Calculate the boiling point (in oC) of an aqueous solution which is 0.126 m in potassium carbonate.

Question

K_b = 0.52 ^oC/m

Answer 1

Δ T = m K_b ι = (0.126 m) (0.52 ^oC/m) (3) = 0.1966 ^oC

Since water has a normal bp, the solution would have a measured bp of ~ (100 + 0.1966) = ~ 100.2 ^oC

NOTE: The value for the van't Hoff Factor, ι = 3 because there are 3 moles of 'particles' formed per mole of CaCl₂.

Answer 2

Colligative properties: Boling point elevation.

∆T = imK

∆T = change in boling point = ?

i = van't Hoff factor = 3 for potassium carbonate (K₂CO₃) which produces 3 particles upon dissociation.

m = molality = 0.126

K = boiling constant = 0.52º/m

∆T = (3)(0.126)(0.52) = 0.197º

This is the CHANGE in the boiling point, so the new boiling point of the solution will be....

100 + 0.197 = 100.197 but since you should limit your answer to 3 sig. figs. it would be 100. degrees C