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10 ml. of N-HCl, 20 ml of N/2 H₂SO₄ and 30 ml of N/3 HNO₃ are mixed together and volume made to one litre. The normality of the resulting solution is
- a)N/2
- b)N/10
- c)N/20
- d)N/40
Correct answer is option 'A'. Can you explain this answer?
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10 ml. of N-HCl, 20 ml of N/2 HSO and 30 ml of N/3 HNO are mixed toget...
Calculation of Normality of the resulting solution
Given:
Volume of N-HCl = 10 ml
Volume of N/2 H2SO4 = 20 ml
Volume of N/3 HNO3 = 30 ml
Total volume after mixing = 1 litre
Let's calculate the normality of each acid first.
Normality of N-HCl:
N1V1 = N2V2
1N x 10ml = N1 x 1000ml
N1 = 0.1N
Normality of N/2 H2SO4:
N1V1 = N2V2
0.5N x 20ml = N1 x 1000ml
N1 = 0.01N
Normality of N/3 HNO3:
N1V1 = N2V2
(1/3)N x 30ml = N1 x 1000ml
N1 = 0.01N
Now, let's calculate the normality of the resulting solution.
Total equivalents of HCl = 0.1N x 10ml = 1 eq
Total equivalents of H2SO4 = 0.5N x 20ml = 5 eq
Total equivalents of HNO3 = (1/3)N x 30ml = 10 eq
Total equivalents of the resulting solution = 1 + 5 + 10 = 16 eq
Total volume of the resulting solution = 1 litre = 1000 ml
Normality of the resulting solution = Total equivalents / Total volume
= 16 eq / 1000 ml
= 0.016N
Therefore, the normality of the resulting solution is N/2 (0.016N ≈ N/2).
Hence, option A is the correct answer.
Note: In this question, we have used the formula N1V1 = N2V2 to calculate the normality of each acid. This formula is based on the principle of equivalence, which states that the number of equivalents of reactants in a chemical reaction is equal to the number of equivalents of products. In other words, the amount of acid in moles (or millimoles) is proportional to the amount of base (or alkali) required to neutralize it.
Given:
Volume of N-HCl = 10 ml
Volume of N/2 H2SO4 = 20 ml
Volume of N/3 HNO3 = 30 ml
Total volume after mixing = 1 litre
Let's calculate the normality of each acid first.
Normality of N-HCl:
N1V1 = N2V2
1N x 10ml = N1 x 1000ml
N1 = 0.1N
Normality of N/2 H2SO4:
N1V1 = N2V2
0.5N x 20ml = N1 x 1000ml
N1 = 0.01N
Normality of N/3 HNO3:
N1V1 = N2V2
(1/3)N x 30ml = N1 x 1000ml
N1 = 0.01N
Now, let's calculate the normality of the resulting solution.
Total equivalents of HCl = 0.1N x 10ml = 1 eq
Total equivalents of H2SO4 = 0.5N x 20ml = 5 eq
Total equivalents of HNO3 = (1/3)N x 30ml = 10 eq
Total equivalents of the resulting solution = 1 + 5 + 10 = 16 eq
Total volume of the resulting solution = 1 litre = 1000 ml
Normality of the resulting solution = Total equivalents / Total volume
= 16 eq / 1000 ml
= 0.016N
Therefore, the normality of the resulting solution is N/2 (0.016N ≈ N/2).
Hence, option A is the correct answer.
Note: In this question, we have used the formula N1V1 = N2V2 to calculate the normality of each acid. This formula is based on the principle of equivalence, which states that the number of equivalents of reactants in a chemical reaction is equal to the number of equivalents of products. In other words, the amount of acid in moles (or millimoles) is proportional to the amount of base (or alkali) required to neutralize it.
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10 ml. of N-HCl, 20 ml of N/2 HSO and 30 ml of N/3 HNO are mixed together and volume made to one litre. The normality of the resulting solution isa)N/2b)N/10c)N/20d)N/40Correct answer is option 'A'. Can you explain this answer?
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