1 g of a sample of NaOH was dissolved in 50 " mL of " 0.33 M alkaline solution of `KMnO_(4)` and refluxed till all the cyanide was converted into `OCN^(ɵ)`. The reaction mixture was cooled and its 5 mL portion was acidified by adding `H_(2)SO_(4)` in excess and then titrated to end point against 19.0 " mL of " 0.1 M `FeSO_(4)` solution. The percentage purity of NaCN sample is

Options

A. `55.95%`
B. `65.95%`
C. `75.95%`
D. `85.95%`

Answer

C

Explanation

Assume medium to be dilute alkaline:
`impliesMnO_(4)^(ɵ)+3e^(-)toMnO_(2)` `(because "n-factor"=3)`
[Although it should have been mentiones clearly but if it as strongly alkaline, it is not possible to solve. Check yourself].
`impliesmEq NaCN=(0.33xx3xx50)-(0.1xx1xx19)xx(50)/(5)=31`
Also, `CN^(ɵ)+H_(2)OtoOCN^(ɵ)+2H^(o+)+2e^(-)`
`impliesmmol NaCN=(31)/(2)=(weight)/(49)xx1000`
`impliesweight=(31)/(2)xx(49)/(1000)impliesNaCN` purity`=75.95%`