Sulphuric acid reacts with sodium hydroxide as follows:

H₂SO₄ + 2NaOH → Na₂SO₄+ 2H₂O

When 1L of 0.1M sulphuric acid solution is allowed to react with 1L of 0.1M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained is

(i) 0.1 mol L-1

(ii) 7.10 g

(iii) 0.025 mol L-1

(iv) 3.55 g

 

Answer:

Correct Answers: (ii) 7.10 g; (iii) 0.025 mol L-1

Explanation:

Given,

0.1 mole of NaOH produces 0.05 mole

Mass of Na₂SO₄​=0.05× (Molar mass of Na₂SO₄) = 7.10 g

Volume of the solution after mixing is 2 litres

Hence, Molarity of Na₂SO₄​is 0.025 mol L−1