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`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:A. `4xx10^(-4)M`B. `2xx10^(-2)M`C. `4xx10^(-3)M`D. `2xx10^(-3)M` |
Answer» Correct Answer - C 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)` |
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