A sample of an alloy weighing `0.50 g` and containing 90% `Ag` was dissolved in concentrated `HNO_(3)`. `Ag` was analysed by volhard method in which `25 mL` of `KCNS` was required for complete neutralisation. Determine the normality of `KCNS`.

Explanation

To determine the normality of `KCNS` used in the Volhard method for analyzing the `Ag` content in the alloy, we can follow these steps:

### Step 1: Calculate the weight of `Ag` in the alloy
The weight of the alloy is given as `0.50 g`, and it contains `90% Ag`.

\[ \text{Weight of } Ag = \text{Weight of alloy} \times \frac{90}{100} = 0.50 \, \text{g} \times 0.90 = 0.45 \, \text{g} \]

### Step 2: Determine the equivalent weight of `Ag`
The equivalent weight of a substance is calculated by dividing its molar mass by its valency factor. The molar mass of `Ag` (Silver) is approximately `108 g/mol`, and since `Ag` has a valency of `1` (as it forms `Ag+` ions), the equivalent weight is:

\[ \text{Equivalent weight of } Ag = \frac{\text{Molar mass of } Ag}{\text{Valency}} = \frac{108 \, \text{g/mol}}{1} = 108 \, \text{g/equiv} \]

### Step 3: Calculate the milliequivalents of `Ag`
Milliequivalents can be calculated using the formula:

\[ \text{Milliequivalents} = \frac{\text{Weight of substance (g)}}{\text{Equivalent weight (g/equiv)}} \]

Substituting the values for `Ag`:

\[ \text{Milliequivalents of } Ag = \frac{0.45 \, \text{g}}{108 \, \text{g/equiv}} = 0.0041667 \, \text{equiv} \text{ or } 4.1667 \, \text{meq} \]

### Step 4: Relate milliequivalents of `Ag` to milliequivalents of `KCNS`
According to the principle of the Volhard method, the milliequivalents of `Ag` will equal the milliequivalents of `KCNS` used for neutralization:

\[ \text{Milliequivalents of } Ag = \text{Milliequivalents of } KCNS \]

### Step 5: Calculate the normality of `KCNS`
Normality (N) is defined as the number of equivalents per liter of solution. We can express this relationship as:

\[ \text{Normality} (N) = \frac{\text{Milliequivalents}}{\text{Volume in liters}} \]

Given that `25 mL` of `KCNS` was used, we convert this to liters:

\[ \text{Volume} = 25 \, \text{mL} = 0.025 \, \text{L} \]

Now substituting the values:

\[ N = \frac{4.1667 \, \text{meq}}{0.025 \, \text{L}} = 166.67 \, \text{N} \]

### Final Answer
The normality of `KCNS` is approximately `166.67 N`.

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