Help with boiling point elevation.?

Question

I know that delta Tb = Kb m is supposed to be used, or at least I think it should be, but I don't know how to use it.
Anyways the question is....
Predict the boiling point of a 0.200m solution of Calcium phosphate in water.

Answer 1

Based on what we've studied, the correct form of the equation for the colligative property which is boiling point elevation is: deltaTb=imKb where i is the van't hoff factor, m= molality and Kb= boiling point elevation constant for water (0.5121 oC/m). The van't hoff factor in this case would =5 (since molar mass of Ca3(PO4)2 should be 5 times that number because it originally dissociated to produce one Ca2+ ion and one (PO4) -3 ion per formula unit of Ca3(PO4)2; the sum of the masses of the five ions should yield the mass of one formula unit of NaCl.)

therefore:

delta Tb = (5)(0.200m)(0.5121 oC/m).
delta Tb= 0.262 degree celsius

thus, the new boiling point of water would be 100 degree C +0.262 degree C = 100.262 degree C

Answer 2

No problem, you would work it like this.

Tb = Kb * m

Kb of water = .512 Celsius

m = molality, not molarity so molality = grams of solute/kg of solvent

so a .2m solution of water and Calcium Phosphate = .2 moles per liter. So assume you have .2 moles of it and 1 liter of water. (1 liter of water = 1 kg of water, nice how metric works like that)

Find the gram formula mass of Ca3(PO4)2 which happens to be 310g/mole. Using dimensional analysis, you have 62g of Calcium Phospate in 1 kg of water.

Tb = .512 * (62/1)

but we are not done yet....

Calcium Phosphate is ionic so you need to multiply that number by number of ions.

Tb = .512 * (62/1) * (5 ions) = 158.72 degrees Celsius elevation.

Answer 3

Oh well !

Just use the unitary method. If 1m causes an elevation of x

then 0.2 m will cause an elevation of 0.2x

and anyway, how can you know a formula and not know how to use it ???