Question

. Calculate the amount of NaOH required to
neutralise 200 ml (0.3 M) H2SO4 solution; it is
reacted as
2NaOH + H2SO4
→ Na2SO4 + 2H2O
(1) 4.8 g
(2) 3.2 g
(3) 1.6 g
(4) 9.2 g​

Answer 1

We have to find the amount of NaOH required to neutralise 200 ml , 0.3 M of H2SO4 solution it is reacted as 2NaOH + H2SO4 => Na2SO4 + 2H2O

solution : concept of neutralisation is ..

no of gram equivalent of acid = no of gram equivalent of base

⇒n - factor of acid × no of moles of acid = n - factor of base × no of moles of Base

here acid is dibasic so the n - factor is 2.

and no of moles = molarity × volume in L

= 0.3 M × 200/1000 L

= 0.3 × 0.2

= 0.06 mol

n - factor of base is 1 because NaOH is monoacid.

now, n - factor of acid × no of moles of acid = n - factor of base × no of moles of Base

⇒2 × 0.06 mol = 1 × no of moles of base

⇒no of moles of base = 0.12 mol

⇒weight of base (NaOH) = 0.12 × molecular weight of NaOH

= 0.12 × (23 + 16 + 1) = 0.12 × 40 = 4.8g

Therefore the amount of NaOH is 4.8 g

Answer 2

Answer:

Ans:4.80g of NaOH

Explanation:

Explanation is in the picture above

Hope this will help!!

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