Transcript 16.4-Colligative calculations and Molality

Calculations & Colligative Properties
Chapter 16.4
Learning Objectives
• Know the difference between molality and molarity
• Be able to calculate molality
• Can use molality to calculate freezing point
depression or boiling point elevation
• Understand how ionic compounds affect colligative
properties
• Can calculate mole fractions
Molarity vs Molality
Molarity (M)
M =
moles of solute
liters of solution
Molality (m)
m =
moles of solute
mass of solvent (kg)
Molarity vs Molality
so 3.2 M means we have 3.2 moles in 1
liter of solution
so 3.2 m means we have 3.2 moles in 1 kg of
solvent
Why do we need to know molality?
Colligative properties …..
We use molality to make calculations
for boiling point elevation or freezing
point depression
Boiling Point Elevation
DTb = Kb
Change in
boiling point
x
m
x
Molality of
solution
Molal boiling point constant
(0.51 oC/m for water)
i
Van’t Hoff
factor
Freezing Point Depression
DTf = Kf
Change in
freezing point
x
m
x
Molality of
solution
Molal freezing point constant
(-1.86 oC/m for water)
i
Van’t Hoff
factor
Important note about these calculations!
The calculations change depending on whether
you have a nonelectrolyte or electrolyte solution
Why?
1 C12H22O11 (s)  1 C12H22O11 (aq) (i =1)
Sugar does not dissociate into ions!
1 Ba(NO3)2 (s)  1 Ba2+(aq) + 2 NO3-1(aq) (i =3)
Barium nitrate will lower the freezing point of its solvent
three times as much as sugar of the same molality
Sample Calculation: Freezing Point Depression:
Electrolyte
What is the freezing point depression of water in a solution
of 62.5 g of barium nitrate, Ba(NO3)2, in 800. g of water?
DTf = Kf •m•i
Let’s calculate m first
m = moles of Ba(NO3)2 / kg solvent
800. g = 0.800 kg
137.3(1)
14(2)
16(6)_____
261.3g/mol
m = 0.239 moles / 0.800 kg water
62.5 g
261.3 g/mol
m = 0.299 mol/kg
DTf = (-1.86 oC/m)(0.299 m)(3)  Van’t Hoff factor (i)
DTf = -1.70 oC
Mole Fraction
• Mole fractions have no units
nA
XA 
n A  nB
nB
XB 
n A  nB
XA + XB = 1
Mole Fraction Calculation
A solution has 1.43 moles of vanadium oxide in
3.45 moles of chloroform. What is the mole
fraction of each component in the solution?
nA
XA 
n A  nB
nB
XB 
n A  nB
XA = 1.43 / 1.43 + 3.45
XA = 0.293
XA = 3.45 / 1.43 + 3.45
XA = 0.707
0.293 + 0.707 = 1