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2.68 x 10-3 moles of a solution containing an ion An+ required 1.61 x 10-3 moles of MnO-4 for the oxidation of An+  to AO_3^- in an acidic medium. What is the value of n?

Option: 1

3


Option: 2

2


Option: 3

4


Option: 4

5


Answers (1)

best_answer

Since in acidic solution:

\mathrm{MnO}_{4}^{-}+8 \mathrm{H}^{+}+5 e^{-} \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H}_{2} \mathrm{O}

\mathrm{MnO}_{4}^{-} \text {gains } 5 e^{-}

\text { So, its equivalent wt. }=\frac{\text { molecular wt. }}{5}

\text { It means } 1.61 \times 10^{-3} \text { moles of } \mathrm{MnO}_{4}^{-} =5 \times 1.61 \times 10^{-3} \text { equivalents }

A^{n+} \rightarrow A \mathrm{O}^{3-} \text { (oxidation number of } A\text { in } A \mathrm{O}_{3}^{-}=5 \text { ) }

\text { loss of electron }=(5-n)

\therefore 2.68 \times 10^{-3} \text { moles of the solution}\text { containing } A^{n+} \text { ions }

=(5-n) \times 2.68 \times 10^{-3} \text { equivalents }

Equivalents of oxidised 'A' =  Equivalents of reduced A

\left(1.61 \times 10^{-3}\right) \times 5=(5-n) \times 2.68 \times 10^{-3}

(5-n)=\frac{1.61 \times 10^{-3} \times 5}{2.68 \times 10^{-3}}=3

n=5-3=2

 

Posted by

Kuldeep Maurya

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