Question

$\left(C\right)$ and $\left(D\right)$ both decolouries acidified $KMn{O}_4$. Identify $\left(D\right)$.

• A. $CO$
• B. Sodium Oxalate
• C. Oxalic acid
• D. $HCOONa$

Answer 1

The correct option is C Oxalic acid

D is formed by acidification of C. Therefore it is an acid and not a sodium salt of an acid. Therefore option B and D cannot be the compound D.

Among oxalic acid and $CO$, oxalic acid decolorizes $KMn{O}_4$ and gets oxidized to $C{O}_2$

$\underset{\left(A\right)}{C}O+NaOH\to HCO\underset{\left(B\right)}{O}Na\underset{-H_2}{\overset{\Delta }{\to }}\left(CO\underset{\left(C\right)}{O}Na{)}_2\stackrel{H^{+}}{\to }\right(CO\underset{\left(D\right)}{O}H{)}_2$

$2Mn{O}_4^{-}+C_2{O}_4^{2-}+16H^{+}\to 10C{O}_2+2M{n}^{2+}+8H_{2}O$

therefore option C is correct.