure of the gas

    k = C/Pgas

    Since 78.0 % of the gas is N2

    P (N2) = 0.78 atm

    k = C/P (N2) = (4.88*10^-4 M) / (0.78 atm)

    k = 6.26 * 10^-4 M / atm

    The value of Henry's law constant for N2 = 6.26 * 10^-4 M / atm
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