Sulphuric acid reacts with sodium hydroxide as follows H2SO4+2NaOH⟶Na2SO4+2H2OWhen 1 L of 0.1 M sulphuric acid solution is allowed to react with 1 L of 0.1 M sodium hydroxide solution, the amount of sodium sulphate formed and its molarity in the solution obtained are

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3.55 g, 0.1M

7.10 g,0.025 M

5.33 g,0.01M

5.33 g,0.25M

answer is B.

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Detailed Solution

Moles of H2SO4=1L×0.1M=0.1molH2SO41 mol+2 NaOH2 mol→Na2SO41 mol+2H2O 0.1 mol 0.1 mol 0.05mol∴ Molarity mass of Na2SO4=0.05×142=7.1g∴ Molar mass of Na2SO4=142 Molarity = moles of Na2SO4 volume of solution =0.051+1=0.025 mol L−1

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