Molarity & Molality | Additional Study Material for NEET PDF Download

CONCENTRATION OF SOLUTION
Concentration of solution can be expressed in any of the following ways.
(a) % by weight:
amount of solute dissolved in 100 gm of solution.
4.9% H₂SO₄ by weight:
100 gm of solution contains 4.9 gm of H₂SO₄.

(b) % by volume: 
volume of solute dissolved in 100 ml of solution.
x% H₂SO₄ by volume:
100 ml of solution contains x ml H₂SO₄

(c) % weight by volume:
wt. of solute present in 100 ml of solution.

(d) % volume by weight:
volume of solute present in 100 gm of solution.

CONCENTRATION TERMS
Molarity (M): No. of moles of solute present in 1000 ml of solution.
Molarity (M) = moles of solute/volume of solution (lit)
M = m.moles of solute/volume of solution (ml)

Fig: Molarity

Molality (m): No. of moles of solute present in 1000 gm of solvent.
m = moles of solute/wt. of solvent in kg  
m = m.moles of solute/wt. of solvent in gm  

Normality (N):
No of gm. equivalents of solute present in 1000 ml of solution.


Number of gram equivalent  of solute  = Mass of solute in gram/equivalent weight of solute
Equivalent weight of solute (E)   = Molar mass of solute/Valence factor
(i) Valence factor for base = acidity of base
(ii) Valence factor for acid = basicity of acid
(iii) Valence factor for element = valency
Thus, if one gram equivalent of a solute is present in one litre of the solution,  the concentration of the solution is said to be one normal.

RELATIONSHIP BETWEEN NORMALITY AND MOLARITY
We know that,
Molarity × Molecular mass = Strength of solution (g/L)  
Similarly,  
Normality × Equivalent mass = Normality of the solution (g/L)  
Hence,  
Molarity × Molecular mass = Normality × Equivalent mass    
So, Normality = n × Molarity

Formality (f): It is the number of formula mass in grams present per litre of solution. In case formula mass is equal to molecular mass, formality is equal to molarity. Like molarity and normality, the formality is also dependent on temperature. It is used for ionic compounds in which there is no existence of molecule. Mole of ionic compounds is called formole and molarity as formality.



Where,
w = weight of solute,
f = formula weight of solute
V= volume of solution
n_f = no. of gram formula weight


Mole fraction: The mole fraction of a particular component in a solution is defined as the number of moles of that component per mole of solution.
If a solution has n_A mole of A & n_B mole of B.

And, X_A + X_B = 1

Parts per million (ppm):
 

Volume strength of H₂O₂: Strength of H₂O₂ is represented as 10V, 20V, 30V etc.
20V H₂O₂ means one litre of this sample of H₂O₂ on decomposition gives 20 lit. of O₂ gas at S.T.P.
Decomposition of H₂O₂ is given as:

1 mole = 1/2 × 22.4 lt O₂ at S.T.P.
⇒ 34 g = 11.2 lt O₂ at S.T.P.
To obtain 11.2 litre O₂ at S.T.P., at least 34 gm H₂O₂ must be decomposed, and for 20 lt O₂, we should decompose at least 34/112 × 20 gm H₂O₂.
1 lt solution of H₂O₂ contains 34/112 × 20  gm H₂O₂
1 lt solution of H₂O₂ contains 34/112 × 20/17 equivalents of H₂O_{2 
(EH2O2 = M/2 = 34/2 = 17)}
Normality of H₂O₂ = 34/112 × 20/17
Normality of H₂O₂ (N) 

IInd Method:

From law of equivalence:
gm eq. of O₂ = gm eq. of H₂O₂
gm eq. of O₂ = moles × n factor of O₂, = 20/22.4 x 4 = 20/5.6
gm. eq. of H₂O₂ = 20/5.6
and the volume of H₂O₂ is 1 lit.
this means 1 lit of H₂O₂ have 20/5.6 gm eq.
i.e. Normality N = 20/5.6
Normality of H₂O₂  

Molarity of H₂O₂ (M) 

Strength (in g/l): Denoted by S.
Strength = molarity × mol. wt.
= molarity × 34
strength = Normality × Eq. weight.
= Normality × 17

Example. A bottle labeled with "12V H₂O₂" contain 700 ml solution. If a student mix 300 ml water in it what is the g/ liter strength & normality and volume strength o final solution.
Solution. N = 12/5.6
Meq. of H₂O₂ = 12/5.6 x 100
let the normality of H₂O₂ on dilution be N.
Meq. before dilution = Meq. after dilution
N × 1000 = 12/5.6 x 100 N = 12/5.6 × 7/10 = 1.5 M = 1.5/2
strength gm/lit = 1.5/2 x 34 = 25.5
volume strength = N × 5.6 =  8.4 V 

Strength of Oleum: Oleum is SO₃ dissolved in 100% H₂SO₄. Sometimes, oleum is reported as more than 100% by weight, say y% (where y > 100). This means that (y - 100) grams of water, when added to 100 g of given oleum sample, will combine with all the free SO₃ in the oleum to give 100% sulphuric acid.

Example 1. Calculate the percentage of free SO₃ in an oleum (considered as a solution of SO₃ in H₂SO₄) that is labeled '109% H₂SO₄ '. 
Solution. 
'109% H₂SO₄' refers to the total mass of pure H₂SO₄, i.e., 109 g that will be formed when 100 g of oleum is diluted by 9 g of H₂O which (H₂O) combines with all the free SO₃ present in oleum to form H₂SO₄.
H₂O + SO₃ → H₂SO₄
1 mole of H₂O combines with 1 mole of SO₃
or 18 g of H₂O combines with 80 g of SO₃
or 9 g of H₂O combines with 40 g of SO₃.
Thus, 100 g of oleum contains 40 g of SO₃ or oleum contains 40% of free SO₃.

Example 2. A 62% by mass of an aqueous solution of acid has specific gravity 1.8. This solution is diluted such that the specific gravity of solution became 1.2. Find the % by wt of acid in new solution.
Solution. 
Density = 1.8
⇒ volume of solution =
Let x gm water be added in solution then
d =0.6 x = 100
⇒ x = 166.67
mass of new solution = 100 + 166.67 = 266.67
266.67 gm solution contains 62 gm of acid
% by mass = 23.24 %

Relationship between Molarity, Molality & density of solution: Let the molarity of solution be 'M', molality be 'm' and the density of solution be d gm/m.
Molarity implies that there are M moles of solute in 1000 ml of solution.
wt. of solution = density × volume
= 1000 d gm wt of solute = MM₁
where M₁ is the molecular wt of solute.
wt of solvent = (1000 d - MM₁) gm.
(1000 d - MM₁) gm of solvent contains M moles of solute.
1000 gm of solvent have

 =  Molality no. of moles of solute present in 1000 gm of solvent
 
= Molality on simplifying


Relationship between Molality & mole fraction: Consider a binary solution consisting of two components A (Solute) and B (Solvent).
Let x_A & x_B are the mole fraction of A & B respectively.

If molality of solution be m then

 =

where M_B is the molecular wt of the solvent B

Example: An aqueous solution is 1.33 molal in methanol. Determine the mole fraction of methanol & H₂O.
Solution. 


⇒ x_A = 0.02394 x_B, x_A x_B = 1
⇒ 1.02394 x_B = 1

 = 0.98, x_A = 0.02

Second Method: Let wt. of solvent = 1000 gm, molality = 1.33

= moles of solute
mole fraction of solute = ,

mole fraction of solute = 0.02

mole fraction of solvent = 1 - 0.02 = 0.98

Example. The density of 3 M solution of sodium thiosulphate (Na₂S₂O₃) is 1.25 g/mL. Calculate
(i) amount of sodium thiosulphate
(ii) mole fraction of sodium thiosulphate
(iii) molality of Na  and S₂O₃^2- ions
Solution.
(i) Let us consider one litre of sodium thiosulphate solution
wt. of the solution = density × volume (mL)
= 1.25 × 1000 = 1250 g.
wt. of Na₂S₂O₃ present in 1 L of the solution
= molarity × mol. wt.
= 3 × 158 = 474 g.
wt. % of Na₂S₂O₃ = 37.92%

(ii) Mass of 1 litre solution = 1.25 × 1000 g = 1250 g
[∵ density = 1.25 g/l]
Mole fraction of Na₂S₂O₃ = Number of moles of Na₂S₂O₃/Total number of moles
Moles of water = 1250 – 158 × 3/18 = 43.1
Mole fraction of Na₂S₂O₃ = 3/3 + 43.1 = 0.065

(iii) 1 mole of sodium thiosulphate (Na₂S₂O₃) yields 2 moles
of Na⁺ and 1 mole of S₂O₃^2-
Molality of Na₂S₂O₃ = 3 × 1000/776 = 3.87
Molality of Na⁺ = 3.87 × 2 = 7.74 m
Molality of S₂O^2-₃ = 3.87 m