Question

Assuming 100% dissociation, calculate the freezing point and boiling point of 0.550 mol of K3PO4 in 1.00 kg of water.

A) freezing point: Answer in Celsius

B) boiling point: Answer in Celsius

Answer 1

The question is based on colligative proprties of solutions.

ΔT_f = T_solvent - T_solution = ikm

ΔT_f = depression in freezing point of the solution.,

T_solvent = freezing point of the solvent. = 0^oC

T_solution = freezing point of the solution

i = vant haff factor. Since dissociation of K₃PO₄ is complete i = 4. Because each formulla unit of K3PO4 gives 4 ions. 3 K+ ions and one PO₄^3-.

k = cryoscopic constant of water = 1.853 C·kg·mol^-1

m = molality = no. of moles of solute / mass of solvent in Kg

m = 0.55 /1 = 0.550m

T_solvent - T_solution = ikm

T_solution =  T_solvent + ikm

T_solution = 0 - 4 X 1.853 X 0.55 = - 4.076 ^oC

ELEVATION IN BOILING POINT (ΔT) = T_solution - T_solvent = ikm

T_solution = boiling point of solution

T_solvent = boiling point of solvent = 100^oC

k = ebullioscopic constant = 0.512°C·kg·mol^-1

m = molality = 0.550m

T_solution - T_solvent = ikm

T_solution = ikm + T_solvent

T_solution = 4 X 0.512 X 0.55 + 100oC = 101.126^oC