Question

What is the strength in g per litre of a solution of H2SO4, 12
ml of which neutralized 15 mL of N/10 NaOH solution?
a. 6.125 g/litre
b. 25 g/litre
c. 275 g/litre
d. 2.5 g/litre​

Answer 1

Answer:

Molarity is 0.6125 gram/litre

Explanation:

For calculating the strength we must know the balanced chemical equation of the reaction

that is

$2NaOH+H_{2}SO_{4} = Na_{2}SO_{4}+2H_{2}O$

as we can clearly see

we need 2 equivalents of NAOH to neutralize One equivalent of H2SO4.

now using the formula of normality

N1V1 = N2V2

where N1= NORMALITY of NAOH

N2 = NORMALITY of H2SO4

V1 = VOLUME of NAOH

V2 = VOLUME of H2SO4

(1/10)× 15 = X×12

→X = 0.125 N

As we know that n-factor of H2SO4 is 2 therefore normality of H2SO4 is twice the molarity since

Normality(N) = n-factor × Molarity(M)

therefore strength/ Molarity of H2SO4 solution is

Molarity = Normality/n-factor

Molarity = 0.125/2

Molarity = 0.6125 gram/ litre

Answer 2

Applying

N1V1(NaOH)=N2V2 (H2SO4 )

1/10×15=N 2×12

N2 = 10×12/15=0.125

Normality×Eq.mass=Strength(g/L)

Strength=0.125×49=6.125 g/L.

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