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638.0 g of `CuSO_(4)` solution is titrated with excess of 0.2 M KI solution. The liberated `I_(2)` required 400 " mL of " 1.0 M `Na_(2)S_(2)O_(3)` for complete reaction. The percentage purity of `CuSO_(4)` in the sample isA. `5%`B. `10%`C. `15%`D. `20%` |
Answer» Correct Answer - B `CuSO_(4)+KItoI_(2)toS_(2)O_(3)^(2-)` `2Cu^(2+)+2e^(-)toCu_(2)^(o+)` `underline(2I^(ɵ)toI_(2)+2e^(-))` `underline(2CuSO_(4)+4KItoCu_(2)I_(2)+2K_(2)SO_(4)+I_(2))` `I_(2)+2S_(2)O_(3)^(2-)to2I^()+S_(4)O_(6)^(2-)` `2 mol CuSO_(4)-=4 mol KI-=1 mol I_(2)` `-=2 mol S_(2)O_(3)^(2-)` `mmol CuSO_(4)-=mmol S_(2)O_(3)^(2-)` `-=400xx1mmol=0.4mol` Weight of `CuSO_(4)-=400xx10^(-3)xx159.5g` `% CuSO_(4)=(400xx10^(-3)xx159.5xx100)/(638)=10%` |
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