\(\mathrm{KMnO}_{4}\) reacts with oxalic acid according to the equation
$$
2 \mathrm{MnO}_{4}^{-}+5 \mathrm{C}_{2} \mathrm{O}_{4}^{2-}+16 \mathrm{H}^{+}
\longrightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO}_{2}+8 \mathrm{H}_{2}
\mathrm{O}
$$
Here, \(20 \mathrm{~mL}\) of \(0.1 \mathrm{M} \mathrm{KMnO}_{4}\) is equivalent to
:
(a) \(120 \mathrm{~mL}\) of \(0.25 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2}
\mathrm{O}_{4}\)
(b) \(150 \mathrm{~mL}\) of \(0.10 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2}
\mathrm{O}_{4}\)
(c) \(25 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2}
\mathrm{O}_{4}\)
(d) \(50 \mathrm{~mL}\) of \(0.20 \mathrm{M} \mathrm{H}_{2} \mathrm{C}_{2}
\mathrm{O}_{4}\)