Henry's law constant for `CO_(2)` in water is `1.67xx10^(8) Pa` at `298 K`. Calculate the quantity of `CO_(2)` in `500mL` of soda water when packed under `2.5atm CO_(2)` pressure at `298 K`.

Explanation

`K_(H) = 1.67 xx 10^(8)Pa`
p = 2.5 atm =`2.5 xx 101.325 xx 10^(3)Pa`
` =2.533 xx 10^(5)Pa`
Now `p = K_(H) x_(2)`
or , `x_(2) = (p)/(K_(H)) = (2.533 xx 10^(5)Pa)/(1.67 xx 10^(8)Pa) = 1.517 xx 10^(-3)`
Moles of water `= (500)/(18) = 27.78`
`x_(2) = (n_(2))/(n_(1) + n_(2)) = (n_(2))/(27.78 + n_(2)) = 1.514 xx 10^(3)`
`n_(2) = 0.0419 + 0.0015 n_(2)`
or `0.9985 n_(2) = 0.0419`
`n_(2) = 0.0420`
Amount of `CO_(2) ` dissolved `= 0.0420 xx 44 = 185 g`.