3/22/12
Math
I. The Nature of Solutions
Solutions
•
•
•
•
•
•
•
Mass Percent
•
Percent (Mass/volume)=
Mass of solute (g)
x100%
Solution volume (mL)
Percent Mass/volume (m/v %)
Percent by Volume (v/v %)
Molarity (M)
Molality (m)
Mole Fraction
Dilution
Stoichiometry
Percent by Volume
•
Percent (volume/volume)=
volume of solute
Solution volume
x100%
Be sure that both volume units are the same.
Molarity (M)
•
Molality (m)
•
Molarity=
moles of solute
Liters of Solution
Molality=
moles of solute
kg of solvent
•
•
•
No need to multiply by 100 to get a %
Convert grams to moles if needed
1 L = 1000mL, so divide mL by 1000 to get to Liters
•
•
•
No need to multiply by 100 to get a %
Multiply kilograms by 1000 to get grams
Divide grams by 1000 to get kilograms
•
You know when moles are mentioned,
stoichiometry is not far behind
•
We’ll need this when we work with colligative
properties.
1
3/22/12
χ
Mole Fraction ( )
moles of substance
Total moles
It is possible to convert between molarity and molality. The only information needed is
density.
Example: If the molarity of a solution is 0.30 M, calculate the molality of the solution
knowing that the density is 3.25 g/mL.
To do this problem we can assume one (1) liter of solution to make the numbers easier.
We need to get from the molarity units of mol/L to the molality units of mol/kg. We work
the problem as follows, remembering that there are 1000 mL in a Liter and 1000 grams
in a kg. This conversion will only be accurate at small molarities and molalities.
0.3 mol
1 L x 1 mL
3.25 g x 1 L
1000 mL x 1000 g
1 kg = 0.09 mols / kg
Conversion of Molarity to Molality
Example
What is the molality of NaCl in an aqueous solution which 4.20 M? The density of
the solution is 1.05 x 103 g/L.
Let's assume we have 1.00 L of solution. This means that the solution will contain
4.20 mol of NaCl. Convert this to mass.
58.44 g NaCl
4.20 mol NaCl
x
---------------=
245 g NaCl
1 mole NaCl
Subtract this from the mass of one liter of solution (obtained from the density) to
obtain the mass of water. mass water = 1050 g solution - 245 g NaCl = 805 g water
4.20 mol NaCl
molality
=
----------------------------=
5.03 m
0.805 kg solvent
Definition
• Physical properties of liquids affected by the
number of solute particles dissolved, not the
identity of the solute
• “depending on the collection”
Colligative Properties
16.4
Conductivity
• Electrolytes-charged
particles
• The more charged
particles, the more
conductive the solution
is.
2
3/22/12
Osmotic Pressure
• Osmosis- diffusion of solvent
through a semi-permeable
membrane
– typically water going from low
to high concentration
• Osmotic Pressure
– The extra pressure on the
membrane from the water
Vapor Pressure
• Vapor pressure is a force exerted by the
gaseous phase of a two phase—gas/liquid or
gas/solid system.
• The more solute particles, the less the vapor
pressure
• The more solute particles,
the more the pressure.
Vapor Pressure
Boiling Point Elevation
Freezing Point Depression
Boiling Point Elevation
• When is this used?
– _________
– _________
• Why do solute particles raise the boiling
point?
3
3/22/12
Freezing Point Depression
• When is this used?
– _________
– _________
The math
• Freezing Point Depression=ΔTf
ΔTf=iKfm
• Why do solute particles lower the freezing
point?
The math
i= number of aqueous solute particles
Kf=freezing point constant for solvents, water = 1.86
m= molality of solution
What is the freezing point depression of a
solution containing 478 g of sodium chloride
in 3202 g of water?
• Boiling Point Elevation=ΔTb
ΔTb=iKbm
i= number of aqueous solute particles
Kb=boiling point constant for solvents, water = 0.512
m= molality of solution
What is the freezing point depression of a
solution containing 478 g of sodium chloride
in 3202 g of water?
A solution of 4.25 grams of solute in 67.0 grams
of a solvent (Kf=7.07;Tf=86.5oC) is found to have
a freezing point of 82.1oC. What is the solute's
molecular weight?
4
3/22/12
A solution of 4.25 grams of solute in 67.0 grams
of a solvent (Kf=7.07;Tf=86.5oC) is found to have
a freezing point of 82.1oC. What is the solute's
molecular weight?
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°
3.5°C?
103.5°C = 100°C + (0.52°C / m) × (? m)
6.73
Assuming 1000g (1kg), the molality
becomes…..
6.73
mole
= ?m
kg
Why is CaCl2 used as rock salt?
mole
×1kg = 6.73mole
kg
250 g
= 6.73mole
? MM
MM = 37 g / mole
5