Sample Problem

(a) Since the solution is 0.750 mol/L and has a density of 1.046 g/mL (or 1.046 kg/L) density, 1.0 L solution contains 0.750 mol (or 73.6 g) H2SO4 and has a mass of 1.046 kg:

Mass of H2O in 1 L solution = 1.046 kg – 0.0736 kg = 0.972 kg0.972 kg H2O = 972 g × 1 mol/18.0 g = 54.0 mol H2OFor H2SO4, X = 0.750 mol H2SO4/(0.750 mol H2SO4 + 54.0

mol H20) = 0.0137

A 0.750 M solution of H2SO4 in water has a density of 1.046 g/mL at 20ºC. What is the concentration in (a) mole fraction, (b) mass percent, (c) molality (MM = 98.086 g/mol) ?

(b) Mass % H2SO4 = 0.0736 kg H2SO4/1.046 kg total = 7.04%

(c) Since 0.972 kg water has 0.750 mol H2SO4 in it, 1 kg water would have 0.772 mol H2SO4 dissolved in it:

1.00 kg H2O × 0.750 mol H2SO4/0.972 kg H2O = 0.772 mol H2SO4

Thus, molality of sulfuric acid is 0.772 m

Sample ProblemA 0.750 M solution of H2SO4 in water has a density of 1.046 g/mL at 20ºC. What is the concentration in (a) mole fraction, (b) mass percent, (c) molality (MM = 98.086 g/mol) ?

Colligative properties are physical properties of solutions that arise because of the number of solute molecules dissolved in solution and not on the kind of solute particles dissolved in solution.

Pure Liquid Pure Liquid with solute

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

Vapor-Pressure Lowering Psolution = Xsolvent P˚ solvent

Boiling-Point Elevation ΔTb = Kb msolution

Freezing-Point Depression ΔTf = -Kf msolution

Osmotic Pressure (π) π = MRT

The Four-Colligative Properties

Vapor-Pressure Lowering: A non-volatile, non-ionic solute added to a pure solvent will lower the vapor pressure of the pure solvent according to Raoult’s Law.

pA = XA·pAº

Vapor pressure of solution

Mole Fraction of Solvent

Vapor Pressure of Pure Solvent

P°solvt vapor

solvent A + nonvolatile solute Bpure solvent A

Nonvolatile solute molecules reduce the number of volatile solvent molecules at the surface of the liquid reducing the # of solvent molecules escaping to the vapor phase.

Psolution = χsolvent P°solvent

At 100ºC what is the vapor pressure of a 50/50 % (v/v) solution of ethylene glycol, C2H6O2 , in water at 1 atm? (MM C2H6O2 = 62.06 g/mol d(C2H6O2) = 1.1155 g/mL, d(H2O) = 1.0000 g/mL, d(50/50) = 1.069 g/mL

Moles C2H6O2 = 500. mL X 1.1155 g C2H6O2

mLX

62.06 g

1 mol =

Moles H2O = 500. mL X 1.000 g H2O

mLX

18.02 g

1 mol =

8.99 mol

27.8 mol

Mole Fraction H2O = 27.8 mol/ (27.8 mol + 8.99 mol) = 0.7556

PH2O = χH2O P°H2O = .7556 X 760. torr = 574. torr

Psolution = χH2O P°H2O

The vapor-pressure lowering can be recast in terms of the mole fraction of solute.

PSoln = χsolvent P°solvent (1)

P Soln = (1 - χSolute ) P°solvent (2)

PSoln = P°solvent - P°solvent (χSolute) (3)

ΔP = (P°solvent - Psoln) = P°solvent (χSolute)

χSolute + χSolvent = 1 χSolvent = 1 - χSolute

substituting

expanding 2

SOLUTION:

PLAN: Find the mol fraction, χ, of glycerol in solution and multiply by the vapor pressure of water.

10.0 mL C3H8O31.26 g C3H8O3

mL C3H8O3

mol C3H8O3

92.09 g C3H8O3= 0.137 mol C3H8O3

500.0 mL H2O 0.988 g H2OmL H2O

mol H2O18.02 g H2O = 27.4 mol H2O

ΔP = 0.137 mol C3H8O3

0.137 mol C3H8O3 + 27.4 mol H2O92.5 torrx

x

x

= 0.461 torr

χ = 0.00498

x

x

ΔP = (P°solvent - Psolvent) = P°solvent (χSolute)

Calculate the vapor pressure lowering, ΔP, when 10.0 mL of glycerol (C3H8O3) is dissolved in 500. mL of water at 50.oC. At this temperature, the vapor pressure of pure water is 92.5 torr and its density is 0.988 g/mL. The density of glycerol is 1.26 g/mL.

2. Boiling Point Elevation

– The addition of a nonvolatile non-ionic solute dissolved in a pure solvent increases the boiling the point of a solution.

ΔTb = Tbp - T˚bp = Kb m

Boiling point elevation (+)

Molal boiling point elevation constant (˚C/m)

Molality (mol/kg)

Pure solvent boiling point Solution

boiling point

3. Freezing Point Depression

– The addition of a non-volatile non-ionic solute dissolved in a pure solvent decreases the freezing point of the solution.

ΔTf = Tf - T˚fp = -Kf m

Freezing point depression (-) Molal freezing

point depression constant (˚C/m)

SolutionMolality (mol/kg)

Pure solvent freezing point

Molal Boiling-Point Eleveation and Freezing Point Depression Constants of Common Liquids

You add 1.00 kg of ethylene glycol antifreeze (C2H6O2) to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the resulting solution?

SOLUTION:

You add 1.00 kg of ethylene glycol antifreeze (C2H6O2) to your car radiator, which contains 4450 g of water. What are the boiling and freezing points of the resulting solution?

PLAN: Find the number of mols of ethylene glycol and m of the solution; multiply by the boiling or freezing point constant; add or subtract, respectively, the changes from the boiling point and freezing point of water.

1.00 x 103 g C2H6O2mol C2H6O2

62.07 g C2H6O2= 16.1 mol C2H6O2

ΔTb = 0.512 oC/m

16.1 mol C2H6O2

4.450 kg H2O= 3.62 m C2H6O2

3.62 mx = 1.85 oC

BP = 101.85 oC

ΔTf = 1.86 oC/m 3.62 mx

FP = -6.73 oC

x

m C2H6O2 =

m C2H6O2 =

Osmosis is the selective passage of solvent molecules through a semipermeable membrane from a dilute solution to a more concentrated one.

dilute moreconcentrated

Osmosis is the diffusion of a solvent (frequently water) through a semi-permeable membrane, from a solution of low solute concentration (high water potential) to a solution with high solute concentration (low water potential), up a solute concentration gradient.

pure solvent

solution

An applied pressure is needed to prevent

volume increase;

this pressure

is the osmotic

pressure!

osmotic pressure

semipermeablemembrane

An pressure difference results from the net movement of solvent from a less-concentrated (hypotonic) to the more-concentrated (hypertonic) solution.

M is the molarity of the solution

R is the gas constant

T is the temperature (in Kelvin)

π = R TnV

π = M R T

For dilute solutions of electrolytes the osmotic pressure is given by:

Remember: The driving force is due to the difference in concentration of the solutions on each side of the membrane.

Δπ = ΔM R T

Calculate molarity of a aqueous solution at 300K which is found to have an osmostic pressure of 3.00 atm.

A solution prepared by dissolving 20.0 mg of insulin in water and diluting to a volume of 5.00 mL gives an osmotic pressure of 12.5 torr at 300K. What is the molecular mass of the insulin?

! = M R TM =

!

R T=

3.00 atm

0.0821L atm mol!1K!1 300 K= 0.122 M

M =12.5 torr ! 1 atm

760 torr0.0821 L atm mol!1 K!1 300 K

=6.68! 10!4 mol insulin

L=

moles = 6.68! 10!4 mol insulin

L! 0.005L = 3.33! 10!6mol

MolarMass = grams/mole = 0.020 g/3.33! 10"6 mol = 5988 g/mol

Suppose we have a 0.020 molar solution of table sugar (sucrose) and a semi-permeable membrane not permeable to sucrose. What osmotic pressure in mm Hg and to what height could this pressure support a column of water (density Hg =13.6 g/mL and water = 1g/mL?

π = M R T

π = 0.02 M x 0.0821 L atm/mol K x 298K

π = .49 atm x 760 torr/1 atm

π = 371 mm Hg x 13.6 = 5.0 meters!

Cell membranes are semi-permeable membranes that are susceptable to diffusion of water and a some ions.

isotonicsolution

hypotonicsolution

hypertonicsolution

Movement of solvent (water) from dilute to concentrated side!

NoOsmoticPressureConcentrationsAre the Same

Osmosis in an Onion Cell

Plasmolyzed cell (cell membrane has shrunk from the cell wall

Now this is wild: Anology to Osmosis

HighVapor Pressure

LowVapor

Pressure

Pure

In a closed container the solution with the highest vapor pressure will completely transfer to the container of lower vapor pressure until the mole fractions of solvent are equal in both! Cool.........

Dialysis and Osmosis

Memb

Pres

Water With High concentration ofdissolved solute

Pure Water

Colligative properties are properties that depend only on the number of solute particles in solution and not on the nature of the solute particles.

Vapor-Pressure Lowering P1 = X1 P˚ 1

Boiling-Point Elevation ΔTb = Kb m

Freezing-Point Depression ΔTf = -Kf m

Osmotic Pressure (π) π = MRT

The Four-Colligative Properties

SOLUTION:

Biochemists have discovered more than 400 mutant varieties of hemoglobin (Hb), the blood protein that carries oxygen throughout the body. A physician studying a form of Hb associated with a fatal disease first finds its molar mass (M). She dissolves 21.5 mg of the protein in water at 5.0 oC to make 1.50 mL of solution and measures an osmotic pressure of 3.61 torr. What is the molar mass of this Hb mutant?

PLAN: We know Π as well as R and T. Convert Π to atm and T to Kelvin. Use the Π equation to find the molarity M and then the amount and volume of the sample to calculate M.

M = Π

RT= 3.61 torr

atm

760 torr

(0.0821 L . atm/mol . K)(278.15 K)

= 2.08 x 10-4 M

2.08 x 10-4 molL 1.50 mL

103 mL

L= 3.12 x 10-7 mol

21.5 mgg

103 mg1

3.12 x 10-7 mol = 6.89 x 104 g/mol

x

x x

xx# mol = g/M

Calculation of Molar mass

We can calculate the Molar Mass of a substance using any four of the colligative properties solutions.

We use the freezing point depression and osmotic pressure normally as both have much larger changes (easier to measure).

If you measure the change in colligative properties for a 1.25 molal sucrose osmotic pressure and freezing point show the largest change and are easiest to measure (especially osmotic pressure).

0.1 m NaCl solution 0.2 m ions in solution

0.1 m nonelectrolytes solution

0.1 m CaCl2 solution 0.3 m ions in solution

0.1 m in solution

We modify the non-ionic colligative equations by multiplying by the van’t Hoff factor, i

Ionic solutes affect colligative properties differently than non-ionic solutes.