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