IntroductionWe now understand colligative properties.
To use this knowledge, we need to be able to predict these colligative properties.
Freezing Point Depression
Boiling Point Elevation
IntroductionWe also need to use a different kind of concentration determination.
Instead of molarity, we will use
molality, m
mole fraction, X
MolalityMolarity - 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
Molalitymsolution = 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
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 x mH2O
nKI = 0.030 mol(0.060)(0.5000) mol
mKI = nKI x MKI = (0.030)(166.0) g = 5.0 g
Example 2
Mole FractionMole 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 CalculationsThe 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 CalculationsThe magnitudes of freezing point depression (∆Tf) and boiling point elevation (∆Tb) are
directly proportional to the molal concentration of all ions in 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
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
SummaryMolality - 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