Assuming 100% dissociation, calculate the freezing point and boiling point of 2.60 m k3po4(aq). constants may be found here.

 to find the freezing point, you use the following formula

fp solution= fp solvent - Δfp solution

fp solvent= 0 C (since the solvent was not stated, you can assume is water which freezing point is 0 degree celsius)

but first, we need to find the depression in freezing point (Δfp solution)

Δfp solution= Kf x i x m, where Kf is constant for the solvent, i is the number of particles produced by the solute and m is the molality of the solution.

Kf= 1.86 for water
i= 4 , becausethe solute breaks into 4 ions, K3PO4---> 3K+ + PO4-3
m= 2.60

Δfp solution= (1.86) (4) (2.60)= 19.3 C

fp solution= 0 - 19.3C= -19.3C


for boiling point, we use a similar formula and way to solve it. 

Δbp solution= bp solvent + Δbp solution

bp solvent= 100 C for water

to find the elevation boiling point (Δbp soln) is equal to:

Δbp solution= Kb x i x m

Kb= 0.512
i= 4
m= 2.60

Δbp solution= (0.512) (4) (2.60)= 5.32 C

bp solution= 100C + 5.32C= 105.32 C

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