2.0 g sample of NaCN is dissolved in 50 " mL of " 0.3 M mild alkaline `KMnO_(4)` and heated strongly to convert all the `CN^(ɵ)` to `OCN^(ɵ)`. The solution after acidification with `H_(2)SO_(4)` requries 500 " mL of " 0.05 M `FeSO_(4)` Calculate the percentage purity of NaCN in the sample.

Explanation

(a). `undersetunderset(x=2)(x-3=-1)(Coverset(+2)(N^(ɵ)))toundersetunderset(x=4)(x-3-2=-1)(OCoverset(+4)(N^(ɵ)))+2e^(-)(n=2)`
(b). `3e^(-)+undersetunderset(x=7)(x-8=-1)(MnO_(4)^(ɵ))toundersetunderset(x=3)(x-4=0)(MnO_(2))`
(c). `5e^+MnO_(4)^(ɵ)toMn^(2+)` `(n=5` in acidic medium)
(d). `Fe^(2+)toFe^(3+)+2e^(-)(n=1)`
`m" Eq of "KMnO_(4)` added in basic medium`=50xx0.3`
(n-factor`=3`)
`=45.0`
`m" Eq of "KMnO_(4)` in acidic medium `(n=5)` left after reaction with `NaCN=m" Eq of "FeSO_(4)` used
`=500xx0.5xx1` (n-factor`=1`)
`=25.0`
`m" Eq of "KMnO_(4)` (n-factor`=3`) left`=(25xx3)/(5)=15`
`m" Eq of "NaCN` in sample `=m" Eq of "KMnO_(4)` added `-m" Eq of "KMnO_(4)` left
`=45-15=30`
`therefore(Weight)/(Ew of NaCN)xx10^(3)=30`
`[Ew of NaCN=(49)/(2)` (n factor`=2`)]
`(W)/((49)/(2))xx10^(3)=30`
`W_(NaCN)=0.735g`.
`% of NaCN=(0.735)/(2.0)xx100=36.75%`