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 1 M acetic acid required to neutralise NaOH obtained above is
A) $1000\ {cm}^{3}$
B) $2000\ {cm}^{3}$
C) $100\ {cm}^{3}$
D) $200\ {cm}^{3}$
Explanation
(c)
Weight of pure NaCl=$6.5 \times 0.9 = 5.85$ g
No. of equivalent of NaCl$= \frac{5.85}{58.5} = 0.1$
No. of equivalent of NaOH obtained=0.1
Volume of 1 M acetic acid required for the neutralisation of
$NaOH = \frac{0.1 \times 1000}{1}$
$= 100\ cm^{3}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.