(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

An element of mass $A_1 \, g$ gives a chloride of mass $A_2 \, g$. Calculate its equivalent mass.

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Calculate the atomic mass (average) of chlorine using the following data:

Isotope $\%$ Natural Abundance Molar Mass $(g/mol)$

$^{35}Cl$ $75.77$ $34.9689$

$^{37}Cl$ $24.23$ $36.9659$

(in $u$)

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The relative abundance of two isotopes of atomic weight $85$ and $87$ is $75\%$ and $25\%$ respectively. The average atomic weight of the element is:

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Accurate determination of atomic masses is done with the instrument called as

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Find out the equivalent mass of $KIO_{3}$ in the given reaction.
$2Cr(OH)_{3} + 4OH^{-} + KIO_{3} \to 2CrO_{4}^{2-} + 5H_{2}O + KI$
(Molecular Mass $= M$)

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Isotope	$\%$ Natural Abundance	Molar Mass $(g/mol)$
$^{35}Cl$	$75.77$	$34.9689$
$^{37}Cl$	$24.23$	$36.9659$

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

An element of mass $A_1 \, g$ gives a chloride of mass $A_2 \, g$. Calculate its equivalent mass.

Calculate the atomic mass (average) of chlorine using the following data: Isotope $\%$ Natural Abundance Molar Mass $(g/mol)$ $^{35}Cl$ $75.77$ $34.9689$ $^{37}Cl$ $24.23$ $36.9659$ (in $u$)

The relative abundance of two isotopes of atomic weight $85$ and $87$ is $75\%$ and $25\%$ respectively. The average atomic weight of the element is:

Accurate determination of atomic masses is done with the instrument called as

Find out the equivalent mass of $KIO_{3}$ in the given reaction.$2Cr(OH)_{3} + 4OH^{-} + KIO_{3} \to 2CrO_{4}^{2-} + 5H_{2}O + KI$(Molecular Mass $= M$)

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Calculate the atomic mass (average) of chlorine using the following data:

Isotope $\%$ Natural Abundance Molar Mass $(g/mol)$

$^{35}Cl$ $75.77$ $34.9689$

$^{37}Cl$ $24.23$ $36.9659$

(in $u$)

Find out the equivalent mass of $KIO_{3}$ in the given reaction.
$2Cr(OH)_{3} + 4OH^{-} + KIO_{3} \to 2CrO_{4}^{2-} + 5H_{2}O + KI$
(Molecular Mass $= M$)