Question

At what partial pressure, nitrogen will have a solubility of 0.05 g `L^(-1)` in water at 293 K ? Given that `k_(H) for N_(2)` at 293 K is 76.48 k bar. Assume that the density of the solution is the same as that of the pure solvent.

Answer 1

Correct Answer - 246.26 bar
Step I. Calculation of `x_(N_(2))`
`"Mass of 1 L of solution"=1000mLxx1" g mL"^(-1)=1000 g`
`"Mass of solution (water)=(1000-0.05)~~1000 g`
`"No. of moles of water "=((1000g))/((18g mol^(-1)))=55.5 mol`
`"No. of moles of nitrogen"=((0.05g))/((28" g mol"^(-1)))~~1.79xx10^(-3)mol`
`X_(N_(2))=(n_(N_(2)))/(n_(N_(2))+n_(H_(2)O))=((1.79xx10^(-3)mol))/((1.79xx10^(-3)+55.5 mol))=3.22xx10^(-5)`
Step II. Calculation os partial pressure of `N_(2)`
`rho_(N_(2))=K_(H)xxX_(N_(2))=(76.48xx10^(3)"bar")xx(3.22xx10^(-5))=246.26"bar"`