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Sulphuric acid reacts with sodium hydroxide as follows : 

H2SO4 + 2NaOH → Na2SO4 + 2H2

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

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(ii), (iii)

(ii) 7.10 g (iii) 0.025 mol L–1

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