1. Methods of Expressing concentration of a Solution
(i) Percentage
(1) For a solid dissolved in a liquid percentage by mass means the mass of the solute in
grams present in 100 g of the solution. Percentage by volume means the mass of solute
in grams present in 100 mL of the solution.
(2) For a liquid dissolved is another liquid.
Percentage by mass has the same meaning as above. Percentage by volume means the
volume of the solute in mL present in 100 mL of the solution.
(ii) Strength.
It is the amount of the solute in grams present I one litre of the solution.
ππ‘πππππ‘β = π΄πππ’ππ‘ ππ π πππ’π‘π ππ π
ππππ’ππ ππ π πππ’π‘πππ ππ πΏ
(iii) Molarity
It is the number of moles of the solute present in one litre of the solution. Its symbol is M.
πππππππ‘π¦ = πππππ ππ π‘βπ π πππ’π‘π
ππππ’ππ ππ π πππ’π‘πππ ππ πΏ
(iv) Formality
It is the number of formula weights of the solute present per litre of the solution.
Its symbols is F.
(v) Normality.
It is the number of gram equivalents of the solute present in one litre of the
solution. Its symbol is N.
ππππππππ‘π¦ = ππ. ππ π ππ. ππ π πππ’π‘π/ππππ’ππ ππ π πππ’π‘πππ ππ πΏ
(πΈπ. π€π‘ ππ ππππ = πππ. πππ π /π΅ππ ππππ‘π¦ )
(πΈπ. π€π‘ ππ πππ π = πππ.πππ π /π΄πππππ‘π¦ )
(πΈπ. π€π‘. ππ π πππ‘ = πππ. πππ π /πππ‘ππ πππ ππ‘ππ£π π£ππππππ¦ ππ πππ‘ππ ππ‘πππ )
(vi) Molality
It is the number of moles of the solute dissolved in 1000 g (1 kg) of the solvent. Its
symbol is m.
πππππππ‘π¦ = ππ. ππ πππππ ππ π πππ’π‘π/πππ π ππ π πππ£πππ‘ ππ ππ
(vii) Mole fraction.
Mole fraction of any component in the solution is the number of moles of that
component divided by the total number of moles of all the components. For a binary
solution containing π2 moles of the solute in π1 moles of the solvent.
Mole fraction of solute (π₯2) = π2/π1+π2
Mole fraction of solvent (π₯1) = π1/π1+π2
Hence, π₯1 + π₯2 = 1
On multiplying mole fraction with 100, we get mole percent.
(viii) Parts per million parts (ppm)
Ppm of any solute in the solution is the mass of the solute present in milli (106)
Parts by mass of the solution
ππππ΄ = πππ π ππ π΄/πππ π ππ π πππ’π‘πππ Γ 106
(ix) Ionic Strength.
The ionic strength I of the solution is a measure of the electrical intensity due to
presence of ions in the solution. It is equal to half of the sum of all the terms obtained
by multiplying the molality of each ion by the square of its valency.
2. Relationship between concentration terms.
(i) Relationship between molality (m) of the solution and mole fraction (ππ) of the solute.
π₯2 = π π1/1 + ππ1
Where π1 = molecular mass of the solvent.
Proof.
Molality (m) means m moles of solute in 1 kg of solvent. But 1 kg of solvent =1/π1 moles
where π1 is molar mass of the solvent in ππ πππβ1
.hence.
π₯2 = π/(π + 1/π1) = π π1/1 + π π1
(ii) Relationship between molality (m) and molarity (M) of the solution.
π = π/(π β ππ2)
Where d = density of solution in KgπΏβ1
π2 = molar mass of solute in ππ πππβ1
Or π = π/(1000 πβππ2)Γ 1000
When d = density of solution in π ππΏβ1
And π2= molar mass of solute in π πππβ1
(iii) Relationship between mole fraction of solute (ππ)and molarity (M) of the solution
π₯2 =ππ1/[π(π1 β π2) + π]
Where π2 = molar mass of solute in ππ πππβ1
π1 =molar mass of solvent in ππ πππβ1
d= density of solution in ππ πΏβ1
(iv) Relationship between molality (m) and solubility.
Solubility of a solute means grams of solute dissolved in 100 g of the solvent to make a
saturated solution at a particular temperature. Thus, solubility βSβ means S grams of solute
in 100 g of solvent.
S grams of solute= π/π2 moles (π2 = Mol. Mass of solute)
β΄ Moles of solute in 100 g of solvent, i.e., molality (m)
= 10π/π2
β΄ π =10π/π2
3. Calculation of Normality/Molarity on Mixing/Titration
(i) Calculation of normality and molarity of solutions on mixing.
If π1 ππ of a solution with normality π1 is mixed with π2 cc of the solution of the same solute
with normality π2, then normality π3 of the final solution can be calculated using the
equation,
π1π1 + π2π2 = π3(π1 + π2)
Similarity, in terms of molarities,
π1π1 + π2π2 = π3(π1 + π2)
(ii) Calculation of normality and molarity in titrations.
For normality, we apply normality
equations viz
π1π1 = π2π2
For molarity, first balanced equation is written for the reaction. If π1 πππ π2 are the
coefficients of the reactants, we apply molarity equation as under:
π1π1/π1 = π2π2/π2
Comparison of 1 M and 1 m solutions
One molar (1M) aqueous solution in more concentrated than one molal aqueous (1m) solution
of the same solute. This is because 1 M solution contains 1 mole of the solute in 1000 cc of the
solution which includes the solute, i.e., mass of solvent is less than 1000 g (as density of π»2π =
1 π/ππ). Thus 1 M solution contains 1 mole of the solute is less than 1000 g of the solvent
whereas 1m solution has 1 mole of the solute in 1000 g of the solvent. Hence, 1 M is more
concentrated than 1 m. However, in case of non-aqueous solution, 1 π > 1 π ππ 1π <
1 π ππ 1 π = 1π depending upon the density of the solution. Solubility of a substance is its maximum amount that can be dissolved in a specified amount of the solvent. It depends upon
the nature of the solute and solvent as well as temperature and pressure.
LETβS TRY OUT!
1. 4 L of 0.02 M aqueous solution of NaCl was diluted by adding one litre of water. The molality of the resultant solution is __________.
a) 0.004 m
b) 0.008 m
c) 0.012 m
d) 0.016 m