20 , mL of a 1 , N solution of KMnO_4 just re | vedclass

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Question

(C) According to the law of equivalence,the number of equivalents of $KMnO_4$ equals the number of equivalents of oxalic acid.$N_1 V_1 = N_2 V_2$Given

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(C) According to the law of equivalence,the number of equivalents of $KMnO_4$ equals the number of equivalents of oxalic acid.$N_1 V_1 = N_2 V_2$Given $N_1 = 1 \, N$,$V_1 = 20 \, mL$,and $V_2 = 20 \, mL$ for oxalic acid.$1 	imes 20 = N_2 	imes 20 \implies N_2 = 1 \, N$Normality $(N) = \frac{	ext{Strength in } g/L}{	ext{Equivalent weight}}$For oxalic acid crystals $(H_2C_2O_4 \cdot 2H_2O)$,the equivalent weight is $63 \, g/eq$.$	ext{Strength} = N 	imes 	ext{Equivalent weight} = 1 	imes 63 = 63 \, g/L$.

$20 \, mL$ of a $1 \, N$ solution of $KMnO_4$ just reacts with $20 \, mL$ of a solution of oxalic acid. The weight of oxalic acid crystals in $1 \, L$ of the solution is $... \, g$

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