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Law of Equivalence
Molarity (M)
Normality (N)
Equivalent Mass
Dilution Effects
Volume strength of H2O2 solution
Percentage Labeling of Oleum
Modifications in law of equivalence
Related Resources
Before heading for the law of equivalents, let us first discuss certain definitions.
It is defined as the number of moles of solute present in one litre of solution.
Molarity (M) =
Let the weight of solute be w g, molar mass of solute be M_{1}g/mol and the volume of solution be V litre.
Number of moles of solute =
Hence M
Hence Number of moles of solute
It is defined as the number of equivalents of a solute present in one litre of solution. Equivalent is also the term used for amount of substance like mole with the difference that one equivalent of a substance in different reactions may be different as well as the one equivalent of each substance is also different.
Normality (N) =
Number of equivalents of solute =
Hence N
Hence Number of equivalents of solute = = N × V (in litre)
Equivalents mass =
Hence Number of equivalents of solute
Hence Number of equivalents of solute = n × number of moles of solute
Also,
N = M × n
Hence
Normality of solution = n × molarity of solution
Where n is n- factor or valency factor here.
When a solution is diluted, the moles and equivalents of solute do not change but molarity and normality changes while on taking out a small volume of solution from a larger volume, the molarity and normality of solution do not change but moles and equivalents change proportionately.
In stoichiometry, the biggest problem is that for solving a problem we need to know a balanced chemical reaction. Since the number of chemical reactions are too many, it is not possible to remember all those chemical reactions. So, there is need to develop an approach which does not require the use of balanced chemical reaction. This approach makes use of a law called law of equivalence.
The law of equivalence provide, us the molar ratio of reactants and products without knowing the complete balanced reaction, which is as good as having a balanced chemical reaction. The molar ratio of reactants and products can be known by knowing the n-factor of relevant species.
According to the law of equivalence, whenever two substances react, the equivalents of one will be equal to the equivalents of other and the equivalents of any product will also be equal to that of the reactant.
Let us suppose we have a reaction, A + B → C + D.
In this reaction, the number of moles of electrons lost by 1 mole of A are x and the number of mole of electrons gained by 1 mole of B are y. Since, the number of mole of electrons lost and gained are not sane, the molar ratio which A & B react cannot be 1 : 1.
Thus, if we take y moles of A, then the total moles of electrons lost by y moles of A would be (x × y).
Similarly, if x moles of B are taken, then the total mole of electrons gained by x moles of B would be (y × x).
Thus, the number of electrons lost by A and number of electrons gained by B becomes equal. For reactant A, its n-factor is x and the number of moles used are y.
So, The equivalents of A reacting = moles of A reacting × n-factor of A = y × x.
Similarly, for reactant B, its n-factor is y and the number of moles used are x, So,
The equivalents of B reacting = moles of B reacting × n-factor of B = x × y
Thus, the equivalents of A reacting would be equal to the equivalent of B reacting. Thus, the balancing coefficients of the reactant would be as
yA + xB → C + D
(n-factor = x) (n-factor = y)
The n-factor of A & B are in the ratio of x : y, and their molar ratio is y : x. Thus, molar ratio is inverse of the n-factor ratio.
In general, whenever two substances react with their n-factors in the ratio of a : b, then their molar ratio in a balanced chemical reaction would be b : a.
To get the equivalents of a substance, its n-factor is to be known.
Let the weight of the substance used in the reaction be w g. Then, equivalents of substance reacted would be (where E and M_{1} are the equivalent mass and molar mass of the substance). Thus, in order to calculate the equivalents of substance, knowledge of n-factor is a must.
When a solution of H_{2}O_{2} is labeled as ‘x volume’, it means that 1 volume (1 ml of 1 litre) of H_{2}O_{2} solution would liberate x volumes (1 ml or 1 litre) of O_{2} at STP on complete decomposition.
H_{2}O_{2} H_{2}O + ½O_{2} …….. (i)
If a H_{2}O_{2} solution (acting as reducing agent) has normality N and its is to be reacted with KMnO_{4} solution (acting as oxidizing agent). Our task is to determine its volume strength. We can say that there are N equivalents of H_{2}O_{2} present in 1 litre of this H_{2}O_{2} solution.
1 ml of H_{2}O_{2} of this solution would contain equivalents.
H_{2}O_{2} + KMnO_{4} O_{2} + Mn^{+2} ……..(ii)
Moles of H_{2}O_{2} in 1 ml of this solution = [from equation (ii)]
When these many moles of H_{2}O_{2} in 1 ml of solution are allowed to decompose according to the reaction, H_{2}O_{2} H_{2}O + ½O_{2}, the volume of O_{2} released (in ml) by them at STP will give the volume strength of H_{2}O_{2} solution.
Moles of O_{2} given by 1 ml of this solution = [from equation (i)]
Volume of O_{2} at STP given by 1ml of this solution =
Volume strength of H_{2}O_{2} = 5.6 × Normality of H_{2}O_{2}
Oleum contains H_{2}SO_{4} and SO_{3} only. When oleum is diluted (by adding water), SO_{3} reacts with H_{2}O to form H_{2}SO_{4}, thus increasing the mass of the solution.
SO_{3} + H_{2}O → H_{2}SO_{4}
The total mass of H_{2}SO_{4} obtained by diluting 100 g of oleum sample with required amount of water, is equal to the percentage labeling of oleum.
Percentage labeling of oleum = Total mass of H_{2}SO_{4} present in oleum after dilution. = mass of H_{2}SO_{4} initially present + mass of H_{2}SO_{4} produced on dilution.
If we have a sample of oleum labeled as 109%, this means that 100 g of oleum and dilution gives 109 g of H_{2}SO_{4}.
Let us calculate the composition of oleum, which is labeled as 109%.
Let the mass of SO_{3} in the sample be x g, then the mass of H_{2}SO_{4} would be
(100 – x) g. On dilution,
Moles of SO_{3} in oleum = = Moles of H_{2}SO_{3} formed on dilution
Mass of H_{2}SO_{4} formed on dilution =
Total mass of H_{2}SO_{4} present in oleum after dilution = + (100 – x) = 109
x = 40.
Thus, oleum contained 40% SO_{3} & 60% H_{2}SO_{4}.
Alternatively,
Let the mass of oleum sample be 100 g, which on dilution 109 g. This implies that 9 g of H_{2}O was added.
Moles of H_{2}O added = Moles of SO_{3} present in oleum sample.
Mass of SO_{3} in oleum = × 80 = 40 g
Thus, oleum sample contained 40% SO_{3} and 60% H_{2}SO_{4}.
There are certain modifications required in order to use law of equivalence. These modifications are
The equivalents of a substance produced and reacted may not necessarily be same. If the n-factor of the substance, in the reaction in which it is produced were different than the n-factor of the same substance, when it is reacting, then the equivalent of the substance produced and reacted would be different.
The equivalents of the same substances can be added or subtracted only when they are of the similar n-factor.
In a reaction, the equivalents of oxidizing agents would always be equal to the equivalents of reducing agents, irrespective of the number of agents used in the reaction.
Question 1: What is the mass of HCl present in 500 ml of 1 molar aqueous solution of HCl ?
a. 1 g
b. 36 g
c. 18 g
d. 22 g
Question 2: Which of the following things not change which a solution is diluted?
a. Molarity
b. Molality
c. Normality
d. Equivalents of solute
Question 3:
What would be molaity of a solution in which 5 moles of a solute is present in 2 L of the solution?
a. 5 M
b. 2 M
c.2.5 M
d. 1.5 M
Question 4: In general, whenever two substances react with their n-factors in the ratio of a : b, then their molar ratio in a balanced chemical reaction would be
a. b : a.
b. a : b
c. 2a : b
d. 2b : a.
Q.1
Q.2
Q.3
Q.4
c
d
a
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