`1` litre solution of unknown molarity is titrated by taking its `50 mL` solution against `KI` solution is strong acidic medium of excess `HCl`. The equivalence point was detected when `10 mL` of `0.1 M KI` was consumed The molarity of `KIO_(3)` solution is:

Options

A. `4xx10^(-4)M`
B. `2xx10^(-2)M`
C. `4xx10^(-3)M`
D. `2xx10^(-3)M`

Answer

C

Explanation

Meq.of `KIO_(3)="Meq.of" KI`
`{:(2I^(5+)+,10e,rarr,I_(2)^(0),),(,2I^(-),rarr,I_(2)^(0)+,2e):}`
`Mxx5xx50=10xx0.1xx1 rArrM=4.0xx10^(-3)`