**H _{2}SO_{4} + 2NaOH → Na_{2}SO_{4}+ 2H_{2}O**

**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 Na_{2}SO_{4}=0.05× (Molar mass of Na_{2}SO_{4}) = 7.10 g

Volume of the solution after mixing is 2 litres

Hence, Molarity of Na_{2}SO_{4}is 0.025 mol L−1