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How many moles of electrons are needed for the reduction of 25 \mathrm{~mL}  of 0.7 \mathrm{M} solution of  \mathrm{KMnO} _ { 4 }  in acid mediums?

Option: 1

 0.0875 


Option: 2

 0.7465 


Option: 3

 0.628 
 


Option: 4

0.0741


Answers (1)

best_answer

\left[\mathrm{Mn}^{7^{+}}+5 e^{-} \rightarrow \mathrm{Mn}^{2+}\right]

\text { Moles of } \mathrm{KMnO}_4 =\mathrm{M} \times \mathrm{V} \\
=0.7 \times 25 \times 10^{-3} \\
=0.0175 \text { moles }

\mathrm{1 \mathrm{KMnO} 4~ require \rightarrow 5 moles}
\mathrm{0.0175 \mathrm{KMnO_4}~ require \longrightarrow 5 \times 0.0175 \Rightarrow 0.0875 ~Moles}.

Posted by

Pankaj

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