Some Basic Concepts of Chemistry - Result Question 77
####15. An aqueous solution of $6.3 g$ oxalic acid dihydrate is made up to $250 mL$. The volume of $0.1 N NaOH$ required to completely neutralise $10 mL$ of this solution is $\quad$ (2001,1M)
(a) $40 mL$
(b) $20 mL$
(c) $10 mL$
(d) $4 mL$
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Solution:
- We know the molecular weight of $C _8 H _7 SO _3 Na$
$$ =12 \times 8+1 \times 7+32+16 \times 3+23=206 $$
we have to find, mole per gram of resin.
$\therefore \quad 1 g$ of $C _8 H _7 SO _3 Na$ has number of mole
$$ =\frac{\text { weight of given resin }}{\text { Molecular, weight of resin }}=\frac{1}{206} mol $$
Now, reaction looks like
$$ 2 C _8 H _7 SO _3 Na+Ca^{2+} \longrightarrow\left(C _8 H _7 SO _3\right) _2 Ca+2 Na $$
$\because 2$ moles of $C _8 H _7 SO _3 Na$ combines with $1 mol Ca^{2+}$
$\therefore 1$ mole of $C _8 H _7 SO _3 Na$ will combine with $\frac{1}{2}$ mol Ca ${ }^{2+}$
$\therefore \frac{1}{206}$ mole of $C _8 H _7 SO _3 Na$ will combine with
$$ \frac{1}{2} \times \frac{1}{206} mol Ca^{2+}=\frac{1}{412} mol Ca^{2+} $$
