calculations involving colligative properties. objectives when you complete this presentation, you...

Calculations Involving Colligative Properties

Objectives• When you complete this presentation, you will

be able too calculate the molality of a solutiono calculate the mole fraction of the components of a

solutiono calculate the freezing point depression of a solution

using the molality or mole fraction of the components of the solution

o calculate the boiling point elevation of a solution using the molality or mole fraction of the components of the solution

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 gives• a 1 m solution.

MolalityExample 1

Find the molality of 87.66 g of sodium chloride dissolved in 2.500 kg of water.

mNaCl = 87.66 gmH2O = 2.500 kgMNaCl = 58.44 g/mol

m = nNaCl/mH2O

nNaCl = mNaCl/MNaCl = 87.66 g/58.44 g/molnNaCl = 1.500 mol

= 1.500 mol/2.500 kg

m = 0.600 mol/kg

First, we write down our known values.This value come from the sum of the atomic mass of Na (22.99 g/mol) and the atomic mass of Cl (35.45 g/mol)

Next, we write down the equation for molality.

We have a value for mH2O, but we need to calculate nNaCl. We know mNaCl and MNaCl, so we can calculate nNaCl.

Now, we can complete our calculation for molality.

Molality

How many grams of potassium iodide must be dissolved in 500.0 g of water to produce a 0.060 molal KI solution?

msolution = 0.060 mmH2O = 500.0 g = 0.5000 kgMKI = 166.0 g/mol

m = nKI/mH2O =nKI = m m∙ H2O = 0.030 mol(0.060)(0.5000) mol

mKI = nKI x MKI = (0.030)(166.0) g = 5.0 g

Example 2

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 X to represent the mole fraction.• Xsolute = nsolute

nsolute + nsolvent

Mole FractionExample 3

Ethylene glycol, 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 ethylene glycol (EG) and 4.00 mol of water?

nEG = 1.25 molnH2O = 4.00 mol

XEG =nEG

nEG + nH2O=

1.25 mol1.25 mol + 4.00 mol

= 0.238

XH2O =nH2O

nEG + nH2O=

4.00 mol1.25 mol + 4.00 mol

= 0.762

Colligative Calculations• The magnitudes of freezing point depression

(∆Tf) and boiling point elevation (∆Tb) areo directly proportional to the molal

concentration of the solute,o if the solute is molecular and not ionic.

Colligative Calculations• The magnitudes of freezing point depression

(∆Tf) and boiling point elevation (∆Tb) areo directly proportional to the molal

concentration of all ions in solution,o if the solute is ionic.

Colligative Calculations• ∆Tf = Kf x mo 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 mo 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

Example 4

What is the freezing point depression of a benzene (C6H6, BZ) solution containing 400 g of benzene and 200 g of the compound acetone (C3H6O, AC). Kf for benzene is 5.12°C/m.

mBZ = 400 g = 0.400 kgmAC = 200 gMAC = 58.0 g/molKf = 5.12°C/m

nAC =mAC

MAC=

200 g58.0 g/mol

3.45 mol

m =nAC

mBZ=

3.45 mol0.400 kg

= 8.62 m

Colligative Calculations

nAC =

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

NaCl produces 2 mols of particles for each mol of salt added;m = 2(1.50 m) = 3.00 m

Example 5

What is the boiling point of a 1.50 m NaCl solution?

m = 1.50 mKb = 0.512°C/mTb = 100.0°C

Colligative Calculations

∆Tb = Kb x m = (0.512°C/m)(3.00 m) = 1.54°C

T = Tb + ∆Tb = 100.0°C + 1.54°C = 101.5°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, and 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, and m is the molal concentration of the solution.

Summary