20 , mL of 1 , N , KMnO_4 solution reacts wit | vedclass

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Question

(C) According to the law of equivalence,the number of gram equivalents of $KMnO_4$ must be equal to the number of gram equivalents of oxalic acid.$N_1

Vedclass · Accepted Answer

$63$

Vedclass · Answer

(C) According to the law of equivalence,the number of gram equivalents of $KMnO_4$ must be equal to the number of gram equivalents of oxalic acid.$N_1 V_1 = N_2 V_2$Given: $N_1 = 1 \, N$,$V_1 = 20 \, mL$,$V_2 = 20 \, mL$.$1 	imes 20 = N_2 	imes 20 \implies N_2 = 1 \, N$.Normality $(N)$ is defined as $\frac{	ext{Number of equivalents}}{	ext{Volume in Liters}}$.Number of equivalents = $N 	imes V(L) = 1 	imes 0.020 = 0.02 \, eq$.For oxalic acid dihydrate $(H_2C_2O_4 \cdot 2H_2O)$,the n-factor is $2$.Equivalent weight = $\frac{	ext{Molar mass}}{n	ext{-factor}} = \frac{126}{2} = 63 \, g/eq$.Weight = $	ext{Number of equivalents} 	imes 	ext{Equivalent weight} = 0.02 	imes 63 = 1.26 \, g$.Wait,re-evaluating the question context: If the question implies the weight of oxalic acid in $1 \, L$ of $1 \, N$ solution,then Weight = $N 	imes 	ext{Equivalent weight} 	imes V(L) = 1 	imes 63 	imes 1 = 63 \, g$.

$20 \, mL$ of $1 \, N \, KMnO_4$ solution reacts with $20 \, mL$ of oxalic acid solution. The weight of oxalic acid crystals $(H_2C_2O_4 \cdot 2H_2O)$ in the $1 \, N$ solution is . . . . . . $g$.

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