Question

Colligative properties are those that depend on the number of solute particles. Because electrolytes dissociate into ions, the concentration of particles in the solution is greater than the formula-unit concentration of the solution. For example, if 1 mol of Na2SO4 totally dissociates, 3 mol of ions are produced (2 mol of Na+ ions and 1 mol of SO42−ions). Thus, a colligative property such as osmotic pressure will be three times greater for a 1 MNa2SO4 solution than for a 1 M nonelectrolyte solution.

However, complete dissociation of electrolytes does not always occur. The extent of dissociation is expressed by the van't Hoff factor, i:

i = moles of particles in solutionmoles of solute dissolved

The equations for colligative properties can be written to include i. For example,

ΔTf = Kf⋅m⋅i

ΔTb = Kb⋅m⋅i

Part A

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.

Express your answer to three significant figures and include the appropriate units.

Answer 1

Volume of water = 275 mL

Freezing point of the solution = - 14.5 ⁰C

Freezing point of the pure solvent = 0⁰C , kf of water = 1.86 ⁰ C / m

We use following formula

Delta Tf= i m kf

Here delta Tf is freezing point depression = Freezing point of the solvent – freezing point of the solution

. i is Vant’hoff factor here for KNO3 = 2 since KNO3 dissociates into two ions.

Kf is freezing point constant of pure solvent and m is molality.

Lets plug given values.

14.5⁰C = 2 x m x 1.86 ⁰C/m

m = 3.898 mol / kg

We have to find moles of solute ( KNO3)

Molality = moles of solute / mass of solvent in kg.

Mass of solvent in kg = Mass of water = ( 275 mL x 1.g/ mL x 1kg / 1000 g ) = 0.275 kg

Lets plug this value in molality.

3.898 mol = moles of KNO3 / 0.275 kg

n KNO3 = 1.07195

mass of KNO3 = moles x molar mass

= 1.07195 mol x 101.1032 g /mol

= 108.38 g

So the mass of KNO3 = 108.32 g