Question

Assuming complete dissociation of the solute, how many grams of KNO3 must be added to 275 mL of water to produce a solution that freezes at −14.5 ∘C? The freezing point for pure water is 0.0 ∘C and Kf is equal to 1.86 ∘C/m ANSWER=109 g PART B NEEDED: If the 3.90 m solution from Part A boils at 103.45 ∘C, what is the actual value of the van't Hoff factor, i? The boiling point of pure water is 100.00 ∘C and Kb is equal to 0.512 ∘C/m.

Answer 1

KNO3 = K+ and NO3- therefore i = 2 ions

V = 275 ml

m = 275 g

Tf normal = 0 °C

Tf = -14.5°

Kf = 1.86 °C/m

molality = mol / kg solution

dTf = -Kf*molality

(0-14.5) = -1.86*mol/0.275

mol = 14.5*0.275/1.86 = 2.14381 mol of ions

Mol = 2.14381/2 = 2.14381/2 = 1.071905 mol of KNO3

MW KNO3 = 101.1032*1.071905 = 108.58397 g

b)

molality = 3.9

Tb = 103.45

Tb normla = 100

Kb = 0.512

then

dTb = i*Kb*molality

(103.45-100) = i*0.512*3.9

i = 3.45/(0.512*3.9) = 1.727

nearest answer is 2 ions;