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 1 M acetic acid required to neutralise NaOH obtained above is

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Detailed Solution

Weight of pure NaCl = 6.5 x 0.9 = 5.85 g

$Number of equivalence of NaCl = \frac{5.85}{58.5} = 0.1$

Number of equivalence of NaOH obtained = 0.1

Volume of I M acetic acid required for the neutralisation of $NaOH = \frac{0.1 \times 1000}{1} = 100 {cm}^{3}$

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