iii. colligative properties
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III. Colligative Properties. Solutions. A. Definition. Colligative Property property that depends on the concentration of solute particles, not their identity. B. Types. Freezing Point Depression ( t f ) f.p. of a solution is lower than f.p. of the pure solvent - PowerPoint PPT PresentationTRANSCRIPT
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
View Flash animation.
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.