Molarity, Dilution
CDURAKOVIC@IUS.EDU.BA
Molarity
molarity =
moles of solute
liter of solution

Molarity is the number of moles of solute dissolved in one liter of solution. The units,
therefore are moles per liter, specifically it's moles of solute per liter of solution.

Rather than writing out moles per liter, these units are abbreviated as M or M. We use a
capital M with a line under it or a capital M written in italics. So when you see M or M it
stands for molarity, and it means moles per liter (not just moles).

You must be very careful to distinguish between moles and molarity. "Moles"
measures the amount or quantity of material you have; "molarity" measures the
concentration of that material. So when you're given a problem or some
information that says the concentration of the solution is 0.1 M that means that it
has 0.1 mole for every liter of solution; it does not mean that it is 0.1 moles. Please be
sure to make that distinction.
Molarity calculations

There are several types of calculations that you need to be able to do with molarity.
First, you should be able to calculate the molarity if you are given the components of the solution.
Second, you should be able to calculate the amount of solute in (or needed to make) a certain
volume of solution.
Third, you might need to calculate the volume of a particular solution sample.
Fourth, you might need to calculate the concentration of a solution made by the dilution of
another solution.
Calculating Molarity from Moles and Volume

Here we are given something to figure out. To get the molarity we
need to divide the number of moles of NaCl by the volume of the
solution. In this case that is 0.32 moles NaCl divided by 3.4 L, and
that gives 0.094 M NaCl.
What is the molarity of a solution containing 0.32 moles of
NaCl in 3.4 liters?
molarity =
0.32 moles NaCl
3.4 L
= 0.094 M NaCl
n=m/M
Calculating Molarity from Mass and Volume
n (amount of substance)
= mass/molar mass
Unit is 1 mol
What is the molarity of a solution made by
dissolving 2.5 g of NaCl in enough water to
make 125 ml of solution?
molarity =
2.5 g NaCl x
moles of solute
liter of solution
1 mole NaCl
58.5 g NaCl
molarity =
= 0.0427 mole
0.0427 mole NaCl
0.125 L
= 0.34 M NaCl
This one is a bit more difficult. To get molarity we still need to divide moles of solute by volume of solution.
But this time we're not given the moles of solute. We have to calculate it from the mass of NaCl. We multiply
2.5 g NaCl by the conversion factor of 1 mole NaCl over the formula weight of NaCl, 58.5 g. That tells us that
we have 0.0427 mole of NaCl. I kept an extra digit here because we are not done with the calculations.
When we are done I'll round off to two digits, the same as in the 2.5 g weight of NaCl.
Now that we know the moles we can calculate the molarity. Moles of solute (0.0427) divided by the volume
of the solution (0.125 L) gives us 0.34 M NaCl.
Calculating Mass of Solute from Molarity
How would you prepare 400. ml of
1.20 M solution of sodium chloride?
1.20 M NaCl =
0.400 L solution x
1.20 moles NaCl
1.00 L solution
1.20 moles NaCl
1.00 L solution
= 0.480 moles NaCl
0.480 moles NaCl x
58.5 g NaCl
1 mole NaCl
= 28.1 g NaCl
Dissolve 28.1 g NaCl in enough water to make 400 mL
of solution.
Algebraic approach:
Set up general equation:
xM=
y moles
zL
Substitute known values:
1.20 M =
y moles
0.400 L
y moles =
1.20 M x
0.400 L
Solve for moles:
n=m/M
n (amount of substance) =
mass/molar mass
Unit is 1 mol
= 0.480
moles
Calculating Moles of Solute from Molarity
How many moles of salt are contained
in 300. mL of a 0.40 M NaCl solution?
0.300 L x
0.40 moles NaCl
1.00 L solution
= 0.12 moles NaCl
This question is a little easier. We do it the same way as the first step of the previous
problem and then we stop.
To find out how many moles of salt are contained in 300. ml of a 0.40 M NaCl solution,
we start with the volume in liters (0.300 L) and multiply it by the number of moles per
liter of solution, which is 0.40 moles over 1.00 L. The answer is 0.12 moles of NaCl.
This could also have been done using algebra by writing down the general equation
relating molarity, moles and liters, substituting the known values, and then solving the
equation for moles.
Molarity Calculations: Practice
a. How would you prepare 100. mL of 0.25 M KNO3 solution?
b. A chemist dissolves 98.4 g of FeSO4 in enough water to make
2.000 L of solution. What is the molarity of the solution?
c. How many moles of KBr are in 25.0 mL of a 1.23 M KBr
solution?
d. Battery acid is generally 3 M H2SO4. Roughly how many
grams of H2SO4 are in 400. mL of this solution?
Dilution
molescon
Mcon x volcon
=
=
molesdil
Mdil x voldil
(Mconc) (Vconc)
=
(Mdil) (Vdil)

Quite often, however, solutions are prepared by diluting a more
concentrated solution. For example, if you needed a one molar
solution you could start with a six molar solution and dilute it.
Consequently, you also need to be familiar with the calculations
that are associated with dilutions.

Calculating new concentrations?

Calculating initial volume?
Calculating New Concentration:
A chemist starts with 50.0 mL of a 0.40 M NaCl solution and dilutes it to
1000. mL. What is the concentration of NaCl in the new solution?
(Mcon) (Vcon) = (Mdil) (Vdil)
(0.40 M) (50.0 mL) = (Mdil) (1000. mL)
(0.40 M) (50.0 mL)
= (Mdil)
(1000. mL)
0.020 M = Mdil
In this example you are asked what the concentration of a solution would be if it were made by diluting 50.0 ml
of 0.40 M NaCl solution to 1000. ml.
As a general procedure, I recommend that you first write down the equation that is given in the first line: the
molarity of the concentrated solution times the volume of the concentrated solution is equal to the molarity of
the dilute solution times the volume of the dilute solution.
Next, substitute the known values into the equation as shown in the next line. Then rearrange the equation to
solve for the unknown value. In this case that is done by dividing both sides of the equation by 1000. ml.
Last, carry out the necessary calculations to get, in this case, .020 M.
Calculating Initial Volume:
A chemist wants to make 500. mL of
0.050 M HCl by diluting a 6.0 M HCl
solution. How much of that solution should
be used?
(Mcon) (Vcon) = (Mdil) (Vdil)
(Mdil) (Vdil)
(Vcon) = -------------(Mcon)
(0.050 M)(500.mL)
(Vcon) = ---------------------(6.0 M)
Vcon = 4.2 mL
The question in this example is different only in that
you are asked to determine a volume instead of a
concentration.
The process used to answer the question is the same
as in the previous example. Write down the
algebraic equation that represents the relationship.
Rearrange the equation to solve for the unknown
quantity.
Substitute numbers (and units) for the known values.
(The second and third steps can be reversed if you
wish.) Calculate the unknown value (which now
becomes known).
Dilution Calculations: Practice
A How much 2.0 M NaCl solution would you need to make 250 mL of
0.15 M NaCl solution?
B What would be the concentration of a solution made by diluting 45.0
mL of 4.2 M KOH to 250 mL?
C What would be the concentration of a solution made by adding 250
mL of water to 45.0 mL of 4.2 M KOH?
D How much 0.20 M glucose solution can be made from 50. mL of 0.50
M glucose solution?
Serial dilution
Useful links:

http://phet.colorado.edu/en/contributions/view/3511

http://dl.clackamas.edu/ch105/lesson4molarity.html

http://education-portal.com/academy/lesson/serial-dilution-inmicrobiology-calculation-method-technique.html

http://mathbench.umd.edu/modules/microbio_serialdilution/page03.htm

https://www.youtube.com/watch?v=HZzpgjGosmg