0.144 g of pure `FeC_(2)O_(4)` was dissolvedin dilute `H_(2)SO_(4)` and the solution was diluted to 100 ml . What volume in ml of 0.1 M `KMnO_(4)` will be needed to oxidise `FeC_(2)O_(4)` solution

Explanation

Step 1 To write the balanced equation for the redox reaction
both the cation and anioic components of `FeC_(2)O_(4)` (ferrous oxalate) i.e `Fe^(2+)` and `C_(2)O_(4)^(2-)` are oxidised by `KmnO_(4)` to `Fe^(3+)` and `CO_(2)` respectively the comlete balanced redox equation is
`5 Fe^(+)+MnO_(4)^(-) +8 H^(+)rarr5 Fe^(3+) rarr 5Fe^(3+)+Mn^(2+)+4H_(2)O`
`5C_(2)O_(4)^(2-)+2 MnO_(4)^(-)+16 H^(+)rarr10CO_(2)+2 Mn^(2+)+8 H_(2)O`
`5FeC_(2)O_(4)+3MnO_(4)^(-)+24H^(+)rarr5Fe^(3+)+10 CO_(2)+3Mn^(2+)+12H_(2)O`
Step 2 To determine the molarity of `FeC_(2)O_(4)` solution
Mol wt of `FeC_(2)O_(4) =56+2xx12+4xx16=144 g`
`"volume" =100 cm^(3)`
`therefore "Molarity" =("weight" )/("mol.wrt")xx(1000)/("volume")=(1.44)/(144xx(1000)/(100)=0.1 M`
Step 3 To calculate of 0.01 M `KMnO_(4)` solution
Applying molarity equatin to balaced redox equation we have
`(M_(12)V_(1))/(n_(1))(FeC_(2)O_(4))=(M_(2)V_(2))/(n_(2))(LMnO_(4))`
substituting the values of `M_(1)(=0.1),V_(1)(=100)n_(1)=5, M_(2)=(=0.01)` and n=3 , we have
`(0.1xx100)/(5)=(0.01xxV_(21))/(3) or V_(2) =(3x0.1xx100)/(5xx0.01)=600 cm^(3)`
thus volume of 0.01 M `KMnO_(4)` solution required =600 `cm^(3)`