What will be the volume strength of 100 mL of H 2 O 2 required to react completely with 61 mL of KMnO 4 in acidic medium ? (Given that 61 mL of KMnO 4 reacts completely with 5 mL of 1 M K 4

Question

What will be the volume strength of 100 mL of H 2 ​ O 2 ​ required to react completely with 61 mL of KMnO 4 ​ in acidic medium ? (Given that 61 mL of KMnO 4 ​ reacts completely with 5 mL of 1 M K 4 ​ [ Fe ( CN ) 6 ​ ] , where it converts into K + , Fe 3 + , CO 3 ​ 2 − and NO 3 ​ − ) :

TestHub · Accepted Answer

To find the volume strength of H₂O₂:

The n-factor of K₄[Fe(CN)₆]: Fe²⁺ → Fe³⁺ ( $Δ = 1$ ), 6C²⁺ → 6CO₃²⁻ ( $Δ = 12$ ), and 6N³⁻ → 6NO₃⁻ ( $Δ = 48$ ), giving a total n-factor = $1 + 12 + 48 = 61$ .

Equivalents of KMnO₄: Meq of KMnO₄ = Meq of K₄[Fe(CN)₆] = $1 M \times 5 m L \times 61 = 305 m e q$ .

Normality of H₂O₂: Since Meq of H₂O₂ = Meq of KMnO₄ = $305$ , we get:

$N or ma l i t y (N) = \frac{305 m e q}{100 m L} = 3.05 N$

Volume Strength:

$V o l u m e St re n g t h = 5.6 \times N or ma l i t y = 5.6 \times 3.05 = 17.08V$

(Option A).

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