The partial pressure of ethane over a solution containing `6.56xx10^(-3)` g of ethane is 1 bar. If the solution contains `5.00xx10^(-2)`g of ethane, then what shall e the partial pressure of the gas?

Answer

`0.76 ba r`

Explanation

According to Henry's law `:`
`p_(gas)=K_(H)xxCHMi_(gas)`
First case
`CHMi_(C_(2)H_(6))=(n_(C_(2)H_(6)))/(n_(C_(2)H_(6))+n_(H_(2)O))=(W_(C_(2)H_(6))//Mw_(C_(2)H_(6)))/((W_(C_(2)H_(6)))/(Mw_(C_(2)H_(6)))+n_(H_(2)O))`
`(n_(C_(2)H_(6)))/(n_(H_(2)O))(` When the solution is dilute `)`
`((6.56xx10^(-2))/(30))/(n_(H_(2)O))=(0.002)/(n_(H_(2)O))`
`p_(gas)=K_(H)xxCHMi_(C_(2)H_(6))`
1 bar `= K_(H)xx(0.002)/(n_(H_(2)O)) ......(i) `
Second case
`CHMi_(C_(2)H_(6))=((5xx10^(-2))/(30))/(n_(H_(2)O))=(0.0016)/(n_(H_(2)O))`
`p_(gas)=K_(H)xxCHMi_(C_(2)H_(6))`
`p_(gas)=K_(H)xx(0.0016)/(n_(H_(2)O)) ................(ii)`
Dividing `Eq. (i),` we get
`(p_(gas))/(1 ba r)=(K_(H))/(K_(H))xx(0.0016xxn_(H_(2)O))/(n_(H_(2)O)xx0.002)`
`p_(gas)=(0.00166)/(0.00218)=0.76ba r`