posted by Sara .
An aqueous solutions is 18.7% by mass Na3PO4. What is the boiling point of this solution? The Kb for water is .512 C/m. Consider if the solute is an electrolyte or a nonelectrolyte.
18.7% means 18.7 g/100 g SOLUTION. Converted to 1000g solution that is 187g/1000 g solution. How much of that is H2O? 1000-187 = about 813g.
mols Na3PO4 = 187/molar mass = about 1.1 mols. So m = mols/kg soln or
m = 1.1/0.8 = about 1.4 or so. You need to go through these calculations and do them more accurately.
Then delta T = i*Kb*m
i = 4
Kb you have
m = your value for m
To find the new boiling point added delta T to 100 C.