Question

Please Help with Mole Fractions?

What are the mole fractions of H3PO4 and water in a solution of 16.4 g of H3PO4 in 132 g of water?

Answer 1

`chi_(H_3PO_4)-=n_(H_3PO_4)/(n_(H_3PO_4)+n_(H_2O))`

Explanation:

And thus...

`chi_(H_3PO_4)-=((16.4*g)/(98.08*g*mol^-1))/((16.4*g)/(98.08*g*mol^-1)+(132*g)/(18.01*g*mol^-1))=0.0223`

And for `chi_(H_2O)` we could go thru the rigmarole again....or note that in a binary solution....`chi_(H_3PO_4)+chi_(H_2O)=1` necessarily.

`chi_(H_2O)-=((132*g)/(18.01*g*mol^-1))/((16.4*g)/(98.08*g*mol^-1)+(132*g)/(18.01*g*mol^-1))=1-0.0113=0.978`

As is typical, the water solvent is present in VAST molar excess, and thus expresses the GREATEST mole fraction.