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

Calculate of mole fractiion of `H_(2)S`
By difinition, 0.195m of the gas means that 0.195 mole of it is dissolved in 1000 g of water.
No. of moles of water in 1000 g= `((1000g))/((18g mol^(-1)))=55.55 mol`
Mole fraction of `H_(2)S(x_(H_(2)S))=(n_(H_(2)S))/(n_(H_(2)S+n_(H_(2)O)))=((0.195 mol))/((0.195+55.55)mol)`
`=((0.195 mol))/((55.745 mol))=0.035`
Calculation of Henery's Law constant
According Henry's Law,
`x_(H_(2)S)=("Partial pressure of"H_(2)S)/(K_(H)"for"H_(2)S)at S.T.P.`
`K_(H)"for"H_(2)S=((0.987 "bar"))/((0.0035))=282"bar"`
`K_(H)=1.67xx10^(8)pa=((1 atm))/((101325 Pa))xx(1.67xx10^(8)Pa)=1.648xx10^(8)atm.`
`x_(CO_(2))=((2.5atm))/((1.648xx10^(3)atm))=1.52xx10^(8),n_(H_(2)O)=((500g))/((18g mol^(-1)))=27.78 mol`
`x_(CO_(2))=(x_(CO_(2)))/(x_(CO_(2))+x_(H_(2)O))=x_(CO_(2))/x_(H_(2)O)=x_(CO_(2))/((27.787 mol))`
`(x_(CO_(2))` has been neglected as the gas teh is very little soluble in water)
`x_(CO_(2))=x_(CO_(2))xx(27.78 mol)=(1.52xx10^(-3))xx(27.78 mol)=0.0422 mol`