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

`MnO_(4)^(-) +3e rarr MnO_(2)` (n-factor =3)
`(0.33 xx 3 xx 50) - (0.1 xx 1xx 19) (50)/(5) = 31`
`CN^(-) +H_(2)O rarr OCN^(-) +2H^(+) +2e^(-)`
`:.` m. moles of `NaCN (31)/(2)`
Weigth of `NaCN = (31)/(2) xx (49)/(1000) = 0.7595g`
% of `NaCN = (0.7595)/(1) xx 100 = 75.95%`