A solution containing 2-68 - 10-3 mol of An- ions requires 1-61 - 10-3 mol of MnO4- for the complete oxidation of An- to AO3- in acidic medium- What is the value of n-

MnO_4^ would convert to Mn^2+. Therefore its 'n' factor would be 5. ∴ Equivalents of MnO_4^ =1.61 × 10^ 3× 5 = 8.05 × 10^ 3 Equivalents of A^n+=8.05 × 10^ 3 'n ...

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Standard XII

Chemistry

Catalytic Hydrogenation

A solution co...

Question

A solution containing $2.68 \times 10^{- 3} mol$ of $A^{n +}$ ions requires $1.61 \times 10^{- 3} mol$ of $M n O_{4}^{-}$ for the complete oxidation of $A^{n +}$ to $A O_{3}^{-}$ in acidic medium. What is the value of $n$ ?

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Solution

$M n O_{4}^{-}$ would convert to $M n^{2 +}$ . Therefore its $^{'} n^{'}$ factor would be $5$ .
$∴$ Equivalents of $M n O_{4}^{-} = 1.61 \times 10^{- 3} \times 5 = 8.05 \times 10^{- 3}$
Equivalents of $A^{n +} = 8.05 \times 10^{- 3}$
$^{'} n^{'}$ factor of $A^{n +} = 5 - n$
$∴ (5 - n) \times 2.68 \times 10^{- 3} = 8.05 \times 10^{- 3}$
$n = 2$

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A solution containing 2.68×10−3 mol of An+ ions requires 1.61×10−3 mol of MnO−4 for the complete oxidation of An+ to AO−3 in acidic medium. What is the value of n?

MnO−4 would convert to Mn2+. Therefore its ′n′ factor would be 5. ∴ Equivalents of MnO−4=1.61×10−3×5=8.05×10−3 Equivalents of An+=8.05×10−3 ′n′ factor of An+=5−n ∴(5−n)×2.68×10−3=8.05×10−3 n=2

A solution containing $2.68 \times 10^{- 3} mol$ of $A^{n +}$ ions requires $1.61 \times 10^{- 3} mol$ of $M n O_{4}^{-}$ for the complete oxidation of $A^{n +}$ to $A O_{3}^{-}$ in acidic medium. What is the value of $n$ ?