Question

The molecular formula of a commercial resin used for exchanging ions in water softening is $C_8{H}_{7}S{O}_{3}Na$ (molecular weight = 206). What would be the maximum uptake of $C{a}^{2+}$ ions by the resin when expressed in mole per gram resin?

• A. $\frac{1}{103}$
• B. $\frac{1}{206}$
• C. $\frac{2}{309}$
• D. $\frac{1}{412}$

Answer 1

The correct option is D

$\frac{1}{412}$

We know the molecular weight of $C_8{H}_{7}S{O}_{3}Na$
$=12\times 8+1\times 7+32+16\times 3+23=206$
we have to find, mole per gram of resin.
$\therefore 1gof{C}_8{H}_{7}S{O}_{3}Na$ has number of mole
$=\frac{weightofgivenresin}{Molecularweightofresin}=\frac{1}{206}mol$
Now, reaction looks like,
$2C_8{H}_{7}S{O}_{3}Na+C{a}^{2+}\to (C_8{H}_{7}S{O}_3{)}_{2}Ca+2Na$
$\because$ 2 moles of $C_8{H}_{7}S{O}_{3}Na$ combines with 1 mol $C{a}^{2+}$
$\because$ 1 moles of $C_8{H}_{7}S{O}_{3}Na$ will combine with $\frac{1}{2}molC{a}^{2+}$
$\because \frac{1}{206}$ moles of $C_8{H}_{7}S{O}_{3}Na$ will combine with $\frac{1}{2}\times \frac{1}{206}molC{a}^{2+}=\frac{1}{412}molC{a}^{2+}$