Calculations Involving Colligative PropertiesFreezing Point Depression and Boiling Point Elevation Calculations

Objectives

• When you complete this presentation, you will be able to• calculate the freezing point depression or boiling point

elevation of a solution when you know its molality• calculate the molality of a solution when you know its

freezing point depression or boiling point elevation

Introduction

• We now understand colligative properties.• To use this knowledge, we need to be able to predict

these colligative properties.• Freezing Point Depression• Boiling Point Elevation

Introduction

• We also need to use a different kind of concentration determination.• Instead of molarity, we will use• molality, m• mole fraction, X

Molality

• Molarity - we measure the number of mols of solute in the volume of the solution.

•

• Msolution = nsolute/Vsolution

•• Molality - we measure the number of mols of solute in

the mass of solvent.•

• msolution = nsolute/msolvent

Molality

• msolution = nsolute/msolvent

• The mass of the solvent is measured in kilograms, kg.• 1 mole of solute in

1,000 g of solvent givesa 1 m solution.

Molality

Example 1:Find the molality of 87.66 g of NaCl dissolved in 2.500 kg of H2O.mNaCl = 87.66 gmH2O = 2.500 kgMNaCl = 58.45 g/mol

nNaCl = mNaCl/MNaCl

m = nNaCl/mH2O

First, we write down the known values.

This value comes from the sum of the atomic masses of Na, 22.99 g/mol, and Cl, 35.45 g/mol.

To calculate molality, m, we need to know the number of mols of NaCl. We only know the mass of NaCl.

Therefore, we calculate the number of mols of NaCl.

We substitute in known values …

= (87.66 g)/(58.44 g/mol)

… and do the calculation.

= 1.500 mol

Now, we can calculate the molality, m of the solution.

We put in the known values …

= (1.500 mol)/(2.500 kg)

… and do the calculation.

= 0.600 mol/kg

Molality

Example Problems:1. Find the molality of 100.0 g of NaOH in 1.500 kg of H2O.

2. Find the molality of 32.00 g of KCl in 0.5000 kg of H2O.

3. Find the molality of 142.5 g of CuCl2 in 2.000 kg of H2O.

4. Find the molality of 180.0 g of H2O in 1.500 kg of ethyl alcohol.

1.667 m

0.8585 m

1.667 m

6.667 m

Molality

Example 2:How many grams of potassium iodide, KI, M = 166.0 g/mol, must be dissolved in 500.0 g of water to produce a 0.060 molal solution?msoln = 0.060 mol/kgmH2O = 0.500 kgMKI = 166.0 g/mol

msoln = nKI/mH2O

mKI = MKInKI

First, we write down the known values.To calculate mKI, the mass of KI, we need to know the number of mols of KI, nKI. We only know the molality of the solution, msoln.

… we substitute in known values …

nKI = (0.060 mol/kg)(0.500 kg)

… and do the calculation.

= 0.030 mol

Now, we can calculate mKI, the mass of KI needed to prepare the solution.

We put in the known values …

= (166.0 g/mol)(0.030 mol)

… and do the calculation.

= 5.0 g

We can rearrange the equation to solve for nKI …

nKI = msolnmH20

Molality

Example Problems:1. Find the mass of NaCl (58.5 g/mol) needed to prepare a

0.600 m solution with a mass of 2.50 kg of water.

2. Find the mass of NaOH (40.0 g/mol) needed to prepare a 1.20 m solutionwith a mass of 0.250 kg of water.

3. Find the mass of CrF2 (90.0 g/mol) needed to prepare a 1.50 m solution with a mass of 0.400 kg of water.

4. Find the mass of KNO3 (101 g/mol) needed to prepare a 1.01 m solution with a mass of 0.100 kg of water.

87.8 g NaCl

12.0 g NaOH

54.0 g CrF2

10.2 g KNO3

Mole Fraction

• Mole Fraction is the ratio of number of mols of the solute to the total number of mols of the solute plus the solvent.• We use the symbol Χ (the Greek letter chi) to represent

the mole fraction.

• Χsolute =nsolute

nsolute + nsolvent

Mole Fraction

Example 3Ethylene glycol (EG), C2H6O2, is added to automobile cooling systems to protect against cold weather. What is the mole fraction of each component in a solution containing 1.25 mols of EG and 4.00 mols of water?nEG = 1.25 molnH2O = 4.00 mol

ΧEG =

ΧH2O =

nEG

nEG + nH2O

nEGnEG + nH2O

1.25 mol(1.25 + 4.00) mol= = 0.238

4.00 mol(1.25 + 4.00) mol= = 0.762

Mole Fraction

Example Problems:1. Find the mole fraction of 1.00 mol of NaCl in 55.6 mols of

water.

2. Find the mole fraction of 5.00 mol of BaCl2 in 10.0 mols of water.

3. Find the mole fraction of 0.240 mol of C6H12O6 in 4.76 mols of ethanol.

4. Find the mole fraction of 0.0320 mol of KNO3 in 10.2 mols of water.

Χ = 0.177

Χ = 0.333

Χ = 0.0480

Χ = 0.00319

Colligative Calculations

• The magnitudes of freezing point depression (∆Tf) and boiling point elevation (∆Tb) are• directly proportional to the molal concentration of the

solute,• if the solute is molecular and not ionic.

Colligative Calculations

• The magnitudes of freezing point depression (∆Tf) and boiling point elevation (∆Tb) are• directly proportional to the molal concentration of all

solute ions in the solution,• if the solute is ionic.

Colligative Calculations

• ∆Tf = −Kf x m• where• ∆Tf is the freezing point depression of the solution.• Kf is the molal freezing point constant for the solvent.• m is the molal concentration of the solution.

Colligative Calculations

• ∆Tb = Kb x m• where• ∆Tb is the boiling point elevation of the solution.• Kb is the molal boiling point constant for the solvent.• m is the molal concentration of the solution.

Colligative Calculations

Example 4What is the freezing point depression, ∆Tf, of a benzene (C6H6, 78.1 g/mol, Bz) solution containing 400 g of Bz and 200 g of the compound acetone (C3H6O, 58.0 g/mol, Ac)? Kf for Bz is 5.12°C/m.mBz = 400 g = 0.400 kg mAc = 200 gMAc = 58.0 g/mol Kf = 5.12°C/m

nAc =

msoln =

mAc

MAc

nAcmBz

200 g58.0 g/mol= = 3.45 mol

3.45 mol0.400 kg= = 8.62 m

∆Tf = −Kf × m = (5.12°C/m)(8.62 m) = −44.1°C

This is a non-ionic compoundThe number of particles in solution is equal to the number of

mols of the compound.

Colligative Calculations

Example 5What is the freezing point depression, ∆Tf, of a sodium chloride (NaCl, 58.4 g/mol) solution containing 87.6 g of NaCl in 1.20 kg water? Kf for water is 1.86°C/m.mH2O = 1.20 kg mNaCl = 87.6 gMNaCl = 58.4 g/mol Kf = 1.86°C/m

nNaCl =

msoln =

mNaCl

MNaCl

nNaClmH2O

87.6 g58.4 g/mol= = 1.50 mol NaCl

3.00 mol1.20 kg= = 2.50 m

∆Tf = −Kf × m = (1.86°C/m)(2.50 m) = −4.65°C

This is an ionic compoundThe number of particles in solution is equal to the number of

ions of the compound.For each mol of NaCl in solution, there is 1 mol of Na+ and 1

mol of Cl− ions. We need to multiply n by 2.

= 3.00 mol

Colligative Calculations

Example Problems: Find the freezing point depression of:1. 58. 44 g of NaCl (M = 58.44 g/mol) in 1.00 kg of H2O. (Kf =

1.86°C/m)

2. 228 g of octane, C8H18 (M = 114 g/mol), in 0.500 kg of benzene. (Kf = 5.12°C/m)

3. 33.5 g of NaC2H3OH (M = 67.0 g/mol) in 1.25 kg of acetic acid. (Kf = 3.90°C/m)

4. 217 g of C5H12 (M = 72.2 g/mol) in 0.600 kg of cyclohexane. (Kf = 20.2°C/m)

∆Tf = −3.72°C

∆Tf = −20.5°C

∆Tf = −3.12°C

∆Tf = −101°C

Colligative CalculationsExample 6What is the boiling point, T, of a benzene (Bz) solution containing 0.600 kg of Bz and 200 g of the compound phenol (C6H5OH, 94.1 g/mol, Ph)? Kb for Bz is 2.53°C/m. Normal boiling point for Bz is Tb = 80.1°C.mBz = 0.600 kg mPh = 200 g Tb = 80.1°CMPh = 94.1 g/mol Kf = 5.12°C/m

nPh =

msoln =

mPh

MPh

nPhmBz

200 g94.1 g/mol

= = 2.13 mol

2.13mol0.600kg

= = 3.54m

∆Tb = Kb × m = (2.53°C/m)(3.54 m) = 8.96°C

T = Tb + ∆Tb = 80.1°C + 8.96°C = 89.1°C

Colligative CalculationsExample 7What is the boiling point, T, of a sodium chloride (NaCl, 58.4 g/mol) solution containing 87.6 g of NaCl in 0.800 kg of acetic acid (AcOH)? Kb for AcOH is 3.07°C/m. The boiling point of AcOH is Tb = 118.5°C.mH2O = 1.20 kg mNaCl = 87.6 g Tb = 118.5°CMNaCl = 58.4 g/mol Kb = 3.07°C/m

nNaCl =

msoln =

mNaCl

MNaCl

nNaClmAcOH

87.6 g58.4 g/mol

= = 1.50 mol NaCl

3.00 mol0.800 kg= = 3.75 m

∆Tb = Kb × m = (3.07°C/m)(3.75m) = 11.5°C

= 3.00 mol particles

T = Tb + ∆Tb = 118.5°C + 11.5°C = 130.0°C

Colligative Calculations

Example Problems: Find the boiling point elevation of:1. 58. 44 g of NaCl (M = 58.44 g/mol) in 1.00 kg of H2O. (Kb =

0.512°C/m)

2. 228 g of octane, C8H18 (M = 114 g/mol), in 0.500 kg of benzene. (Kb = 2.53°C/m)

3. 33.5 g of NaC2H3OH (M = 67.0 g/mol) in 1.25 kg of acetic acid. (Kb = 3.07°C/m)

4. 217 g of C5H12 (M = 72.2 g/mol) in 0.600 kg of cyclohexane. (Kb = 2.79°C/m)

∆Tb = 1.02°C

∆Tb = 10.1°C

∆Tb = 2.46°C

∆Tb = 14.0°C

Summary

• Molality - we measure the number of mols of solute in the mass of solvent.• msolution = nsolute/msolvent

• The mass of the solvent is measured in kilograms, kg.• Mole Fraction is the ratio of number of mols of the solute

to the total number of mols of the solute plus the solvent.• We use the symbol X to represent the mole fraction.• Xsolute = (nsolute)/(nsolute + nsolvent)

Summary

• ∆Tf = Kf x m• ∆Tf is the freezing point depression of the solution

• Kf is the molal freezing point constant for the solute• m is the molal concentration of the solution.

• ∆Tb = Kb x m

• ∆Tb is the boiling point elevation of the solution

• Kb is the molal boiling point constant for the solute• m is the molal concentration of the solution.