Some Basic Concepts of Chemistry 2 Question 15
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15. An aqueous solution of $6.3 \mathrm{~g}$ oxalic acid dihydrate is made up to $250 \mathrm{~mL}$. The volume of $0.1 \mathrm{~N} \mathrm{NaOH}$ required to completely neutralise $10 \mathrm{~mL}$ of this solution is $\quad$ (2001,1M)
======= ####15. An aqueous solution of $6.3 \mathrm{~g}$ oxalic acid dihydrate is made up to $250 \mathrm{~mL}$. The volume of $0.1 \mathrm{~N} \mathrm{NaOH}$ required to completely neutralise $10 \mathrm{~mL}$ of this solution is $\quad$ (2001,1M)
3e0f7ab6f6a50373c3f2dbda6ca2533482a77bed (a) $40 \mathrm{~mL}$
(b) $20 \mathrm{~mL}$
(c) $10 \mathrm{~mL}$
(d) $4 \mathrm{~mL}$
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Solution:
- Oxalic acid dihydrate $\mathrm{H}{2} \mathrm{C}{2} \mathrm{O}{4} \cdot 2 \mathrm{H}{2} \mathrm{O}$ : $\mathrm{mw}=126$
It is a dibasic acid, hence equivalent weight $=63$
$$ \begin{array}{llrl} \Rightarrow & & \text { Normality } & =\frac{6.3}{63} \times \frac{1000}{250}=0.4 \mathrm{~N} \ \Rightarrow & & N_{1} V_{1} & =N_{2} V_{2} \ \Rightarrow & 0.1 \times V_{1} & =0.4 \times 10 \ \text { Hence, } & & V_{1} & =40 \mathrm{~mL} \end{array} $$
