A 0.5 g sample containing `MnO_(2)` is treated with HCl liberating `Cl_(2)` is passed into a solution of KI and 30.0 " mL of " 0.1 M `Na_(2)S_(2)O_(3)` are required to titrate the liberated iodine. Calculate the percentage of `MnO_(2)` is the sample.

Explanation

We know,
`MnO_(2)overset(HCl)rarrCl_(2)overset(KI)rarrI_(2)overset(Na_(2)S_(2)O_(3))rarrNaI+Na_(2)S_(4)O_(6)`
Redox changes are: `2e+I_(2)^(0)rarr2I^(-)`
`2S_(2)^(2+)rarrS_(4)^((5//2)+)+2e`
`2e+Mn^(4+)rarrMn^(2+)`
The reactions suggest that, Meq.of `MnO_(2)="Meq.of" Cl_(2) "formed" =
"Meq.of" I_(2)`
`"liberated" = "Meq.of" Na_(2)S_(2)O_(3)` used
`therefore (w)/(M//2)xx1000=0.1xx1xx30`
`[ because N_(Na_(2)S_(2)O_(3))=M_(Na_(2)S_(2)O_(3))` since valency factor `=1` see redox changes for `Na_(2)S_(2)O_(3)`]
or , `w=(0.1xx1xx30xxM)/(2000)=(0.1xx1xx30xx87)/(2000)`
`( because M_(MnO_(2))=87)`
`w_(MnO_(2))=0.1305`
`therefore` Purity of `MnO_(2)=(0.13005)/(0.5)xx100=26.1%`