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9 May, 09:07

If 0.278g of argon dissolves in 1.5 l of water at 62 bar, what quantity of argon will dissolve at 78 bar

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  1. 9 May, 10:04
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    When P1/P2 = C1/C2

    and C is the molarity which = moles/volume

    so, P1/P2 = [ (mass1/mw) / volume] / [ (mass2/mw) / volume]

    P1/P2 = (mass1/mw) / 1.5L / (mass2/mw) / 1.5L

    so, Mw and 1.5 L will cancel out:

    ∴P1/P2 = mass1 / mass2

    ∴ mass 2 = mass1 * (P2 / P1)

    = 0.278g * (78 bar / 62 bar)

    = 0.35 g

    ∴ the quantity of argon that will dissolve at 78 bar = 0.35 g
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