`0.24g` of a gas dissolves in 1 L of water at `1.5` atm pressure. Calculate the amount of dissolved gas when the pressure is raised to `6.0` atm at constant temperature.

Explanation

`P _("solute") = K_(H)x _("solute in solution")`
At pressure `1.5` atm, `P_(1) = K_(H) x _(1)" "…(1)`
At pressure `6.0` atm,` P _(2) = K _(H) x _(2)" "…(2)`
Dividing equation (1) by (2)
We get `(p _(1))/( p _(2)) = (x _(1))/( x _(2))`
`(1.5)/(6.0) = (0.24)/(x _(2))`
Therefore `x _(2) = (0.24 xx 6. 0)/(1.5) =0.96 g //L`