Question:
$10.0 \mathrm{~mL}$ of $0.05 \mathrm{M} \mathrm{KMnO}_{4}$ solution was consumed in a titration with $10.0 \mathrm{~mL}$ of given oxalic acid dihydrate solution. The strength of given oxalic acid solution is ........ $\times 10^{-2} \mathrm{~g} / \mathrm{L}$.
(Round off to the nearest integer)
Solution:
$\mathrm{n}_{\mathrm{eq}} \mathrm{KMnO}_{4}=\mathrm{n}_{\mathrm{eq}} \mathrm{H}_{2} \mathrm{C}_{2} \mathrm{O}_{4} \cdot 2 \mathrm{H}_{2} \mathrm{O}$
or,$\frac{10 \times 0.05}{1000} \times 5=\frac{10 \times \mathrm{M}}{1000} \times 2$
$\therefore$ Conc. of oxalic acid solution $=0.125 \mathrm{M}$
$=0.125 \times 126 \mathrm{~g} / \mathrm{L}=15.75 \mathrm{~g} / \mathrm{L}$
$=1575 \times 10^{-2} \mathrm{~g} / \mathrm{L}$