Assuming 100% dissociation, calculate the freezing point and boiling point of 1.30 m k3po4(aq). constants may be found here.

Freezing point = -9.6°C
 Boiling point = 102.7°C 
 Freezing point depression is modeled by the equation:
 Î”TF = KF · b · i,
 where
 Î”TF = Change in freezing point
 KF = Cryoscopic constant of solvent (water is 1.853 K*kg/mol)
 b = molarity of solution
 i = van't Hoff factor (number of ions per molecule of solute. Ideal for K3PO4 is 4) 
 So plugging in the given values and calculating gives:

 Î”TF = KF · b · i,
 Î”TF = 1.853 K*kg/mol · 1.30 mol/kg · 4
 Î”TF = 9.6356 K 
 So the freezing point will be -9.6°C 
 For boiling point elevation the formula is
 Î”Tb = Kb · bB
 where
 Î”Tb =Boiling point elevation
 Kb = Ebullioscopic constant of solvent (water = 0.512 K*kg/mol)
 bB = Molarity of solution after taking into account the dissociation. 
 So substitute the known values and calculate:
 Î”Tb = Kb · bB
 Î”Tb = 0.512 K*kg/mol · 1.30 m · 4
 Î”Tb = 2.6624 K 
 So the new boiling point will be 100°C + 2.7°C = 102.7°C

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