Concentration Measurement:
Molarity
Mr. Shields
Regents Chemistry
U12 L02
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Molarity
Let’s start our discussion of solution concentration
Measurement with
MOLARITY
Molarity: the most common solution concentration
measurement is defined as follows:
Capital M
M=
moles of solute
LITERS of solution
Which means Molarity = # of moles per liter of solution
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Molarity
Let’s Consider the following Problem:
What is the molarity of a solution in which 50g of
CuSO4 is dissolved in water to make a 2 liter solution.
Step 1. What’s the formula for MOLARITY?
M = moles of solute
liters of solution
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Molarity
Step 2. We need to determine how many moles we have
in 50g of CuSO4 .
How do we do that? Remember the mole Hole?
- 1. Determine the molar mass of CuSO4
(159.5g/mol)
- 2. Calculate the # of moles of CuSO4
Since we’re going into the mole hole we need
to DIVIDE # of grams by molar mass
(50/159.5) = 0.31 moles
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Molarity
Step 3. Calculate the Molarity of the solution using
the formula for MOLARITY
M = moles of Solute
liters of solution
How many moles do we have?
How many liters do we have?
So
M = 0.31 moles
2 liters
0.31 mol
2L
M = 0.155 mol/liter
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Making a 1M NaCl Solution
1. Weigh out and add 1 mole of NaCl to a volumetric Flask
2. Add a small amount of solvent to dissolve the salt
3. Once the solute
is dissolved add
additional solvent
up to the 1 liter
mark
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Problem
What is the molarity of a solution made by dissolving 2.1
liters of NH3 (g) in sufficient water to make 2 liters of
solution?
What do we need to begin calculating the Molarity of the solution?
1.Calculate how many moles of ammonia there are
# of moles = volume = 2.1 liters
= 0.094 mol
molar vol
22.4 l/mol
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Problem
We now know the number of moles of NH3 we have (0.094) so …
Step 2. Calculate the molarity of the solution
using the molarity equation
M = # moles
= 0.094 mol = 0.047M
# liters of sol’n
2 liters
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Molarity by dilution
Now what if we had a solution already prepared but
Wanted to decrease the molarity to some lower precise
Value.
How could we use an already prepared higher molarity
solution to accomplish this?
Note that it is only possible to go from a more
conc. (higher molarity) solution to a less conc.
(lower molarity) solution.
conc
dil
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Molarity by dilution
To make something more dilute we’re going to need to
add more solvent to the solution.
Recall that Molarity (M) = # moles of solute
# liters of solution
So we need to dilute the solution by adding solvent
But how much more solvent do we need to add?
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Molarity by dilution
When we add more solvent to a solution do we
change the number of moles of solute in the
solution?
NO! But what does change?
We change the # of moles
per unit volume of Solvent because
the same # of moles are present but
now there’s more liquid
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Molarity by Dilution
Let’s see how this works …
Let’s call the Molarity of the original solution M1 and the
New Molarity M2
The number of liters of the original solution and the new
Solution will then be designated by V1 and V2
So the molarity of the old solution is…
M1 = #moles of solute = #moles
#liters of solution
V1
So…
#moles = M1 x V1
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Molarity by Dilution
The Molarity of the new solution will then be …
M2 = #moles of solute
#liters of solution
# moles = M2 x V2
Remember that the # of moles does not change when
We dilute solutions.
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Molarity by dilution
So…
M1 x V1 = M2 x V2
If we know three values we can calculate the 4th.
Let’s see how we use this relationship
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Dilution Problem
How much of a 3M solution of NiCL2 is needed to
Prepare 100 ml of a 1.75M solution of NiCl2?
OK how do we solve this? What’s the equation
we’re going to use?
Right!
M1V1 = M2V2
M1 = 3M; M2 = 1.75M;
V2 = 0.1L (100ml)
So … 3 x V1 = 1.75 x 0.1
V1 = 0.058L or 58ml
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Molarity by dilution
How do we make 5L of
A 1.5M sol’n of KCL
From 12.0M stock sol’n?
M1V1 = M2V2
12 x V1 = 1.5 x 5
12V1= 7.5
V1 = 7.5/12 = 0.625L
Add 4.375L H20 to a
0.625L of a 12M sol’n
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