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The
mole concept and solutions
Francine
Taylor-Campbell, Contributor
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A
glimpse of the 2005 National
Junior and Senior Athletic Championships
at the National Stadium on Sunday,
June 26, 2005. - Ricardo Makyn
Photo
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POINTS
TO NOTE
Concentration is expressed as the
amount of solute in a given volume
of solution.
Concentration is expressed in units
of moles of solute in 1dm3 of solution
(mol/dm³).
Mass concentration is the mass of the
solute in 1dm3 of solution (g/dm³).
A
standard solution is a solution, the
concentration of which, in mol/dm³
is known or can be calculated.
Eg.
Calculate the mass of sodium carbonate
needed to make up 500cm³ of a
0.20M solution.
M
(molarity) is the number of moles
in 1dm³, that is mol/dm3.
Thus,
0.2M = 0.2 mol in 1000cm³ (1dm³)
X
mol == 500cm³
X
= (500x0.2)/1000 = 0.1 mol
Mr
of Na2CO3
= 106g, thus the mass of 0.1 mol =
106x0.1 = 10.6g
10.6g
of Na2CO3
dissolved in 500cm³ has a concentration
of 0.2M
ALTERNATIVE
METHOD
What
mass of NaOH is needed to make up
250cm³ of a 2M solution.
2M
= 2 mol in 1000cm³ (1dm³)
Mr
of NaOH = 40g then 2 mol = 80g
80g
= 1000cm³
Xg
= 250cm³
X
= (250x80)/1000 = 20g of NaOH
CONCENTRATION
CALCULATION
As
stated earlier, concentration can
be expressed in mol/dm³ and g/dm³.
Eg.
What is the concentration of 20cm³
of sulphuric acid containing 0.25
mol H2SO4?
0.25
mol = 250 cm³
x mol = 1000cm³ (1dm³)
x
= (0.25x1000)/250 = 1 mol
Concentration
= 1 mol/dm³
Calculate
the concentration in g/dm³ of
the same solution?
Since,
0.25 mol H2SO4
is present in 250cm³
Mr
of H2SO4
= 98g mass of 0.25 mol = 98x0.25 =
24.5g
Thus,
24.5g = 250cm³
X
g = 1000cm³
X
= (1000x24.5)/250 = 98g
Concentration
= 98g/dm³
ALTERNATIVE
METHOD
Since
the Concentration = 1 mol/dm³
Concentration
in g/dm³ = 1 mol/dm³ x Mr
= 1 mol/dm³ x 98 g/mol = 98g/dm³
Let
us examine some questions on titration
which require application of the mole
concept.
Consider
the following: 24 cm³ of H2SO4
of concentration 5.20g/dm³ was
titrated against NaOH solution and
required 25 cm³ of NaOH for complete
neutralization. Calculate the concentration
of the NaOH solution.
First
of all a balanced equation is needed
to determine the mole ratio in which
the reactants combine
2NaOH
(aq) + H2SO4(aq)
= Na2SO4(aq)
+ 2H2O(l)
2
mol of NaOH react with 1 mol of H2SO4.
Ratio is 2:1
Concentration
of H2SO4
in mol/dm³ = (5.20g/dm³)/98g/mol
= 0.053mol/dm³
Next
calculate the number of moles of H2SO4
in 24cm³
0.053mol
H2SO4
= 1000cm³ (1dm³)
x
mol = 24cm³
x
= (24x0.053)/1000 = 0.00127 mol
Since
NaOH reacts with H2SO4
in the ratio 2:1
Then,
the number of moles of NaOH that would
have reacted = 0.00127 x 2= 0.00254mol
0.00254
mol NaOH is present in 25cm³
x
mol = 1000cm³
x
= (1000x0.00254)/25 = 0.102 mol
Concentration
= 0.102 mol/dm³
Concentration
in g/dm³ = 0.102 mol/dm³
x Mr = 0.102 x 40 = 4.07g/dm³
*
Francine Taylor-Campbell is
an independent contributor.
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