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?

Explanation

According to Henry.s law for a gas at two different concentrations and pressures,
`(m_(1))/(p_(1))=(m_(2))/(p_(2))" "…(1)`
According to the given data:
`m_(1)=6.56xx10^(-2)g , m_(2)=5.0xx10^(-2)g`
`p_(1)=1" bar ", p_(2)=`?
Substituting the values in equation (1),
`(6.56xx10^(-2))/(1) = (5.0xx10^(-2))/(p_(2))`
or `p_(2)=(5.0xx10^(-2))/(6.56xx10^(-2))=0.762` bar
Alternatively,
`""^(p)C_(2)H_(6)=K_(H)xx` mole fraction of gas in solution
`=K_(H) xx` conc. of gas in solution
or 1 bar `=K_(H) xx 6.56 xx 10^(-2)g, K_(H)=(1)/(6.56xx10^(-2))`
Again `""^(p)C_(2)H_(6)=K_(H) xx ` Conc. of gas in sol
`=(1)/(6.56xx10^(-2)) xx5.00 xx 10^(-2) = (5)/(6.56)=0.762` bar