Applying Henry's law: At `1.0 atm` pressure, `24 g` of acetylene `(C_(2)H_(2))` dissolves in `1 L` of acetone. If the partial pressure of acetylene is increased to `12 atm`, what is the solubility in acetone?
Strategy: Let `S_(1)` be the solubility of the gas at partical pressure `P_(1)`, and let `S_(2)` be the solubility at partial pressure `P_(2)`. Writing Henry's law for both pressure, we have
`S_(1)=KP_(1)`
`S_(2)=KP_(2)`
Dividing the second equation by the first, we get
`S_(2)/S_(1)=(KP_(2))/(KP_(1))`
or `S_(2)/S_(1)=P_(2)/P_(1)`
we can use this relation to find the solubility at one pressure given the solubility at another.

Explanation

At 1.0 atm partial pressure `(P_(1))` of acetylene, the solubility `(S_(1))` is `24 g C_(2)H_(2)` per litre of acetone. For a partial pressure of 12 atm `(P_(2))`,
`S_(2)/(24 g C_(2)H_(2)//L "acetone")=(12 atm)/(1.0 atm)`
or `S_(2)=(24 g C_(2)H_(2))/(L "acetone")xx12/1.0`
`=(2.9xx10^(2) g C_(2)H_(2))/(L "acetone")`
Thus, at 12 atm partial pressure of acetylene, `1 L` of acetone dissolves `2.9xx10^(2) g` of `C_(2)H_(2)`.