Question

60 ml of N/5 h2so4,10 ml of N/2 hno3 and 30 ml of n/10 hcl are mixed together.the strength of reulting mixture is

Answer 1

We know

Meq= N × V

where Meq is the equivalent molarity

N is normality of the solution

V is volume of solution

Applying the formula to each of the solution separately

1) HCl

Meq= 1× 5 = 5M

2) H2SO4

Meq=1÷2× 20 = 10M

3) HNO3

Meq= 1÷3×30 = 10 M

For total solution

Meq = M1 + M2 + M3

Meq = 5+10+10 = 25 M

and volume = 1000 mL

Thus applying the same formula

Meq= N × V

25 = N × 1000

N= 25÷1000 N

= N÷40

Thus normality of the solution would be N÷40. Hence correct answer is d.

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Answer 2

Answer:

The strength of resulting mixture of H₂SO₄, HNO₃ and HCl is equal to 0.2N.

Explanation:

Given:

The volume of H₂SO₄ in solution $V_{1}=60ml$

The volume of HNO₃ in the solution $V_{2}=10ml$

The volume of HCl in solution $V_{3}=30ml$

Normality of  Sulphuric acid $N_{1}=\frac{N}{5}$

Normality of Nitric acid $N_{2}=\frac{N}{2}$

Normality of hydrochloric acid $N_{3}=\frac{N}{10}$

The strength of resulting solution is given by;

Strength $=\frac{N_{1}V_{1}+N_{2}V_{2}+N_{3}V_{3}}{V_{1}+V_{2}+V_{3}}$

               $=\frac{60*\frac{N}{5}+10*\frac{N}{2} +30*\frac{N}{10} }{60+10+30}$

                $=\frac{12N+5N+3N}{100}$

                $=\frac{20N}{100}$

                $=0.2N$

Therefore, the strength of resulting mixture will be equal to 0.2N.