1.

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

Answer»


SOLUTION :`30.0 mL 0.1 M Na_(2)S_(2)O_(3)-=30.0 mL 0.1 N Na_(2)S_(2)O_(3)`
`-=30.0 mL 0.1 N I_(2)`
`-=30.0 mL 0.1 N Cl_(2)`
`-=30.0 mL 0.1 N MnO_(2)`
Amount of `MnO_(2)` present `=(NxxExxV)/(1000)`
`=(1)/(10)xx(87)/(2)xx(30)/(1000)`
`% MnO_(2)=(87xx30xx100)/(10xx2xx1000xx0.5)=26.1`


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