Ratio Of Mols Of Fe(II) Oxidised By Equal Volumes Of Equilmolar KMnO4and K2Cr2O7 solutions In Acidic Medium Will Be

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Question

Ratio of mols of Fe(II) oxidised by equal volumes of equilmolar KMnO₄and K₂Cr₂O₇ solutions in acidic medium will be

5 : 3
1 : 1
1 : 2
5 : 6

medium

Solution

Correct option is

5 : 6

Eq. mass of Fe(II) = molar mass

Eq. mass of KMnO₄ = molar mass/5

Eq. mass K₂Cr₂O₇ = molar mass/6

Eqiv. of KMnO₄ in VL of x M =5xV =mol of Fe(II)

Eqiv. of K₂Cr₂O₇ in VL of x M = 6xV =mol of Fe(II)

Hence, ratio of moles of Fe(II) oxidized by KMnO₄ and K₂Cr₂O₇ = 5 : 6

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