Part II. Concentrations of solutions
All solutions contain a solute (what is being dissolved) and a solvent (what is doing the dissolving).
Depending on the amount of solute and amount of solvent, all solutions have a concentration. Have you ever made
lemonade? If you don’t add enough sugar, it may seem a little tart. If you add too much sugar, it may be too sweet.
By varying the amount of sugar you add, you are effectively changing the concentration of the solution.
There are three methods of measuring solution concentration. These are listed below, with their equations
used to find each.
% by mass = mass of solute x 100
mass of solution
Molarity (M) = moles solute
Liters of solution
molality (m) = moles solute
Kg solvent
Let’s first look at % by mass:
1. What is the % by mass of a solution prepared by dissolving 4 g of acetic acid in 35 g of water? (10%)
Equation)
Substitution)
solution)
2. Determine the % by mass of a solution prepared by dissolving 32.5 g of glucose in 155 g water? (17.3%)
Equation)
Substitution)
solution)
3. If you have 50 ml (50 g) of a vinegar solution that is 5% acetic acid in water, what is the mass of acetic acid?
Equation)
Substitution)
solution)
4. Find the mass of solute present in each of the following solution:
a)
50.0 g of a solvent in a 10.0 % by mass NaCl solution. (5.55 g)
Equation)
Substitution)
solution)
b) 8.5 g of solvent in a 2.4 % by mass KOH solution. (0.209 g)
Equation)
Substitution)
5.
solution)
What is the % of sugar in 1 cup of kool-aid with 2 oz of sugar?
Equation)
Substitution)
solution)
Percent by mass is a nice way to determine the concentration of a solution when a comparison is not
needed. What if you were to compare the concentration of some acids or bases (very common solutions that are
studied a lot in Chemistry)? Because the mass of atoms varies, it is important to use another concentration
determination method that allows us to relate concentration to the number of dissolved ions. Thes methods are
Molarity or molality.
Let’s try some Molarity problems:
1. What is the molarity of 3.50 L of solution that contains 90.0 g of NaCl?
Equation)
M = moles solute/ Liters solution
Substitution) M = (90.0 g NaCl x 1 mol
58.5 g =
3.50 L
1.53 moles
3.50 L
Solution) M = 0.440 M
2. What is the Molarity of a solution composed of 5.85 g of Potassium Iodide dissolved in enough water to make
0.125 L of solution? (0.282 M KI)
Equation)
Substitution)
Solution)
3. How many moles of H2SO4 are present in 0.500 L of a 0.150 M H2SO4 solution? (0.0750 moles)
Equation)
Substitution)
Solution)
4. What is the Molarity of a solution prepared by dissolving 9.94 g of Cobalt II sulfate in 2.50 x 10 2 cm3 of
solution? (0.256 M)
Equation)
Substitution)
Solution)
5. How many grams of Silver I fluoride would be needed to make 0.5 dm3 solution of 1.50 M AgF? (95.25 g)
Equation)
Substitution)
Solution)
6. What is the Molarity of a solution prepared by adding 5.23 g of Iron II nitrate in 100 ml of solution? (0.29 M)
Equation)
Substitution)
Solution)
7. How many grams of NiCl2 would be needed to make 1 L of 3.00 M solution of Nickel II Chloride? (387 g)
Equation)
Substitution)
Solution)
8. What is the Molarity of 500 ml of sugar solution that is 28 % sugar in water? (assume the total mass of the sugar
water is the same as the total volume of the solution.)
(work)
Now that you are able to determine calculate the Molarity of a solution, and now that you are able to write
dissociation equations showing the break up of ionic solutes when solutions are formed, let’s see if you are able to
calculate the concentration of the ions in solutions??
1. Calculate the Molarity of zinc ions in a 0.5 M zinc chloride solution.
ZnCl2(s)  Zn+2 + 2Cl-(aq)
0.5 M
1 (0.5 M)
2 (0.5 M)
= 0.5 M
= 1.0 M
total concentration of ions = 3 total ions x (0.5 M) = 1.5 M
2. Calculate the concentration of sulfate ions in a 0.25 M solution of potassium sulfate?
3. Calculate the concentration of ammonium ions in a solution prepared by adding 45 g of ammonium nitrate in
enough water to make 500 ml of solution?
4. Calculate the Molarity of hydroxide ions prepared by dissolving 10 g of calcium hydroxide in enough water to
make 250 ml?
Molality is the last method we will discuss regarding the concentration of solutions. This concentration is
used when discussing colligative properties of solutions. Colligative properties are properties of solutions that are
determined by the number of solute particles present, not on the type of solute particles. For example, a 1 molal
solution of NaCl has the same effect on the freezing point of water as does a 1 molal solution of LiCl.
Let’s try some molality calculations:
1.
Calculate the molality of a solution prepared by adding 199 g of Nickel II Bromide in 500 g of water?(1.82 m)
Equation)
Substitution)
Solution)
2. Calculate the molality of a solution prepared by adding 98.0 g of Rubidium Bromide in 824 g of water?(0.719 m)
Equation)
Substitution)
3.
If you were to add 630 g of nitric acid (HNO 3) to 6500 g of water, what would the molality of the solution?
(1.54 m)
Equation)
Substitution)
4.
Solution)
What mass of solute would be needed to add to 250 g of water to make a 0.525 m solution of potassium
chloride? (9.78 g)
Equation)
Substitution)
5.
Solution)
Solution)
Calculate the molality of a solution when 15.5 g of silver I nitrate is added to 375 g of water? (0.243 m)
Equation)
Substitution)
6.
Solution)
What mass of lithium sulfate is needed to be added to 500 ml of water to make a 0.384 m solution?
Equation)
Substitution)
Solution)
7.
If you added 25 g of calcium phosphate to 50 ml of water, but only 1.2 g of the salt dissolved in the water, what
is the molality of the solution?
Equation)
Substitution)
Solution)
Earlier, we discussed colligative properties, properties of solutions determined by the amount of solute
particles dissolved, not on what those solute particles are. What are the colligative properties? These are listed
below:
1) Vapor pressure lowering (of the solvent) when a solute is dissolved in a solvent – the amount of
lowering depends on the molality of the solution formed.
2) Freezing point lowering – making a solution out of a solvent lowers the freezing point of the solvent,
again, how much depends on the molality.
3) Boiling point elevation – making a solution out of a solvent raises the boiling point of the solvent,
again, how much depends on the molality.
See the vapor pressure curves below to see what is meant by vapor pressure changes mentioned above. If
you recall, boiling point is the temperature at which the air pressure = the vapor pressure of the liquid. If you lower
the vapor pressure, the boiling point increases and the freezing point decreases. How much depends on the molality
of the solution according to the equation found on the next page.
To calculate the amount of boiling point of freezing point change requires the equations listed below.
Delta Tb = Kb x molality
Change in b.p.
Boiling point elevation
Constant
(depends on solvent)
(see table below)
Delta Tf = Kf x molality
Change in f.p.
Freezing point depression
Constant
(depends on solvent)
(see table below)
Now try some calculations:
1. What is the freezing point depression of water in a solution of 17.1 g of sucrose and 200 g of water? (-0.465 oC)
Equation)
Substitution)
2.
Solution)
A solution consists of 10.3 g of glucose (C6H12O6), dissolved in 250 g of water. By how much does the
freezing point of water change? (-0.426 oC)
Equation)
Substitution)
3.
Solution)
In a laboratory experiment, the freezing point of a water solution of glucose is found to be -0.325 oC. What is
the molality of the solution? (0.175 m)
Equation)
Substitution)
Solution)
4
What is the boiling point of a 2.5 molal sucrose solution? (101.275 oC)
Equation)
Substitution)
5.
Solution)
What is the boiling point of a 2.5 molal glucose solution? (101.275 oC)
You will notice in all of the above problems, the solute was a sugar (sucrose or glucose). The reason for
these is that they are non-electrolytes, meaning they dissolve in water, but do NOT produce ions. When a solute
produces ions (an electrolyte), we need to consider the number of ions in solution. When calculating the freezing
point or boiling point changes, we need to multiply the changes by the number of ions to get a truer estimate of the
expected temperature changes. For this reason, equal concentrations of NaCl and CaCl 2 do not have the same effect
on freezing point or boiling point. The calcium chloride has a greater effect as it breaks up into more ions.
You try:
1) 1.20 moles of a non-electrolyte is dissolved in 1000 g of water. How much does the freezing point of water
change? (-2.32 oC)
Equation)
Substitution)
Solution)
2a) 0.400 moles of a non-electrolyte is dissolved in 2000 g of phenol. Calculate the freezing point depression.
(1.48 oC)
Equation)
Substitution)
Solution)
2b) What is the freezing point of the above solution? __________________
3a) 78.1 g of a non-electrolyte that has a molar mass of 60.3 g/mol is dissolved in 10.5 Kg of phenol. How much
does the boiling point go up?
Equation)
Substitution)
Solution0
3b) What is the boiling point of the above solution? ___________________
4.
What is the boiling point of a solution prepared by dissolved 12.5 g of a non-electrolyte with a molar mass of
34.5 g/mol into 1500 g of ether?
If you were to algebraically manipulate the equations used to answer the above questions, you can get an
equation to find the Molar Mass of the solute. If you know the Molar Mass, you could use the periodic table to
help identify any unknown solutes. We will show the derivation of this equation in class. The space below is
for you to write in this derivation.
Now try to use the above equation to find Molar Mass or expected temperature changes.
1.
When 1.56 g of an unknown, non volatile solute is dissolved in 200. g of water, the freezing point
depression of the solvent is -0.453 oC. Determine the Molar Mass of the solute. (32.0 g/mol)
2.
If 1.84 g of a molecular solute is dissolved in 150 g of acetic acid, the boiling point of the solvent is
elevated 0.60 oC. What is the MM of the solute? (63 g/mol)
3.
What is the Molar Mass of a non-electrolyte, given that 151 g of the substance depresses the freezing point
of 3500 g of water by 2.53 oC?
4.
What is the MM of a solute if 465 g of the solute added to 4500 g of carbon disulfide raises the boiling
point to 53.3 oC?