Calculate vapor pressure of an aqueous solution...HELP!!?

Question

Calculate the vapor pressure of an aqueous solution of 1.50 M Cu(NO3)2 at 25 degrees Celsius given that the vapor pressure of pure water is 23.8 Torr at 25 degrees Celsius. Assume the ideal van't Hoff factor for Cu(NO3)2 knowing that this salt is a strong electrolyte. Assume also that this solution has a density of 1.150 g/mL.

I really don't know where to go with this... I'll give 10 pts for best!!! thanks!!

Answer 1

1.50 mol/L x 1 mL/1.150 g x 1 L/1000 mL=0.00130 mol Cu(NO3)2/gram of solution

So assume we have 1 gram of solution then this means we have 0.00130 moles of Cu(NO3)2.

0.00130 mol Cu(NO3)2 x 187.5 g/mol Cu(NO3)2=0.245 g Cu(NO3)2

1.00 g-0.245 g=0.755 g H2O

0.245 g Cu(NO3)2=0.00130 mol Cu(NO3)2

0.755 g H2O x 1 mol/18 g H2O=0.0420 mol H2O

mole fraction H2O=0.0420/(0.0420+0.00130)=0.970

By Raoult's law

P=0.970 x 23.8 torr=23.1 torr

EDIT: Ah, I made a silly mistake. Vapor pressure lowering is a colligative property so we have to take into consideration the number of solute ions.

Ok, here's the correct solution.

0.00130 mol Cu(NO3)2=0.00390 mol ions

mole fraction H2O=0.0420/(0.0420+0.00390)=0.915

P=23.8 torr x 0.915=21.8 torr

That's the correct answer.