Verified answer

This is a colligative property so it is only affected by how much solute is in the solution. Take note right away that NaClO will dissociate to Na^+ and ClO^-. So for every mole of NaClO you put in the solution, 2 moles particles will be found in the solution: the i factor is 2.

The vapor pressure decreases with the addition of solute and it equals the mole fraction of water times the original vapor pressure

So start by finding the number of moles of solute that will be in solution:

45.24g NaClO (mol NaClO/74.44g NaClO) ( 2 moles of particles/1 mol of NaClO)=1.215 mol of particles.

Now find the number of moles of water:

191.2g(mol/18g)=10.62 mol of water

Mole Fraction of Water:

10.62 mol of water/ (10.62 mol + 1.215 mol total) = .8973

Vapor Pressure= (.8973)(760)

Vapor Pressure= 682.0mmHg

mole NaOCl = 45.24 g NaOCl74.44 g NaOCl/mole = 0.6077 mole NaOCl

moles ions = 0.6077 mole x 2 ions/mole = 1.215 moles

mole H2O = 191.2 g H2O / 18.01 g H2O/mole = 10.62 mole

mole fraction H2O = 10.62 / (10.62 + 1.22) = 0.897

vp = 760 mm x 0.897 = 681 mm Hg