I am sure that you have some experience
with the concept of solution concentrations. If you ever made or
drank a liquid made from a powdered mix, such as ice tea or hot cocoa, you probably are
familiar with the difference between what is called a "weak" solution or a
"strong" solution. Even if you don't drink coffee, you may have heard a
relative complain about his or her coffee being too strong or to weak. Molarity
is simply a measure of the "strength" of a solution. A solution that we
would call "strong" would have a higher molarity than one that we would call
"weak".
Now, a solution is made
up of two parts. The solute is what gets mixed into the solution,
like powdered drink mix. The solvent is that which does the
dissolving, like water in the case of ice tea. Let us suppose that we have a
powdered ice tea mix that calls for 4 scoops of powder for every 2 quarts of water.
We might say that the normal recipe for the ice tea mix looks like:
4 scoops of powder
To make ice tea of "normal" strength =
-----------------------
2 quarts of water
If we wanted to make the ice tea twice
as strong as normal, we could use 8 scoops of powder with 2 quarts of water, or 4 scoops
of powder with 1 quart of water:
8 scoops of powder 4 scoops of powder
To make ice tea twice the "normal" strength = -------------------- or
----------------------
2 quarts of water 1 quart of water
Molarity is just like the strength of
the solution above, except that it is more exact. "Scoops" are not very
accurate measuring devices, and they don't need to be when you are making ice tea.
However, when you are doing any type of quantitative analysis in the lab, you want to be
as accurate as possible. Therefore, the formula for Molarity will be more exact than
the formula for ice tea mix.
# of moles of solute
Molarity = ----------------------
Liters of solution
The unit for molarity is M and is read as "molar".
(i.e. 3 M = three molar)
Molarity problems vary quite a bit.
Pay careful attention to the wording of the problem, and focus on what you are
given and what the problem is asking for. Start with the original
molarity formula
shown above, but be prepared to modify it when the need arises. I show you several
examples of molarity problems in order to introduce you to the methods involved in solving
different types of problems.
I. Basic molarity problems where the molarity
is the unknown.
Example 1. What is the molarity of a 5.00 liter solution
that was made with 10.0 moles of KBr ?
Solution: We can use the original formula. Note that in this particular
example, where the number of moles of solute is given, the identity of the solute (KBr)
has nothing to do with solving the problem.
# of moles of solute
Molarity = ----------------------
Liters of solution
Given: # of moles of solute = 10.0 moles
Liters of solution = 5.00
liters
10.0 moles of KBr
Molarity = -------------------------- = 2.00 M
5.00 Liters of solution
Answer = 2.00 M
Example 2. A 250 ml solution is made with 0.50 moles of
NaCl. What is the Molarity of the solution?
Solution: In this case we are given ml, while the formula
calls for L. We must change the ml to Liters as shown below:
250 ml 1 liter
x
-------- = 0.25 liters
1000 ml
To avoid confusion, I will usually make the unit change right in
the original question:
Example 2. A 250 ml 0.25 L
solution is made with 0.50 moles of NaCl. What is the Molarity of the solution?
Now, solve the problem as you solved example 1.
# of moles of solute
Molarity = ----------------------
Liters of solution
Given: Number of moles of solute = 0.50 moles of NaCl
Liters of solution =
0.25 L of solution
0.50 moles of NaCl
Molarity = --------------------- = 2.0 M solution
0.25 L
Answer = 2.0 M solution of NaCl
II. Basic molarity problems where volume is
the unknown.
This is similar to when we studied
density, we have a formula with three possible unknowns. When the molarity of the
solution and the number of moles of solute are given, but the volume is unknown, we must
adjust our original formula to isolate the unknown variable. Observe:
# of moles of solute
Molarity = ----------------------
Liters of solution
# of moles of solute
Molarity x Liters of solution = ---------------------- x Liters of solution
Liters of solution
Molarity x Liters of solution = # of moles of solute
----------
--------------------
Molarity
Molarity
# of moles of solute
Liters of solution = --------------------
Molarity
Example 1. What would be the volume of a 2.00 M solution
made with 6.00 moles of LiF?
Solution:
# of moles of solute
Liters of solution = --------------------
Molarity
Given: # of moles of solute = 6.00 moles
Molarity = 2.00 M (moles/L)
Liters of solution = 6.00 moles
-----------
2.00 moles/L
Answer = 3.00 L of solution
Now, you must also be prepared for the fact that the number of
moles is not always given to you. Sometimes you will be given the mass of the solute
and you will need to determine the number of moles by dividing the mass given by the Molar
mass of the solute. In these cases, use the formula below.
mass given
# of moles = -----------------
Molar mass
Example 2. What is the volume of 3.0 M solution of NaCl
made with 526g of solute?
Solution:
First find the molar mass of NaCl.
Na = 23.0 g x 1 ion per formula unit = 23.0 g
Cl = 35.5 g x 1 ion per formula unit = 35.5 g
----------
58.5 g
Now find out how many moles of NaCl you have:
mass of sample
# of moles = -----------------
Molar mass
Given: mass of sample = 526 g
Molar mass = 58.5 g
526 g
# of moles of NaCl = ------------
58.5 g
Answer: # of moles of NaCl = 8.99 moles
Finally, go back to your molarity formula to solve the problem:
# of moles of solute
Liters of solution = --------------------
Molarity
Given: # of moles of solute = 8.99 moles
Molarity of the solution =
3.0 M (moles/L)
8.99 moles
# of Liters of solution = -------------
3.0 moles/L
Final Answer = 3.0 L
III. Basic molarity problems where the number of
moles is the unknown.
Of course, the total number of moles
used in the creation of a solution might be unknown to you. However, given the
molarity and the volume of the solution, you can determine the number of moles of solute.
Observe:
# of moles of solute
Molarity = ----------------------
Liters of solution
# of moles of solute
Molarity x Liters of solution = ---------------------- x Liters of solution
Liters of solution
# of moles of solute = Molarity x Liters of solution.
Example 1. How many moles of CaCl2 would be used
in the making of 5.00 x 102 cm3 of a 5.0M solution?
Notice that the volume is given in cm3. Since
there are 1000 cm3 in 1 liter, 500 cm3 must be equal to 0.500
liters. Make that change right in the problem.
Example 1. How many moles of CaCl2 would be used
in the making of 5.00 x 102 cm3 0.500 L of a 5.0M
solution?
Now you are ready to solve.
Solution:
# of moles of solute = Molarity x Liters of solution.
Given: Molarity = 5.0 M (moles/L)
Volume = 0.500 L
# of moles of CaCl2 = 5.0 moles/L x
0.500 moles
Answer = 2.5 moles of CaCl2
Notice that the identity of the solute does not work into the
math of the problem. However, if the wording was different, it would. Observe
example # 2.
Example 2. How many grams of CaCl2 would be used
in the making of 5.00 x 102 cm3 of a 5.0M solution?
In this case, what they are looking for is different. You
could start to solve this problem the same way you did example 1, but the end would
require you to change the number of moles of CaCl2 to the mass of CaCl2.
You would use the formula below.
mass of sample
# of moles = -----------------
Molar mass
mass of sample
# of moles x Molar mass = ----------------- x Molar mass
Molar mass
mass of sample = # moles of solute x Molar mass
Given: # of moles of solute = 2.5 moles (from our answer to example 1.)
Molar mass of solute (CaCl2)
= 111 g/mole (from the periodic table)
Mass of CaCl2 = 2.5 moles x 111 g/mole
Answer: Mass of CaCl2 = 280 g (when rounded correctly)
Now that you have seen some of the types
of problems that can be solved with these formulas, go on and practice these calculations
using the worksheets below. Make sure that you can adjust each formula to fit a
given situation. Avoid memorizing formulas and trying to work the problems
mechanically. Use the correct units as a way to check your answer. When you
feel ready, take the random online quiz to see how you are doing.
Please forward all questions, comments and criticisms to Gregory L. Curran.
© Copyright 2004 Fordham Preparatory School, All Rights Reserved.
Last Modified February 07, 2008 |
