Students $A$ and $B$ derived equivalent weights of $K H C_{2} O_{4}$ quite different .This is because one of the students (say $A$ ) would have derived equivalent weight based on $H^{+}$ which is $=$ molecular weight. Another student (say $B$ ) would have derived equivalent weight based on $C_{2} O_{4}^{2 -}$ (oxalate).
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Solution

$K H C_{2} O_{4} ⇌ K^{+} + H C_{2} O_{4}^{-}$
$H C_{2} O_{4}^{-} ⇌ H^{+} + C_{2} O_{4}^{2 -}$
One of the students (say $A$ ) would have derived equivalent weight based on $H^{+}$ which is $=$ Molecular weight.
Another student (say $B$ ) would have derived equivalent weight based on $C_{2} O_{4}^{2 -}$ (oxalate) which is $\frac{1}{2} \times$ Molecular weight.

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