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

`MnO_(2) overset(HCl) to Cl_(2) overset(Kl) to l_(2) overset(Na_(2)S_(2)O_(2)) to Nal + Na_(2)S_(4)O_(6)`
Redox changes are :
`l_(2)^(0) + 2e to 2l^(-)`
`2 (S^(2+) ) _(2) to (S^(5//2+))_(4) + 2e`
`Mn^(4+) + 2e to Mn^(2+)`
Meq. of `MnO_(2)` = Meq. of `Cl_(2)` formed = Meq. of `l_(2)` liberated = Meq. of `Na_(2)S_(2)O_(3)` used
` = 0.1 xx 1xx 30 `
`W_(MnO_(2))/(M//2) xx 1000 = 0.1 xx 1 xx 30`
`rArr W_(MnO_(2)) = (0.1 xx 1 xx 30 xx M)/(1000 xx 2) = (0.1xx1xx30 xx 87)/2000`
`W_(MnO_(2)) = 0.1305`
Purity of `MnO_(2) = (0.1305)/(0.5) xx 100 = 26.10 %`