Chapter 12 Worksheet 2 (ws12.2)
Concentration Units
molarity (M)
moles of solute / liter of solution (most commonly used unit of concentration)
molality (m)
moles of solute / kg of solvent
Compare to molarity and notice differences in the denominator. (L vs kg; solution vs. solvent)
Molality is independent of temperature so it is used for calculating boiling point elevation
and freezing point depression.
mole fraction (X) and mole %
moles of component / total moles of all components
A component is any substance in the solution so it can be a solute or the solvent.
mole % = mole fraction x 100
mass fraction and mass (or weight) percent (%)
mass fraction = mass of component / total mass of soln
mass % = mass fraction x 100
parts per million/billion/trillion (ppm/ppb/ppt)
ppm = (mass fraction) x 106
ppb = (mass fraction) x 109
ppt = (mass fraction) x 1012
Used for very dilute solutions (for example pollutants in drinking water). Notice that mass percent could also be called parts per hundred.
When an ionic compound dissolves in water, it dissociates into its component ions. So how do we specify the concentration of such a solution? Is it the concentration of the cation? The concentration of the anion? The sum of these two concentrations? IT IS NONE OF THESE. Unless stated otherwise, the concentration of a solution of an ionic compound is the concentration of the original undissociated solute.
1. In a 2 M solution of magnesium phosphate:
a. What is the molarity of magnesium ions?
Mg3(PO4)2 (s) → 3Mg2+(aq) + 2PO43-(aq)
b. What is the molarity of phosphate ions?
[PO43-] = 4 M
c. In this chapter you will learn about properties of solutions that depend on the total concentration of solute particles. What is the total molarity of particles in this solution (the sum of the two concentrations above)?
[particles] = 10 M
2. What is the molarity of an ammonium carbonate solution if the concentration of ammonium ions is 2 M? What is the concentration of carbonate ions and what is the total concentration of solute particles?
[CO32-] = 1M
[particles] = 3 M
3. A solution was made by dissolving 800.0 g of NaOH in 2.00 L of water. Calculate the molality, mole fraction, mass % and ppm of NaOH in this solution. (You will need the density of water.)
Molality
(800 g NaOH)/(40.0 g/mol) = 20.0 mol NaOH
(2.00 L water)(1 kg/L water) = 2.00 kg water
Molality = mol solute/kg solvent = 20.0 mol/2.00 kg = 10.0 m
Mole fraction (X)
X = moles NaOH / total moles = moles NaOH / (moles NaOH + moles H2O)
(2000 g water)/(18.0 g/mol) = 111.1 mol H2O
X = 20.0 moles / (20.0 moles + 111.1 moles) = 0.153 (no unit)
Notice that the answer does not depend on the identity of the solute. The mole fraction of solute in any 10 m aqueous solution is 0.153!
Mass % and ppm
[800 g NaOH/(800 g NaOH + 2000 g water)] x 100 = 28.6%
Ppm = mass fraction x 106 = 0.286 x 106 = 2.86 x 105 ppm
4. Consider a 0.300 M solution of NaOH?
a. Calculate the molality of NaOH. (Assume the density of the solution is the same as the density of pure water.)
Assume that you have 1 L of solution:
Mass of water = mass of solution – mass of NaOH = 1000 g – (0.300 mol)(40 g/mol)
= 1000 g – 12 g = 988 g
Molality = 0.300 mol/0.988 kg = 0.304 m
b. Notice that the molality is a little bigger than molarity. Why are these two numbers similar and why is molality always greater than molarity? Why can’t one convert molality to molarity without additional information?
They are similar because the mass of the water (in kg) will be similar to the volume of the solution (in L) as long as there is not too much solute.
The volume (in liters) and mass (in kg) of pure water are about the same (density is 1 kg/L). Molality is greater than molarity because the volume increases when the solute dissolves in water so now the volume of the solution is greater than the mass of the water. If there is very little solute (dilute solution) then it is a good assumption that the volume does not change and molarity equals molality.
5. What is the molarity of pure water at 4oC? (Hint: Use the density and molar mass of water.) What happens to the concentration of water when a solute is dissolved in it?
(1000 g/L) / 18.0 g/mol = 55.6 mol/L
The concentration of water decreases as a the concentration of a solute increases because the solute occupies some of the volume and the water molecules are pushed apart.
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