Some Basic Concepts of Chemistry - Result Question 13
15. The molecular formula of a commercial resin used for exchanging ions in water softening is $\mathrm{C}_8 \mathrm{H}_7 \mathrm{SO}_3 \mathrm{Na}$ (molecular weight $=206$ ). What would be the maximum uptake of $\mathrm{Ca}^{2+}$ ions by the resin when expressed in mole per gram resin?
(2015 JEE Main)
(a) $\frac{1}{103}$
(b) $\frac{1}{206}$
(c) $\frac{2}{309}$
(d) $\frac{1}{412}$
Show Answer
Answer:
Correct Answer: 15. (d)
Solution:
- We know the molecular weight of $\mathrm{C}_8 \mathrm{H}_7 \mathrm{SO}_3 \mathrm{Na}$
$=12 \times 8+1 \times 7+32+16 \times 3+23=206$
we have to find, mole per gram of resin.
$\therefore \quad 1 \mathrm{~g}$ of $\mathrm{C}_8 \mathrm{H}_7 \mathrm{SO}_3 \mathrm{Na}$ has number of mole
$=\frac{\text { weight of given resin }}{\text { Molecular, weight of resin }}=\frac{1}{206} \mathrm{~mol}$
Now, reaction looks like
$2 \mathrm{C}_8 \mathrm{H}_7 \mathrm{SO}_3 \mathrm{Na}+\mathrm{Ca}^{2+} \longrightarrow\left(\mathrm{C}_8 \mathrm{H}_7 \mathrm{SO}_3\right)_2 \mathrm{Ca}+2 \mathrm{Na}$
$\because 2$ moles of $\mathrm{C}_8 \mathrm{H}_7 \mathrm{SO}_3 \mathrm{Na}$ combines with $1 \mathrm{~mol} \mathrm{Ca}^{2+}$
$\therefore 1$ mole of $\mathrm{C}_8 \mathrm{H}_7 \mathrm{SO}_3 \mathrm{Na}$ will combine with $\frac{1}{2} \mathrm{~mol} \mathrm{Ca}^{2+}$
$\therefore \frac{1}{206}$ mole of $\mathrm{C}_8 \mathrm{H}_7 \mathrm{SO}_3 \mathrm{Na}$ will combine with
$\frac{1}{2} \times \frac{1}{206} \mathrm{~mol} \mathrm{Ca}^{2+}=\frac{1}{412} \mathrm{~mol} \mathrm{Ca}^{2+}$
BETA
