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(D) For $KHC_2O_4 \cdot H_2C_2O_4$ acting as an acid,the number of replaceable $H^+$ ions is $3$ ($1$ from $KHC_2O_4$ and $2$ from $H_2C_2O_4$).Theref

Vedclass · Accepted Answer

$KHC_2O_4 \cdot H_2C_2O_4$ (As acid) $\rightarrow E = \frac{M}{4}$

Vedclass · Answer

(D) For $KHC_2O_4 \cdot H_2C_2O_4$ acting as an acid,the number of replaceable $H^+$ ions is $3$ ($1$ from $KHC_2O_4$ and $2$ from $H_2C_2O_4$).Therefore,the equivalent weight $E = \frac{M}{3}$.Option $D$ states $E = \frac{M}{4}$,which is incorrect.For $KHC_2O_4 \cdot H_2C_2O_4$ acting as a reducing agent,the total change in oxidation state of carbon is $2 	imes (4-3) + 3 	imes (4-3) = 5$ (or calculation based on total electrons lost),resulting in $E = \frac{M}{4}$ is also context-dependent but $D$ is definitely incorrect as an acid.

Which of the following is not the correct match for equivalent weight :-

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