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 be the partial pressure of the gas ?

Explanation

Molar mass of ethane `(C_(2)H_(6))`
`= 2xx12+6xx1`
= 30 g `mol^(-1)`
`therefore` Number of molar present in `6.56xx10^(-3)` g of ethane `= (6.56xx10^(-2))/(1)`
`= 2.187xx10^(-4)`
Let the number of moles of the solvent be 55.55 assuming solvent is water.
According to Henry.s low,
`p = K_(H)X`
1 bar `= KH(2.187xx10^(-4)//55.455)`
[Assuming dilution condition i.e., moles of solvent `gt gt` moles of solute]
`K_(H)=1` bar / `(0.039xx10^(-4))`
If mass of ethane = 0.05 g then moles of ethane
= 0.05/30 mol
= 0.00166 mol
So `p=K_(H)X`
`p=1//0.039xx10^(-4)xx(0.00166//55.55)`
= 7.66 bar