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`.

Answer

(b) `K_(H)=1.67xx10^(a)Pa`.
`P_co_(2))=2.5` atm `=2.5xx101325Pa`
Applying Henry's law, `P_(co_(1))=K_(H)xxH_(co_(2))`
`therefore X_(co_(2))=(P_(co_(2))/(K_(H))=(2.5xx101325)(1.67xx10^(8))=1.517xx10^(-3)`
Since `X_(co_(2)=(n_(co_(2))/(n_(H_(2)+n_(co_(2))=(n_(co_(2))/(n_(H_(2)O)=1.517xx10^(-3)`
For `1L` of soda water
Water present `-1L=1000g=(1000)/(18)=55.55 moles.
i.e. `n_(H_(2))o=55.55` moles
`therefore (n_(co_(2))/)(n_(H_(2)O)=1.517xx10^(-3)` or `(n_(co_(2))/(55.55)=1.517xx10%^(-3)`
`therefore n_(co_(2))=84.26xx10^(-3)` mole `=84.26xx10^(-3)xx44g=3.707g`.