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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)

Molarity & Molality | Additional Study Material for NEETFig: 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.
Molarity & Molality | Additional Study Material for NEET
Molarity & Molality | Additional Study Material for NEET
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.

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What does the term "molarity" represent?
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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.
Molarity & Molality | Additional Study Material for NEET
Molarity & Molality | Additional Study Material for NEET
Molarity & Molality | Additional Study Material for NEET
Where,
w = weight of solute,
f = formula weight of solute
V= volume of solution
nf = no. of gram formula weight

Molarity & Molality | Additional Study Material for NEET

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 nA mole of A & nB mole of B.

Molarity & Molality | Additional Study Material for NEET
And, XA + XB = 1

Parts per million (ppm):
 Molarity & Molality | Additional Study Material for NEET


Volume strength of H2O2Strength 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:
Molarity & Molality | Additional Study Material for NEET

Molarity & Molality | Additional Study Material for NEET

Fig: Chemical reaction


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 H2O2 
(EH2O2 = M/2 = 34/2 = 17)
Normality of H2O2 = 34/112 × 20/17
Normality of H2O(N) 

Molarity & Molality | Additional Study Material for NEET
IInd Method:
Molarity & Molality | Additional Study Material for NEET
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 & Molality | Additional Study Material for NEET
Molarity of H2O(M) 
Molarity & Molality | Additional Study Material for NEET
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.

Molarity & Molality | Additional Study Material for NEET

Fig: Structure of Oleum


Example 1. Calculate the percentage of free SO3 in an oleum (considered as a solution of SO3 in H2SO4) that is labeled '109% H2SO'. 
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
Molarity & Molality | Additional Study Material for NEET
 =  Molality no. of moles of solute present in 1000 gm of solvent
Molarity & Molality | Additional Study Material for NEET 
= Molality on simplifying
Molarity & Molality | Additional Study Material for NEET

Relationship between Molality & mole fraction: Consider a binary solution consisting of two components A (Solute) and B (Solvent).
Let xA & xB are the mole fraction of A & B respectively.
Molarity & Molality | Additional Study Material for NEET
If molality of solution be m then

Molarity & Molality | Additional Study Material for NEET = Molarity & Molality | Additional Study Material for NEET

where MB is the molecular wt of the solvent B
Molarity & Molality | Additional Study Material for NEET

Molarity & Molality | Additional Study Material for NEET


Example: An aqueous solution is 1.33 molal in methanol. Determine the mole fraction of methanol & H2O.
Solution. 
Molarity & Molality | Additional Study Material for NEET
Molarity & Molality | Additional Study Material for NEET
⇒ xA = 0.02394 xB, xA xB = 1
⇒ 1.02394 xB = 1

Molarity & Molality | Additional Study Material for NEET = 0.98, xA = 0.02

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

= moles of solute
mole fraction of solute = Molarity & Molality | Additional Study Material for NEET, Molarity & Molality | Additional Study Material for NEET

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 S2O32- 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 S2O32-
Molality of Na2S2O3 = 3 × 1000/776 = 3.87
Molality of Na+ = 3.87 × 2 = 7.74 m
Molality of S2O2-3 = 3.87 m

The document Molarity & Molality | Additional Study Material for NEET is a part of the NEET Course Additional Study Material for NEET.
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FAQs on Molarity & Molality - Additional Study Material for NEET

1. What is molarity and how is it calculated?
Ans. Molarity is the measure of the concentration of a solution expressed as the number of moles of solute per liter of solution. It is calculated by dividing the number of moles of solute by the volume of the solution in liters.
2. What is the difference between molarity and molality?
Ans. Molarity is the measure of the concentration of a solution expressed in moles of solute per liter of solution, while molality is the measure of the concentration of a solution expressed in moles of solute per kilogram of solvent.
3. How is molality calculated and what is its significance?
Ans. Molality is calculated by dividing the number of moles of solute by the mass of the solvent in kilograms. It is significant because it is independent of temperature and does not change with the addition or removal of solvent.
4. How do you convert between molarity and molality?
Ans. To convert from molarity to molality, you need to know the density of the solution and the molar mass of the solvent. To convert from molality to molarity, you need to know the density of the solution and the molar mass of the solute.
5. What are some practical applications of molarity and molality in industry?
Ans. Molarity and molality are used in a variety of industrial applications, including the production of pharmaceuticals, the manufacturing of chemicals, and the analysis of environmental samples. They are also used in the food and beverage industry to measure the concentration of ingredients in recipes and to ensure consistency in product quality.
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