Transcript XI. Solutions and Colligative Properties

Chap13 - Solutions and
Colligative Properties
A solution is a solute (A) dissolved
into a solvent (B).
Solute A
Solvent B
A. Concentration

1. Mass Percent
=
mass component
x 100
total mass solution
=
grams A
x 100
grams A + grams B

2. Parts per million (ppm)
= mass component
total mass solution
ppt = x 103 (
ppb = x 109 (
x 106
Parts per thousand
Parts per billion
)
)

3. Mole fraction (X)
= mole component
total moles
XA =
mole A
mole A + mole B

4. Molarity (M)
= moles of solute
Liters of solution
M=
mole A
L solution
*Remember*
1L = 1000mL
Look at Problem #1
3.16 g MgBr2
1 mol
184.1 g MgBr2
= 0.0172 mol MgBr2
0.859 L
= 0.0200 M

5. Molality (m)
= moles solute
kg solvent
m = mole A
kg B
*Remember*
1g = 1mL for H2O
1000g = 1 kg
Look at Problem #4
4.8 g NaCL
1 mol
58.5 g NaCl
= 0.0821 mol NaCl
0.5 kg
= 0.164 m
Dilution of Solution

M1V1 = M2V2
Answer to Worksheet
3a – 50 mL diluted
3b – 1.28 mL diluted
Conversion between Units

For H2O only, Molarity = molality.

Why? Because the density of H2O is equal
to 1.00 g/mL.

Therefore, 1000mL = 1000g
1 L = 1 kg
Conversion between Units
For any other solution other than an
aqueous solution YOU MUST
USE THE DENSITY!!!!!
Use the density to convert mass to
volume.
Conversion between Units
m = mole/kg
mass solvent
+ add
mass solute
molar
mass
moles solute
mass solution
density
volume solution
M = mole/L
Conversion between Units
1000 g = 1 kg
mass solvent
m = mole/kg
+ add
mass solute
molar
mass
moles solute
mass solution
density
volume solution
1000 mL = 1 L
M = mole/L
Conversion between Units
1000 g = 1 kg
+ add
molar
mass
density
1000 mL = 1 L
Your homework/classwork is
worksheet –
concentration conversions
Convert mass % to …..

5% HC2H3O2 5 g x 1mol/60g =
0.0833 mol

95% H2O  95 g x 1mol/18g =
5.28 mol
Mole fraction

X = 0.0833mol / (0.0833 mol + 5.28 mol)
molality

m = 0.0833 mol / 0.095 kg
convert 95 g
Convert M to m

1.13
mol to
L solution
mol
kg solvent

1000 mL x 1.05 g/ml = 1050 g solution
density

1.13 mole KOH x 56.1 g/mol = 63.4 g solute
1050 g solution
– 63.4 g KOH
986.6 g solvent


m = 1.13 mol KOH = 1.15 mol
0.9866 kg
kg
What about mass percent?
g solute
Mass % 
*100
g solution
Mass % 
63.4g KOH
*100
1050g solution
63.4 g KOH
1050 g
solution
Dimensional Analysis


What is the molarity of concentrated HCl?
39.0% HCl by mass and 1.13 g/mL density
39.0 g HCl 1Mole 1.13g 1000m L
*
*
*
 12.1M
100g so ln 36.5 g 1m L
1L
Solution Calculations

What is the molarity
of a 1.11 ppm solution
of Zn2+ ions?

1.11mg Zn2
1g
1 moleZn
*
*
 1.70X 105 M
1 liter
1000mg 65.39g
Solid Calculations


Chemical analysis showed 1.23 mg Fe in a
15.67 g sample of soil.
What is the Fe concentration in ppm?
1.23m g Fe
1g
*
*106  78.5 ppm
15.67g sam ple 1000m g
Unusual concentration units

How many nano moles of Cu are present in 12.3
µL of 25 ppm CuSO4?
1L
25 mg 1 m mole 106 n mole
12.3L * 6 *
*
*
 1.93 n mole
10 L
L
159.65mg 1 m mole
B. Colligative Properties

1. Boiling Point Elevation
ΔTb = kb • m • i
for an aqueous solution
Tb = 100oC + (0.52 oC/m) •(m)
Normal B.P.
Kbfor water
* note that as molality increases
ΔTb increases as well
B. Colligative Properties

2. Freezing Point Depression
ΔTf = kf • m • i
for an aqueous solution
Tf = 0oC - (1.86 oC/m) •(m)
Normal F.P.
Kf for water
* note that as molality increases
ΔTf increases as well
Ex. Non-electrolyte (i=1)
Antifreeze is made at 25% C2H6O2 by mass.
What is the Tb and the Tf?
 Make your life easy and assume 1000g.
 Why? Because molality is based upon kg of
solvent
 Mass percent  250 g C2H6O2
 750 g H2O

molality
1m ole 4.03m ole
m  250g 

 5.37m
62.1g 0.750kg
C2H6O2
H2O
Boiling and Freezing Point
Tb  100C  (0.52C / m)  (5.37m)  102.8C
Tf  0C  (1.86C / m)  (5.37m)  10C
Ex. Molecular Weight of Unknown

What is the MM of a sample if 250grams of
the sample is placed into 1000grams of
water and the temperature rose by 3.5°C?
103.5C  100C  (0.52C / m)  (? m)
m ole
6.73
 ?m
kg

Assuming 1000g (1kg), the molality
becomes…..
m ole
6.73
1kg  6.73m ole
kg
250 g
 6.73mole
? MM
MM  37g / m ole
Ex. Electrolyte (i = ?)
IMPORTANT – the colligative properties of
freezing point and boiling point are
proportional to the number of particles
present in the solution.
 van Hoft factor, i
 NaCl  i = 2 moles
Ie. 1m = 2m
 CaCl2  i = 3 moles
Ie. 1m = 3m
 Al2(SO4)3  i = 5 moles Ie. 1m = 5m

increasing
colligative
effect