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

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