InterviewSolution
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InterviewSolution
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1.0 gm of moist sample of mixture of potassium chlorate (KClO_(3)) and potassium chloride (KCl) was dissolved in water and solution was made upto 250 ml. This solution was treated with SO_(2) to reduce all ClO_(3)^(-) " to " Cl^(-) and excess of SO_(2) was removed by boiling . The total chloride was precipitated as silver chloride . The weight of precipitate was found to be 0.1435 gm. In another experiment , 25 ml of original solutions was heatod with 30 ml 0.2 N FeSO_(4) and unused FeSO_(4)required 37.5ml of 0.08 N KMnO_(4) solutions . Calculate the molar ratio of the ClO_(3)^(-) to the Cl^(-) in the given mixture Given that , ClO_(3)^(-) + 6l^(-) e^(2+) + 6H^(+) to Cl^(-) + 6Fe^(3+) + 3H_(2)O 3SO_(2) + ClO_(3)^(-) + 3H_(2)O to Cl^(-) + 3SO_(4)^(2-) + 6H^(+) |
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Answer» Solution :`ClO_(3)^(-)` is reduced to `Cl^(-) " by " SO_(2) " and " ClO_(3)^(-)` is also reduced to `Cl^(-) " by " FE^(2+) ` , hence AgCl is formed due to TOTAL `Cl^(-)` Meq. of `Fe^(2+) `initially taken ` = 30 xx 0.2 = 6` Meq. of `Fe^(2+)` unused` = 37 . 5 xx 0.08 = 3` ` :. ` Meq of `Fe^(2+) " used " = 6.0 - 3.0 = 3.0` Thus, Meq. of `ClO_(3)^(-) " in " 25 ml= 3.0` Moles of `ClO_(3)^(-) " in " 25 ml = (3.0)/(1000 xx 6) = 0.0005` `.^(+5)ClO_(3)^(-) to .^(-1)Cl^(-) `(n-factor 6) 0.N `5 .... -1` Thus , moles of `ClO_(3)^(-)` in 25 ml solution ` = 0.0005` `ClO_(3)^(-)` is also reduced to `Cl^(-)` by `SO_(2)` in first experiment andprecipitated as AgCl. Thus , `Cl^(-)` formed from `ClO_(3)^(-) =AgCl` from `ClO_(3)^(-) = 0.0005` Total AgCl formed both from actual and `Cl^(-) " from " ClO_(3)^(-) = 0.1435` GM ` = (0.1435)/(143.5)=0.0010` mol Thus, AgCl formed due to `Cl^(-) ` only = `0.0010 - 0.0005 = 0.0005` mol Thus , `ClO_(3)^(-) " and " Cl^(-)` are in molar ratio ` = 1 : 1 `. |
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