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12 g of impure cyanogen undergoes hydrolysis by two different paths. (i). `(CN)_(2)+4H_(2)Oto(NH_(4))_(2)C_(2)O_(4)` (ii). `(CN)_(2)+2H_(2)OtoNH_(2)CONH_(2)` When 11.52 g of pure ammonium carbonate `[(NH_(4))_(2)CO_(3)]` was heated, the exact amount of urea was obtained. 20 " mL of " 1.6 M acidic `KMnO_(4)` is required to completely oxidise `(NH_(4))_(2)C_(2)O_(4)`. Q. The percentage purity of cyanogenA. `86.67%`B. `76.67%`C. `66.67%`D. `56.67%` |
Answer» Correct Answer - A `underset(1mol)(NH_(4))_(2)CO_(3)toH_(2)N-overset(O)overset(||)underset(1mol)C-NH_(2)+2H_(2)O` (Mw of `(NH_(4))_(2)CO_(3)=96)` Moles of urea`=(11.56)/(96)=0.12` `MnO_(4)^(ɵ)-=(NH_(4))_(2)C_(2)O_(4)(n=2)` `[C_(2)O_(4)^(2-)to2CO_(2)+2e^(-)]` `20xx1.6xx5mEq-=mEq]` `160mEq-=160mEq` `-=(160)/(2)mmol=80mmol=0.08mol` Therefore, moles of `(NH_(4))_(2)C_(2)O_(4)=0.08` Let a and b " mol of "`(CN)_(2)` react in reactions (i) and (ii), respectively. (i). `underset(a=0.08 mol)((CN)_(2)+4H_(2)O)tounderset(a=0.08mol)((NH_(4))_(2)CONH_(2))` `thereforea=0.08,b=0.12` Total moles of `(CN)_(2)=0.08+0.12=0.2` Weight of `(CN)_(2)=0.2xx52=10.4` `%` purity of `(CN)_(2)=(10.4)/(12)xx100=86.67%` |
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