An aqueous solution containing 6.5gm 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:

A. 100 cm3

B. 200 cm3

C. 1000 cm3

D. 2000 cm3

Solution: (A)

Let us calculate the weight of pure NaCl = 6.5 × 0.9 = 5.85 g

No. of equivalence of NaCl = 5.85 / 58.5 = 0.1

No. of equivalence of NaOH obtained = 0.1

Thus, the volume of 1M acetic acid required for the neutralisation of NaOH = 0.1 × 1000 / 1

= 100 cm3

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