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How many gram of `I_(2)` are present in a solution which requires `40 mL` of `0.11N Na_(2)S_(2)O_(3)` to react with it, `S_(2)O_(3)^(2-)+I_(2)rarrS_(4)O_(6)^(2-)+2I`A. `12.7 g`B. `0.558 g`C. `25.4 g`D. `11.4 g` |
Answer» Correct Answer - B Meq.of `I_(2)="Meq.of" Na_(2)S_(2)O_(3)` `=40xx0.11` `:. (w)/(254//2)xx1000=40xx0.11` `w_(I_(2))=0.558g` |
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