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

# 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

### 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  M =5xV =mol of Fe(II)

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

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

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