This document provides information on various concepts related to solutions including vocabulary terms, molarity, molality, and percent composition. For molarity, it defines the term, provides examples of how to calculate molarity given different variables, and discusses dilution problems. For molality, it notes some of the challenges in calculating molality for hydrated solutes. It also defines percent composition and provides sample problems calculating density, molality and molarity for different solutions.

1. SOLUTIONS
Vocabulary
Molarity
Molality
Percent composition
L. Allen 1/29/12

2. Vocabulary
• Heterogeneous
• Solution equilibrium
• Homogeneous
• Hydration process
• Suspension
• “Like dissolves like”
• Colloid
• Miscible
• Tyndall effect
• Henry’s law
• Solute
• Gas v. solid solubility
• Solvent
• Enthalpy of solution
• Electrolyte
• Parts of solvation
• Saturated
• Concentration

3. Molarity, designated with M
• Moles of solute / liters of solution
• This is the simplest of the ways to identify the
concentration of a solution. If I have one liter of 1M
solution, I have one mole of the solute in there. I can
measure simple volumes, and know the moles of solute.
• Note that if you are mixing a solution that is a hydrate, you
must include the mass of the water when you figure out
how many grams of solute to include.
• Also, if you can’t write a chemical formula, you end up
with the wrong molar mass, and the wrong molarity, and
you feel caught in the current of a huge FLUSH. So make
a point of getting the right chemical formula.

4. Check out page 420 in the blue Holt Modern Chemistry.
• Given volume of
solution and mass of
solute, find the
molarity.
• How to?
• Convert grams of
solute to moles.
• Moles / liters = M
• Check sig figs, confirm
labels, confirm with
common sense.

5. Sample problem B
• Given volume and
molarity, find moles.
• Moles / Liters = M
• Rearrange with
algebra, find that M x
liters = moles.
• Plug in values with
labels, calculate, check
sig figs, confirm that it
makes sense.

6. Sample problem C
• Given a load of
information, some of
which I can’t possibly
want, how much volume
do I need to get this
mass of solute?
• Begin with desired
mass. Calculate moles
you want. (gmol)
• M = moles/liters
• Rearrange this to get
liters – beware possible
fraction errors.
• Divide the moles you
want by the molarity.
• Check sig figs and
common sense.

7. One more thing about molarity!
• Molarity x volume = moles
• This is useful for looking at dilution problems. If I put a
certain number of moles of solute into solution, I get
molarity… so if those moles came from another molarity
solution… I can identify the original molarity as M1…
• I can say the original number of moles could be
represented as M1 x V1. That is how many moles I am
adding to the new solution.
• The new molarity is that number of moles divided by the
new volume.
• (M1 x V1) / V2 = M2

8. Here’s an example problem
• If I take .20 liters of 8.0 molar solution, and dilute it to a full
liter, what is the new molarity?
• The molarity of the new solution can be calculated like
this (M1 x V1) / V2 = M2
• The new molarity is 1.6 M.
• Does that make sense? Are the sig figs right?
M1 x V1 = M2 x V2
• This is the easiest form of this to remember. It is
extremely useful in calculating dilutions.

9. A well-deserved break from text

10. Molality, designated with m
• This is the least favorite of the ways to identify
concentration for a couple of reasons.
• First, if you are calculating to make a certain molality
solution from a hydrated solute, you have to include the
mass of the water in figuring out how many grams of
solute to include, but you have to include the water of
hydration in with the mass of the solvent, too.
• You can’t just avoid molality because there is a whole
phenomenon (colligative properties) that are directly
proportional to molality, and not molarity.
• Your book conveniently ignores water of hydration in this
math. 

11. Check out the problem on page 422
• Okay, first they calculated the molar mass of the hydrate,
•
•
•
•
•
as I did. The anhydrous mass is 159.61 g/mol, and the
water weighs 90.10 g/mol, for a total of 249.71 g/mol.
Half a mole of this weighs 124.8 grams. Fine. Dump it
into your beaker. (I prefer 124.9, but I won’t argue)
We should add 1000 grams of water next. But we actually
already added some water, in with that hydrate.
Calculate this via percent composition, or however you
prefer. It looks like 45.05 grams of water to me.
Add (1000 – 45.05) or 955 grams of water to your
hydrate.
NOW you have your .5000 m solution!

12. The easiest way to describe concentration
• Percent by mass
• Overall, this is not the most useful way to describe
concentrations, but it has nice links to density, which
make it practical.
• Percent by mass = grams of solute / grams of solution
• A 20% HCl solution has 20 g of HCl and 80 g of water in
it.

13. Phosphoric acid is usually obtained as an 87.0% phosphoric acid solution. If it is
13.0 M, what is the density of this solution? What is its molality?
Problem stolen from online sources and fiddled with for clarity. 
Solution for density:
1) Determine the moles of H3PO4 in 100.0 grams of 87.0%
solution:

14. Phosphoric acid is usually obtained as an 87.0% phosphoric acid solution. If it is
13.0 M, what is the density of this solution? What is its molality?
Problem stolen from online sources and fiddled with for clarity. 
Solution for density:
1) Determine the moles of H3PO4 in 100.0 grams of 87.0%
solution:
87.0 g of the 100.0 g is H3PO4
Moles H3PO4 = 87.0 g / 97.9937 g/mol = 0.8878 mol
2) Calculate the volume of 13.0 M solution which contains
0.8878 mol of H3PO4:
13.0 mol/L = 0.8878 mol / x
x = 0.0683 L = 68.3 mL
3) Determine the density of the solution:
100.0 g / 68.3 mL = 1.464 g/mL
= 1.46 g/mL (to three sig fig)

15. Phosphoric acid is usually obtained as an 87.0% phosphoric acid solution. If it is
13.0 M, what is the density of this solution? What is its molality?
Problem stolen from online sources and fiddled with for clarity. 
Solution for molality:
1) Let us assume 100.0 grams of solution. Therefore:
87.0 g is H3PO4 (13.0 g is H2O)
2) Calculate the molality:
Moles H3PO4 = 87.0 g / 97.9937 g/mol = 0.8878 mol
kg of water = 0.0130 kg
Molality = 0.8878 mol / 0.0130 kg = 68.3 molal

16. An aqueous solution of hydrofluoric acid is 30.0% HF, by mass, and has a
density of 1.101 g cm-3. What are the molality and molarity of HF in this solution?
Solution for molality:

17. An aqueous solution of hydrofluoric acid is 30.0% HF, by mass, and has a
density of 1.101 g cm-3. What are the molality and molarity of HF in this solution?
Solution for molality:
1) Let us assume 100.0 grams of solution. Therefore:
30.0 g is HF
70.0 g is H2O
2) Calculate the molality:
Moles HF = 30.0 g / 20.0059 g/mol = 1.49956 mol
mass of water = 0.0700 kg
Molality = 1.49956 mol / 0.0700 kg
= 21.4 molal (3 sig figs)

18. An aqueous solution of hydrofluoric acid is 30.0% HF, by mass, and has a
density of 1.101 g cm-3. What are the molality and molarity of HF in this solution?
Solution for molarity:
1) Determine moles of HF in 100.0 g of 30.0% solution:
1.49956 mol
2) Determine volume of 100.0 g of solution:
density = mass / volume, volume = mass / density
= 100.0 g / 1.101 g/mL
= 90.8265 mL
3) Determine molarity:
1.49956 mol / 0.0908265 L
= 16.5 M (chop to 3 sig figs at the end)