MOLAR SOLUTION
The molarity of a solution is defined as the number of moles of solute per liter of solution. The
mathematical expression for molarity is given in Eq. 1.
Eq. 1:
Molarity (M) =
Moles of solute
Liters of solution
The mathematical expression for the number of moles is given in Eq. 2.
Eq. 2:
Moles =
Grams of solute
Molecular weight of solute
For example, to find the moles of ten grams of sodium hydroxide, we divide the grams of sodium hydroxide
10
by the molecular weight of sodium hydroxide (that is,
.25 )
40
Liters is equal to milliliters of solution divided by 1,000. The mathematical expression is given in Eq. 3.
Eq. 3:
Liter =
ml of solution
1000
For example, to find the liters in 400 ml of solution, we divide 400 by 1000 which will give us .4 i.e.
400
.4 l.
1000
Substituting Eq 2 and 3 in Eq 1 we obtain Eq 4.
Eq. 4:
grams of solute
molecular weight of solute
Molarity =
ml of solution
1000
Rearranging Eq 4 we obtain Eq 5.
Eq. 5:
Molarity =
(1000) (Grams of solute)
( molecular wt. of solute) ( ml of solution
Solving Eq 5 for grams Eq 6 is obtained.
Eq 6:
Grams of solute =
( M) (molecular wt of solute (ml of solution)
1000
It is important to remember the following points:
(1) the molecular weight of any compound is obtained by adding the atomic weight of the atoms in the
formula. (2) One mole of any compound contains 6.02  1023 molecules, that is Avogodro’s number of
molecules. (3) For our purpose in this discussion, 1 mole is the same as 1 gram molecular weight which is
the same as 1 gram formula weight.
EXAMPLE 1
Calculate the number of moles in 15 grams of calcium hydroxide.
SOLUTION
Mole =
gram
molecular wt.
Mole =
15
 .2
74
EXAMPLE 2
Calculate the molarity of a solution containing 50 grams of sodium hydroxide in 2 liters of solution.
SOLUTION
Using Eq. 2
Mole =
=
grams of solute
molecular wt. of solute
50
40
= 1.25
Using Eq 1
the molarity =
M=
moles of solute
liter of solution
125
. m
 .625
2
An alternate method would be to change the 2 liters to 2000 ml and substitute the values in Eq. 5.
M=
(1000)(50)
= .625
(40)(2000)
EXAMPLE 3
Calculate the molarity of a solution containing 5 grams of sodium chloride in 50 grams of solution.
The density of the sodium chloride solution is 1.1 grams per milliliter.
SOLUTION
V=
M
D
volume of solution =
M=
50 g
= 45.5 ml
11
. g / ml
(1000)(5)
= 1.88
(58.5)(455
.)
EXAMPLE 4
How many grams of sodium hydroxide are contained in 350 ml of .15 molar sodium hydroxide
solution?
SOLUTION
( M)( ml)( molecular wt.)
1000
g=
=
(.15)(350)(40)
= 2.1
1000
From Eq. 1
M=
moles of solute
liters of solution
Divide numerator & denominator by 1000
moles of solute
1000
M=
liters of solution
1000
Eq. 7 M =
millimoles
ml
Problems on Molar Solutions
1.
How many grams of NaOH will be required to prepare 1 liter of .5M NaOH?
2.
How many grams of Na2SO4 will be required to prepare 500 ml of .4M Na2SO4?
3.
10 g NaOH are dissolved in enough water to give 450 ml of solution, calculate the molarity of the
solution.
4.
If 300 ml of .2 M Na2SO4 are evaporated to dryness, how many grams of dry Na2SO4 will be
obtained?
5.
20 ml of H2SO4 of density 1.6 g/ml and contains 60% H2SO4 were dissolved in enough water to
give 50 ml of solution. Calculate the molarity of the final solution.
6.
4.9 grams of H2SO4 are dissolved in enough water to give 500 ml of solution. Calculate the
molarity of the solution.
ANSWERS
1.
2.
3.
4.
5.
6.
20 g NaOH
28.4 g Na2SO4
0.56 M
8.5 g Na2SO4
3.92 M
0.10 M