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Q. An aqueous solution containing 6.5 g of $ NaCl $ of 90 purity was subjected to electrolysis. After the complete electrolysis, the solution was evaporated to get solid $ NaOH $ . The volume of 1M acetic acid required to neutralise $ NaOH $ obtained above is

MGIMS WardhaMGIMS Wardha 2014

Solution:

Weight of pure $ NaCl=6.5\times 0.9=5.85\text{ }g $ Number of equivalence of $ NaCl=\frac{5.85}{58.5}=0.1 $ Number of equivalence of $ NaOH $ obtained $ =0.1 $ Volume of 1 M acetic acid required for the neutralisation of $ NaOH=\frac{0.1\times 1000}{1}=100\text{ }c{{m}^{3}} $