The amount of KMnO_4 required to prepare 100 | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) In an alkaline medium,$KMnO_4$ acts as an oxidizing agent as follows:$MnO_4^- + e^- 	o MnO_4^{2-}$Here,the change in oxidation state of $Mn$ is f

Vedclass · Accepted Answer

$1.58$

Vedclass · Answer

(A) In an alkaline medium,$KMnO_4$ acts as an oxidizing agent as follows:$MnO_4^- + e^- 	o MnO_4^{2-}$Here,the change in oxidation state of $Mn$ is from $+7$ to $+6$,so the $n$-factor is $1$.Therefore,the equivalent weight of $KMnO_4 = \frac{	ext{Molar mass}}{n	ext{-factor}} = \frac{158}{1} = 158 \ g/eq$.Using the formula for normality: $	ext{Normality} = \frac{	ext{Mass}}{	ext{Equivalent mass} 	imes 	ext{Volume in } L}$.$0.1 = \frac{	ext{Mass}}{158 	imes (100/1000)}$.$	ext{Mass} = 0.1 	imes 158 	imes 0.1 = 1.58 \ g$.

The amount of $KMnO_4$ required to prepare $100 \ mL$ of $0.1 \ N$ solution in alkaline medium is ............... $g$.

Similar Questions

If $50 \ mL$ of $0.5 \ M$ oxalic acid is required to neutralize $25 \ mL$ of $NaOH$ solution, the amount of $NaOH$ in $50 \ mL$ of the given $NaOH$ solution is . . . . . . $g$.

If $V \ mL$ of a substance at $NTP$ weighs $W \ g$,then the molecular weight of the substance is....

$510 \ mg$ of a liquid on vaporisation in Victor Mayer's apparatus displaces $510 \ cm^3$ of dry air (at $NTP$). The molecular weight of the liquid is:

For the complete combustion of $1 \, mol$ of a compound with the molecular formula $C_xH_yO_z$,the number of moles of $O_2$ required is:

What is the volume in $dm^3$ occupied by $3 \ mol$ of ammonia gas at $STP$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module