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2-3-2011

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Change in Freezing Change in Freezing PointPoint Common Applications of Common Applications of

Freezing Point Freezing Point DepressionDepression

Propylene glycol

Ethylene glycol – deadly to small animals

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Freezing point depression

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Freezing-Point Depression

Tf = T f – Tf

0

T f > Tf

0 Tf > 0

T f is the freezing point of the pure solvent

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Tf is the freezing point of the solution

Tf = Kf m

m is the molality of the solution

Kf is the molal freezing-point depression constant (0C/m)

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What is the freezing point of a solution containing 478 g of ethylene glycol (antifreeze) in 3202 g of water? The molar mass of ethylene glycol is 62.01 g.

Tf = Kf m

m =moles of solute

mass of solvent (kg)= 2.41 m=

3.202 kg solvent

478 g x 1 mol62.01 g

Kf (water) = 1.86 0C/m

Tf = Kf m = 1.86 0C/m x 2.41 m = 4.48 0C

Tf = T f – Tf0

Tf = T f – Tf0 = 0.00 0C – 4.48 0C = -4.48 0C

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Osmotic Pressure ()

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

A semipermeable membrane allows the passage of solvent molecules but blocks the passage of solute molecules.

Osmotic pressure () is the pressure required to stop osmosis.

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dilutemore

concentrated

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Higher P Lower P

Osmotic Pressure ()

= MRTM is the molarity of the solution

R is the gas constant

T is the temperature (in K)

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hypertonicsolution

A cell in an:

isotonicsolution

hypotonicsolution

Dilute Soln

Conc. SolnConc

Dilute Soln

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Colligative Properties of Nonelectrolyte Solutions

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 10

Boiling-Point Elevation Tb = Kb m

Freezing-Point Depression Tf = Kf m

Osmotic Pressure () = MRT

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Molecular mass from colligative properties

From MolalityFrom Molarity

1. Get molality

2. Get no. of moles

Mol = m * kg solvent

3. FW = (g solute/no. mol)

1. Get molarity

2. Get no. of moles

Mol = M * VL

3. FW = (g solute/no. mol)

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Molecular Weight of Unknown

What is the molecular mass of a sample if 250 grams of the sample is placed into 1000 grams of water and the temperature rose by 3.5°C?

Tb = Kb m3.5 oC = (0.52 oC/m)* mm = 6.73

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Find the number of moles in 1000 g

No. mol = 6.73 mol

No. mol = mass/FW

FW = 250/6.73 = 37 g/mol

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A 5.50 g of a newly synthesized compound was dissolved in 250 g of benzene )kf = 5.12 oC/m) and the freezing point depression was found to be 1.2 oC. Find the molecular mass of the compound.

 

 Tf  = kf m

1.2 oC 5.12 = oC/m * m

Molality = 0.199 m

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Mol = m * kg solvent

Mol solute = (0.199 mol/kg) * (0.250 kg)

mol solute = 0.0498 moles

mol solute = Wt solute/FW

FW = 5.50 g/0.0498 mol = 111 g/mol

It should be realized that a solvent with high kf is an advantage for such experimental calculations of molecular masses.

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Example

A solution of 0.85 g of a compound in 100 g of benzene has a freezing point of 5.16 oC. What are the molality and the molar mass of the solute? The normal freezing point of benzene is 5.5 oC, kf = 5.12 oC/m.

Tf  = kf m

m = (5.5 – 5.16)/5.12 = 0.066 mol/kg benzene

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Mol = m * kg solvent

Number of moles of compound = (0.066 mol/kg benzene)* 0.100 kg benzene = 6.6*10-3

FW = g compound/mol

FW = 0.85 g/(6.6*10-3 mol) = 128 g/mol

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Molecular mass from osmotic pressure

A solution is prepared by dissolving 2.47 g of an organic polymer in 202 mL of benzene. The solution has an osmotic pressure of 8.63 mmHg at 21 0C. Find the molar mass of the polymer.

Can calculate the molarity, where:

= MRT

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Get the number of moles of polymer where 1 L of polymer solution contains 4.53*10-4 mol

Mol polymer = (4.53*10-4 mol/1L soln) * 0.202 L soln = 9.15*10-5 mol

Molar mass = g polymer/no. molMolar mass = (2.47 g / 9.15*10-5 mol) = 26,992

g/mol

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Remember that osmotic pressure can be easily and accurately measured in mmHg. However, it is not as easy to measure freezing point depression when the concentration of solute is very small. Therefore, osmotic pressure is better suited for calculation of molar mass of solutes dissolved in solutions.

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Colligative Properties of Electrolyte Solutions

0.1 m NaCl solution 0.1 m Na+ ions & 0.1 m Cl- ions

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

0.1 m NaCl solution 0.2 m ions in solution

van’t Hoff factor (i) = actual number of particles in soln after dissociation

number of formula units initially dissolved in soln

nonelectrolytesNaCl

CaCl2

i should be

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3

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Boiling-Point Elevation Tb = i Kb m

Freezing-Point Depression Tf = i Kf m

Osmotic Pressure () = iMRT

Colligative Properties of Electrolyte Solutions

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At what temperature will a 5.4 molal At what temperature will a 5.4 molal solution of NaCl freeze? Assume i = solution of NaCl freeze? Assume i = 22

SolutionSolution

∆∆TTff = K = Kff • m • i • m • i

∆ ∆TTff = (1.86 = (1.86 ooC/m) • 5.4 m • 2C/m) • 5.4 m • 2

∆ ∆TTf f = 20.1= 20.1 ooCC

FP = 0 – 20.1 = -20.1FP = 0 – 20.1 = -20.1 ooCC

Freezing Point Freezing Point DepressionDepression

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What is the freezing point depression of a 0.15 m aqueous solution of Al2(SO4)3? (kf = 1.86 oC/m)?

Solution Al2(SO4)3 = 2 Al3+ + 3 SO4

2-

 Therefore, the overall effective molality of the

solution is:m = 0.15 * 5 = 0.75, neglecting interionic

attractions Tf  = {1.86 oC/m}*0.75 m = 1.4 oC

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Vapor pressure lowering

A solution of CaCl2 (FW = 111 g/mol) was prepared by dissolving 25.0 g of CaCl2 in exactly 500 g of H2O. What is the expected vapor pressure at 80 oC (Po

water at 80 oC = 355 torr)? What would the vapor pressure of the solution be if CaCl2 were not an electrolyte?

  Psoln = xsolvent Po

solvent

Therefore, calculate the number of moles of water and CaCl2

nwater = 500g/{18.0 g/mol} = 27.8

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nCaCl2 = 25.0 g/{111 g/mol} = 0.225nspecies = 3 * 0.225 = 0.675xwater = 27.8/{27.8 + 0.675} = 0.975 Psoln = 0.975 * 355 torr = 346 torr If CaCl2 were not an electrolyte, CaCl2 will not

dissociate and the number of moles of all species of solute will be just 0.225

Xwater = 27.8/{27.8 + 0.225} = 0.993 Psoln = 0.993 * 355 torr = 352 torr

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Boiling point elevation 

Find the boiling point elevation of a 0.100 m MgSO4  aqueous solution (kb = 0.51 oC/m, i = 1.3).

  

Tb = ikb * m Tb = 1.3 * 0.51 oC/m * 0.100m = 0.066 oC If MgSO4 was not an electrolyte Tb = 0.51 oC/m * 0.100 m = 0.051 oC.

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Selected Problems: 6, 7 – 13, 15 - 17, 20 - 24, 27, 29, 31, 35, 39 – 44, 49, 50 - 52, 54 – 56, 60 - 64, 70, 71, 73, 75 -77, 79, 80.