0.2828 g of iron wire was dissolved in excess dilute `H_(2)SO_(4)` and the solution was made upto 100mL. 20mL of this solution required 30mL of `N/(30)K_(2)Cr_(2)O_(7)` solution for exact oxidation. Calculate percent purity of Fe in wire.

Answer

`99%`

Explanation

Redox changes are :
`FetoFe^(+2)+2r^(-) " " ("in" H_2SO_4)`
`Fe^(+2)to Fe^(+3)+e^(-) " " ("with" K_2Cr_2O_7)`
`6e+Cr_2^(+8)to2Cr^(+3)`
m.eq. of `Fe^(+2)` in 20 mL =m.eq. of `K_2Cr_2O_7=30xx1/30=1`
`:.` m.eq of `Fe^(+2)` in 100 mL `=(1xx100)/20=5`
`:.` m.moles of `Fe^(+2)=(meq)/(v.f.)=5/1=5`=m.moles of Fe
`:.` Mass of pure Fe in wire =`0.28/0.2828xx100=99%`