(B) The equivalent weight $(E)$ of $KMnO_4$ in an acidic medium is calculated as $E = \frac{\text{Molar Mass}}{n\text{-factor}}$.For $KMnO_4$ in an ac

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(B) The equivalent weight $(E)$ of $KMnO_4$ in an acidic medium is calculated as $E = \frac{\text{Molar Mass}}{n\text{-factor}}$.For $KMnO_4$ in an acidic medium,the $n\text{-factor}$ is $5$ (as $MnO_4^-$ is reduced to $Mn^{2+}$).Molar mass of $KMnO_4 = 39 + 55 + (4 \times 16) = 158 \ g/mol$.Therefore,$E = \frac{158}{5} = 31.6 \ g/eq$.For a $1 \ N$ solution in $1 \ L$,the required mass is $N \times E \times V(L) = 1 \times 31.6 \times 1 = 31.6 \ g$.

Class 11 Chemistry Some Basic Concepts of Chemistry Atomic, Molecular and Equivalent masses

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In order to prepare one litre of a $1 \ N$ (normal) solution of $KMnO_4$,how many grams of $KMnO_4$ are required if the solution is used in an acidic medium for oxidation?

A
$158$
B
$31.6$
C
$790$
D
$62$

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Some Basic Concepts of Chemistry

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Atomic, Molecular and Equivalent masses

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In order to prepare one litre of a $1 \ N$ (normal) solution of $KMnO_4$,how many grams of $KMnO_4$ are required if the solution is used in an acidic medium for oxidation?

Similar Questions

$0.126 \ g$ of an acid requires $20 \ mL$ of $0.1 \ N$ $NaOH$ for complete neutralization. The equivalent weight of the acid is:

In a Victor Meyer's experiment,$116 \, mg$ of a compound is vaporized,and it displaces $44.8 \, mL$ of air at $S.T.P.$. The molar mass of the compound is .....

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Sulfur forms two chlorides,$S_2Cl_2$ and $SCl_2$. The equivalent mass of sulfur in $SCl_2$ is $16$. What is the equivalent mass of sulfur in $S_2Cl_2$?

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