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 & 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%`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Understand the Reaction
The reaction involves the conversion of cyanide ions (CN⁻) to cyanate ions (OCN⁻) using KMnO₄ in an alkaline medium. The relevant half-reaction for KMnO₄ in alkaline solution is:
\[ \text{MnO}_4^- + 3e^- \rightarrow \text{MnO}_2 + 4\text{OH}^- \]
This indicates that 1 mole of KMnO₄ can react with 3 moles of electrons.

### Step 2: Calculate the Moles of KMnO₄
We need to calculate the number of moles of KMnO₄ used in the reaction:
- Molarity (M) = 0.33 M
- Volume (V) = 50 mL = 0.050 L

Using the formula:
\[ \text{Moles of KMnO}_4 = \text{Molarity} \times \text{Volume} \]
\[ \text{Moles of KMnO}_4 = 0.33 \, \text{mol/L} \times 0.050 \, \text{L} = 0.0165 \, \text{mol} \]

### Step 3: Calculate the Equivalent of KMnO₄
The n-factor for KMnO₄ in this reaction is 3 (as it gains 3 electrons):
\[ \text{Equivalents of KMnO}_4 = \text{Moles} \times \text{n-factor} \]
\[ \text{Equivalents of KMnO}_4 = 0.0165 \, \text{mol} \times 3 = 0.0495 \, \text{eq} \]

### Step 4: Calculate the Equivalent of NaCN
The 5 mL portion of the reaction mixture was titrated with 19.0 mL of 0.1 M FeSO₄. First, calculate the equivalents of FeSO₄ used:
\[ \text{Moles of FeSO}_4 = \text{Molarity} \times \text{Volume} = 0.1 \, \text{mol/L} \times 0.019 \, \text{L} = 0.0019 \, \text{mol} \]
Since Fe²⁺ reacts in a 1:1 ratio with OCN⁻, the equivalents of OCN⁻ will also be:
\[ \text{Equivalents of OCN}^- = 0.0019 \, \text{eq} \]

### Step 5: Relate OCN⁻ to NaCN
From the stoichiometry of the reaction, 1 equivalent of OCN⁻ corresponds to 1 equivalent of NaCN. Therefore, the equivalents of NaCN in the 5 mL portion is also 0.0019 eq.

### Step 6: Calculate Total Equivalents in 50 mL
Since the 5 mL portion is part of the 50 mL, we can scale up the equivalents:
\[ \text{Total equivalents of NaCN} = 0.0019 \, \text{eq} \times \frac{50 \, \text{mL}}{5 \, \text{mL}} = 0.019 \, \text{eq} \]

### Step 7: Calculate the Moles of NaCN
Using the n-factor for NaCN, which is 1 (since it donates 1 electron):
\[ \text{Moles of NaCN} = \text{Total equivalents} = 0.019 \, \text{mol} \]

### Step 8: Calculate the Mass of NaCN
The molar mass of NaCN is approximately 49 g/mol. Therefore, the mass of NaCN can be calculated as:
\[ \text{Mass of NaCN} = \text{Moles} \times \text{Molar Mass} \]
\[ \text{Mass of NaCN} = 0.019 \, \text{mol} \times 49 \, \text{g/mol} = 0.931 \, \text{g} \]

### Step 9: Calculate Percentage Purity of NaCN
The percentage purity of NaCN in the original 1 g sample can be calculated as:
\[ \text{Percentage Purity} = \left( \frac{\text{Mass of NaCN}}{\text{Total mass of sample}} \right) \times 100 \]
\[ \text{Percentage Purity} = \left( \frac{0.931 \, \text{g}}{1 \, \text{g}} \right) \times 100 = 93.1\% \]

### Conclusion
The percentage purity of the NaCN sample is approximately **93.1%**.