Colligative Properties: A Comprehensive Guide with Solved Problems, Slides of Law

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Colligative

Properties

Colligative Properties

Are physical properties that depend upon the quantity or concentration of the solution They do not depend upon the identity of the solute

Colligative properties -Vapor

Volatile liquids are easily evaporated and nonvolatile liquids do not evaporate easily. Examples of volatile liquids are acetone and gasoline Examples of nonvolatile liquids oils and glycerin

Vapor Pressure lowering

A volatile liquid is mixed with a non- volatile substance The nonvolatile liquid or particles will get in the way of the evaporation of the liquid This action lowers the vapor pressure since less gas is formed Raoult’s law is used to predict the vapor pressure of a system

Problem

Glycerin, C 3 H 8 O 3 , is a nonvolatile nonelectrolyte with a density of 1. g/mL. Calculate the vapor pressure of 25°C of a solution made by adding 50. mL glycerin to 500.0 mL of water. The vapor pressure of pure water at 25°C is 23.8 torr.

Problem

1 st^ determine the mole fraction

1.26 g/ml *50ml = 63 g glycerin

63g glycerin* (1mol/92.09g) = 0.684mol

glycerin

500.0 g H 2 O * (1mol/18.01 g) =

0.684 glycerin/ (0.684 +27.76) = 0.024 Xglycerin

2 nd^ plug in numbers and solve

23.8 torr *.024 = 0.57 torr

Problem

PA = Xglycol P°A 1.0 atm = 760 torr

760 torr = Xglycol * 1070 torr 760 torr/1070 torr = Xglycol Xglycol = 0.71 Xglycol

Boiling point elevation

By adding a solute to a liquid, the boiling temperature is increased

4.7 m solution of NaCl

How many ions formed? And how does that change the molality?

2

4.7 * 2

9.4 m is plugged into the formula

0.7 m of CaCl 2

How many ions formed? And how does that change the molality?

3

0.7m * 3 = 2.1 m

Freezing point depression

The lowering of the freezing point by adding solute

Tf = kf * m

Tf = change in freezing point

kf = molar freezing-point depression constant (solvent dependent)

m = molality of solution

• multiply the molality of an ionic solution by the number of ions formed in solution

Solution

Step 1: determine the molality of solution

Assume 100g, we have 25g of ethylene glycol

and 75 g of solvent (water)

Convert 25 g of ethylene glycol to mol

25g*(1 mol/ 62.07g)= 0.403mol ethylene glycol

The density of water is about 1g/mL with means

we have 75ml or 0.075L

0.403 mol / 0.075 L = 5.37m

Solution

Step 2: Plug into freezing point formula then

subtract from normal freezing point of water

5.37m * 1.86°C/m = 9.99°C

0°C-9.99°C = -9.99°C

Step 3: plug into boiling point formula then

add to normal boiling point of water

5.37m*0.52°C/m = 2.79°C

100°C + 2.79°C = 102.8°C