Calculations & Colligative PropertiesChapter 16.4

Learning Objectives

• Know the difference between molality and molarity• Be able to calculate molality• Can use molality to calculate freezing point

depression or boiling point elevation• Understand how ionic compounds affect colligative

properties• Can calculate mole fractions

Molarity vs Molality

M =moles of solute

liters of solution

Molarity (M)

Molality (m)

m =moles of solute

mass of solvent (kg)

Molarity vs Molality

so 3.2 M means we have 3.2 moles in 1 liter of solution

so 3.2 m means we have 3.2 moles in 1 kg of solvent

Why do we need to know molality?

Colligative properties …..

We use molality to make calculations for boiling point elevation or freezing point depression

Boiling Point Elevation

Tb = Kb x m x i

Change in boiling point

Molality of solution

Van’t Hoff factor

Molal boiling point constant (0.51 oC/m for water)

Freezing Point Depression

Tf = Kf x m x i

Change in freezing point

Molal freezing point constant (-1.86 oC/m for water)

Molality of solution

Van’t Hoff factor

Important note about these calculations!

The calculations change depending on whether you have a nonelectrolyte or electrolyte solution

Why?1 C12H22O11 (s) 1 C12H22O11 (aq) (i =1)Sugar does not dissociate into ions!

1 Ba(NO3)2 (s) 1 Ba2+(aq) + 2 NO3-1(aq) (i =3)

Barium nitrate will lower the freezing point of its solventthree times as much as sugar of the same molality

Sample Calculation: Freezing Point Depression: Electrolyte

What is the freezing point depression of water in a solution of 62.5 g of barium nitrate, Ba(NO3)2, in 800. g of water?

Tf = Kf •m•i

Let’s calculate m first

m = moles of Ba(NO3)2 / kg solvent

m = 0.239 moles / 0.800 kg water

800. g = 0.800 kg

62.5 g261.3 g/mol

137.3(1) 14(2) 16(6)_____261.3g/mol

m = 0.299 mol/kg

Tf = (-1.86 oC/m)(0.299 m)(3) Van’t Hoff factor (i)

Tf = -1.70 oC

Mole Fraction

• Mole fractions have no units

BA

BB

BA

AA nn

nX

nn

nX

XA + XB = 1

Mole Fraction Calculation

BA

BB

BA

AA nn

nX

nn

nX

A solution has 1.43 moles of vanadium oxide in 3.45 moles of chloroform. What is the mole fraction of each component in the solution?

XA = 1.43 / 1.43 + 3.45XA = 0.293

XA = 3.45 / 1.43 + 3.45XA = 0.707

0.293 + 0.707 = 1