Q. An aqueous solution of 6.3 g oxalic acid dihydrate is made up to 250 mL. The volume of 0.1 N sodium hydroxide required to completely neutralise 10 mL of this solution is
A) 40 mL
B) 20 mL
C) 10 mL
D) 4 mL
Explanation
(a)
$\ W = \ \frac{NE\ V}{1000}\$
$N = \frac{W \times 1000}{E \times V}$
=$\frac{6.3 \times 1000}{63 \times 250}$ =0.4N
$\ \ N_{1}V_{1} = N_{2}V_{2}$
$\ 0.1 \times \ V_{1}\ = \ 0.4\ \times \ 10\$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ V_{1} = \frac{\ 0.4\ \times \ 10}{0.1}$
$V_{1}$ = 40 mL
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.