iii. colligative properties

Solutions

Colligative PropertyColligative Property

property that depends on the concentration of solute particles, not their identity

Freezing Point DepressionFreezing Point Depression (tf) f.p. of a solution is lower than f.p. of the pure

solvent

Boiling Point ElevationBoiling Point Elevation (tb) b.p. of a solution is higher than b.p. of the

pure solvent

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Freezing Point Depression

Solute particles weaken IMF in the solvent.

Boiling Point Elevation

Applicationssalting icy roadsmaking ice creamantifreeze

cars (-64°C to 136°C) fish & insects

t: change in temperature (°C)k: constant based on the solvent (°C·kg/mol)m:molality (m)n: # of particles

t = k · m · n

# of Particles# of Particles

Nonelectrolytes (covalent) remain intact when dissolved 1 particle

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

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

m = 3.2mn = 1tb = kb · m · n

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

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

m = 4.8m

n = 2

tf = kf · m · n

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

Percent Solutions If both solute & solvent are liquids

Percent by volume (% v/v) = volume of solute × 100% solution volume

If a solid is dissolved in a liquidPercent (mass/volume) (%(m/v)) = mass of solute (g)

× 100%solution volume (mL)

Must be the same unit: mL or L

Must be this unit

Example 1What is the percent by volume of ethanol

(C2H6O) or ethyl alcohol, in the final solution when 85 mL of ethanol is diluted to a volume of 250 mL with water?

Volume of solute = 85 mLVolume of solution = 250 mL

% (v/v) = 85 mL ethanol × 100% 250 mL solution

= 34% ethanol

% (v/v) = volume of solute × 100% volume of solution

Example 2How many grams of glucose (C6H12O6)

would you need to prepare 2.0 L of 2.8% glucose (m/v) solution?

Solution volume = 2.0 L → change to mLPercent by mass = 2.8%

Percent (mass/volume) (%(m/v) = mass of solute (g) × 100%solution volume (mL)

2.8% = mass of solute (g) × 100% 2,000 mL

2.0L

100% 100%

0.028 = X 2,000 mL

X = 56 g of solute

1L1000mL= 2,000 mL

1. What is the concentration, in percent (m/v), of a solution with 75g K2SO4 in 1500mL of solution?

2. A bottle of hydrogen peroxide antiseptic is labeled 3.0% (v/v). How many mL H2O2 are in a 400.0 mL bottle of this solution?

3. Calculate the grams of solute required to make 250 mL of 0.10% MgSO4 (m/v).

Percent Solution ProblemsYou do not have to write the problem. You MUST show your work.