Colligative Properties (solutions)

• A. Definition

• Colligative PropertyColligative Property

– property that depends on the concentration of

solute particles, not their identity

4 Colligative Properties

• Vapor-pressure

• Freezing point

• Boiling point

• Osmotic pressure

Vapor pressure

• vapor pressure is the measure of the tendency of molecules to escape from a liquid

• volatile- high vapor pressure• non-volatile- low vapor pressure• vapor pressure lowering depends on the

concentration of a nonelectrolyte solute and is independent of solute identity, it is a colligative property

• because vapor pressure is lowered, this lowers the freezing point and raises the boiling point.

Freezing Point DepressionFreezing Point Depression

• (tf) - is the difference between the freezing points of the pure solvent, and it is directly proportional to the molal concentration of the solution.

– f.p. of a solution is lower than f.p. of the pure solvent

• the freezing point of a 1-molal solution of any nonelectrolyte solute in water is found by experiment to be 1.86°C lower than the freezing point of water.

• Molal freezing-point constant, K, is the freezing point depression of the solvent in a 1- molal solution of a nonvolatile, nonelectroyte solute

• Each substance has a different molal freezing point constant (p. 438)

• Applications– salting icy roads– making ice cream– antifreeze

• cars (-64°C to 136°C)

• ∆tf = Kf mi

∆tf = freezing point depression (°C)

Kf = °C/ m m = mol solute/kg of solvent

i = # of particles

Boiling Point ElevationBoiling Point Elevation

• (tb)– b.p. of a solution is higher than b.p. of the pure

solvent– When the vapor pressure is equal to the atmospheric

pressure boiling will occur.

– molal boiling point constant (kb ) is the boiling-point elevation of the solvent in a 1-molal solution of a nonvolitle, nonelectrolyte solute. (0.51 °C/m)

– Boiling-point elevation,∆tb , is the difference between the boiling points of the pure solvent and a nonelectrolyte solution of that solvent, and is directly proportional to the molal connection of the solution.

• ∆tb = Kb m i

∆tb = boiling-point elevation (°C)

Kb = °C/ m

m = mol solute/kg of solvent

i = # of particles

Osmotic pressure:

• Semipermeable membranes allow the movement of some particles while blocking the movement of others.

• Osmosis: the movement of solvent through a semipermeable membrane from the side of lower solute concentration to the side of higher solute concentration.

• Osmotic pressure is the external pressure that must be applied to stop osmosis.

• Electrolytes: remember dissociation of ionic compounds

• NaCl lowers the freezing point twice as much as sucrose C12H22O11– NaCl Na+ + Cl-

• CaCl2 lowers the freezing point three times as much as C12H22O11 due to the dissociation of the ionic compounds– CaCl2 → Ca2 + + 2Cl-

C. Calculations

t: change in temperature (°C)k: constant based on the solvent (°C·kg/mol) (different depending on freezing or boiling)m:molality (m)i: # of particles

t = k · m · i

• # of Particles# of Particles

– Nonelectrolytes (covalent)• remain intact when dissolved • 1 particle

– Electrolytes (ionic)• dissociate into ions when dissolved• 2 or more particles

When we dissolve most ionic substances in water they break up into their individual ions.

NaCl(s) Cl- (aq) + Na+ (aq)

Most molecules don’t break up into ions. For example sugar.C12 H24O12 (s) + H2O (aq) C12 H24O12 (aq)

But some molecules do break up into ions. Acids and bases are examples.

HCl(l) + H2O (aq) Cl- (aq) + H3O + (aq)

Dissociation Equations ( ionic Dissociation Equations ( ionic ))

NaCl(s)

AgNO3(s) MgCl2(s)

Na2SO4(s)

AlCl3(s)

Na+(aq) + Cl-(aq)

Ag+(aq) + NO3-(aq)

Mg2+(aq) + 2 Cl-(aq)

2 Na+(aq) + SO42-

(aq)Al3+(aq) + 3 Cl-(aq)

i = 2

i = 2

i = 3

i = 3

i = 4

C. Calculations

• At what temperature will a solution that is composed of 0.73 moles of glucose in 225 g of phenol boil?

m = 3.2mi = 1tb = kb · m · i

WORK:

m = 0.73mol ÷ 0.225kg

GIVEN:b.p. = ?tb = ?

kb = 3.60°C·kg/moltb = (3.60°C·kg/mol)(3.2m)(1)

tb = 12°C

b.p. = 181.8°C + 12°C

b.p. = 194°C

C. Calculations

• Find the freezing point of a saturated solution of NaCl containing 28 g NaCl in 100. mL water.

m = 4.8m

i = 2

tf = kf · m · i

WORK:

m = 0.48mol ÷ 0.100kg

GIVEN:

f.p. = ?

tf = ?

kf = -1.86°C·kg/mol

tf = (-1.86°C·kg/mol)(4.8m)(2)

tf = -18°C

f.p. = 0.00°C - 18°C

f.p. = -18°C

What is the boiling point of a solution made by dissolving 1.20 moles of NaCl in 750 g of

water?What is the boiling point?

Δtb = kb .m.iMolality=1.20moles/.750 kg

molality= 1.6 mΔtb = (1.86°C/m)(1.6m)(2)

Δtb = 6.0°C

Find molality!

Ex: What is the freezing point depression of water in a solution of 17.1 g of sucrose, C12 H22O11 , and 200. g of water? What is the actual freezing point of the solution?

• Δtf = kf .m.i• Molar mass of sucrose is 342.34 g/mol• Find moles:

• = 0.04995 mol• Molality= 0.04995 moles/.200 kg• molality= 0.2498 m• Δtb = (1.86°C/m)(0.2498m)(1)• Δtb = 0.465°C

g

molg

34.342

11.17

• Ex: A water solution containing an unknown quantity of a nonelectrolyte solute is found to have a freezing point of

-0.23°C. What is the molal concentration of the solution?

Δtf = kf .m.i

-0.23 °C = (-1.86 °C/m)(x)(1)

-0.23 °C = x

-1.86 °C

0.12 m = x

• Ex: What is the boiling point elevation of a solution made from 20.0 g of a nonelectrolyte solute and 400.0 g of water? The molar mass of the solute is 62.0g/mol.

• Find molality first: 20.0 g / 62.0 g/mol

• Molality = 0.3226 m

• Δtb = kb .m.i

= (0.51 °C/m)(0.3226 m)(1)

= 0.165 °C

• Ex: What is the expected change in the freezing point of water in a solution of 62.5 g of barium nitrate, Ba(NO3 )2 , in 1.00 kg of water?

• Molality= 62.5 g/261.36 g/mol =0.2487 m

• Δtf = kf .m.i

= (-1.86 °C/m)(0.2487 m)(3)

= 0.746 °C

Assignment:

• Do practice problems 1-4 on page 440

• Do practice problems 1-4 on page 441.

• Do practice problems 1-3 on page 445