InterviewSolution
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InterviewSolution
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1.0 g a of moist sample of a mixture of KCl and `KClO_3` was dissolved in water and made up to `250mL`. 25 " mL of " this solution was treated with `SO_2`. The chlorate was reduced to chloride and excess of `SO_2` was removed by boiling. The total chloride was precipitated as `AgCl` . The weight of the precipitate was `0.1435 g`. In another experiment, 25 " mL of " the original solution was heated with 30 " mL of " 0.2 N solution of ferrous sulphate, and the unreacted ferrous sulphate required 37.5 " mL of " 0.08 N solution of an oxidising agent for complete oxidation. Calculate the molar ratio of the chlorate to the chloride in the given mixture `Fe^(2+)` reacts with `ClO_3^(ɵ)` according to the equation. `ClO_3^(ɵ)+6Fe^(2+)+6H^(o+)toCl^(ɵ)+6Fe^(3+)+3H_2O` |
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Answer» Total `Cl^(ɵ)` (from `KCl+KClO_3)(ClO_3^(ɵ)toCl^(ɵ))` 25 " mL of " the mixture`=0.1435 g of AgCl` `=(0.1435)/(143.5)` " mol of "AgCl `=10^(-3) " mol of "Cl^(ɵ)` ions Second experiment: Total volume of `FeSO_4=30mL` Excess volume of `FeSO_4=(37.5xx0.08)/(2)=15mL` Volume of `FeSO_4` used `=30-15=15mL` `15 mL` of `0.2N FeSO_4` used up `=15xx0.2xx10^(-3) " Eq of "FeSO_4` `=3xx10^(-3) " mol of "FeSO_4` Since 1 mol `ClO_3^(ɵ)` uses 6 " mol of "`FeSO_4`, `3xx10^(-3)` " mol of "`FeSO_4=(1)/(6)xx3xx10^(-3)` `=0.5xx10^(-3) " mol of "ClO_3^(ɵ)` `Cl^(ɵ)+ClO_3^(ɵ)=10^(-3)` `thereforeCl^(ɵ)=10^(-3)-0.5xx10^(-3)=0.5xx10^(-3)` `thereforeCl^(ɵ)=0.5xx10^(-3),ClO_3^(ɵ)=0.5xx10^(-3)` |
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