A 10 gm sample of NaCN is dissolved in 50 ml of 0.3 M mild alkaline solution of KMnO 4 and heated strongly to convert all CN − to OCN − . The left solution is then acidified with H 2 SO 4 which

Question

A 10 gm sample of NaCN is dissolved in 50 ml of 0.3 M mild alkaline solution of KMnO 4 ​ and heated strongly to convert all CN − to OCN − . The left solution is then acidified with H 2 ​ SO 4 ​ which require 500 ml of 0.05 M FeSO 4 ​ . Calculate percentage purity of NaCN in sample.

TestHub · Accepted Answer

For II titration milligram equivalents of FeSO₄ = KMnO₄ (in acidic medium).

$0.05 \times 500 \times 1 = n_{K M n O_{4}} \times 5$

So, $n_{K M n O_{4}} = 5$ . used Total millimoles of KMnO₄ initially taken = 15,

So millimoles of KMnO₄ used in I titration = $15 - 5 = 10$ .

For I titration milligram equivalents of NaCN = KMnO₄ (in basic medium).

$n_{N a CN} \times 2 = 10 \times 3$

$n_{N a CN} = 15$

$% N a CN = \frac{15 \times 1 0 ^{- 13} \times 49}{10} \times 100 = 7.35$

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JEE Advanced 2028 (Class 11)