3.0 g of a sample of impure ammonium chloride were boiled with excess of caustic soda solution. Ammonia gas so evolved was passed into 120 mL of `N/2 H_(2)SO_(4)` mL of `N/2 NaOH` were required to neutralise excess of the acid. Calculate the percent purity of the given sample of ammonium chloride.

Answer

To calculate the percent purity of the given sample of ammonium chloride (NH4Cl), we will follow these steps:

### Step 1: Identify the reactions involved
When ammonium chloride is boiled with caustic soda (NaOH), ammonia gas (NH3) is produced along with sodium chloride (NaCl) and water. The reaction can be represented as:
\[ \text{NH}_4\text{Cl} + \text{NaOH} \rightarrow \text{NH}_3 + \text{NaCl} + \text{H}_2\text{O} \]

### Step 2: Determine the amount of sulfuric acid (H2SO4) consumed
The ammonia gas produced is passed into 120 mL of \( \frac{N}{2} \) sulfuric acid (H2SO4). The reaction between ammonia and sulfuric acid produces ammonium sulfate:
\[ 2 \text{NH}_3 + \text{H}_2\text{SO}_4 \rightarrow \text{(NH}_4)_2\text{SO}_4 \]

### Step 3: Calculate the excess sulfuric acid
After the reaction, 28 mL of \( \frac{N}{2} \) NaOH is used to neutralize the excess sulfuric acid. We can use the neutralization equation:
\[ N_1V_1 = N_2V_2 \]
Where:
- \( N_1 \) = Normality of H2SO4 = \( \frac{1}{2} \) N
- \( V_1 \) = Volume of excess H2SO4 (unknown)
- \( N_2 \) = Normality of NaOH = \( \frac{1}{2} \) N
- \( V_2 \) = Volume of NaOH = 28 mL

Substituting the known values:
\[ \frac{1}{2} \cdot V_1 = \frac{1}{2} \cdot 28 \]
Cancelling \( \frac{1}{2} \):
\[ V_1 = 28 \, \text{mL} \]

### Step 4: Calculate the consumed sulfuric acid
The total volume of sulfuric acid used is 120 mL. Therefore, the amount of sulfuric acid consumed in the reaction is:
\[ \text{Consumed H2SO4} = 120 \, \text{mL} - 28 \, \text{mL} = 92 \, \text{mL} \]

### Step 5: Calculate the moles of consumed sulfuric acid
To find the number of moles of sulfuric acid consumed, we first convert the volume from mL to L:
\[ 92 \, \text{mL} = 0.092 \, \text{L} \]
Using the formula for normality:
\[ \text{Normality} = \text{Molarity} \times \text{Basicity} \]
Since sulfuric acid has a basicity of 2:
\[ \text{Molarity of H2SO4} = \frac{N}{B} = \frac{\frac{1}{2}}{2} = \frac{1}{4} \, \text{M} \]

Now, using the molarity to find moles:
\[ \text{Moles of H2SO4} = \text{Molarity} \times \text{Volume (L)} = \frac{1}{4} \times 0.092 = 0.023 \, \text{moles} \]

### Step 6: Calculate moles of ammonia produced
From the stoichiometry of the reaction, 1 mole of H2SO4 reacts with 2 moles of NH3:
\[ \text{Moles of NH3} = 2 \times \text{Moles of H2SO4} = 2 \times 0.023 = 0.046 \, \text{moles} \]

### Step 7: Calculate the mass of NH4Cl that produced the ammonia
From the reaction between NH4Cl and NaOH, we know that 1 mole of NH4Cl produces 1 mole of NH3. Therefore, the moles of NH4Cl that reacted is also 0.046 moles.

The molar mass of NH4Cl is calculated as follows:
- N: 14 g/mol
- H: 4 g/mol (4 H atoms)
- Cl: 35.5 g/mol
Total molar mass = 14 + 4 + 35.5 = 53.5 g/mol

Now, we can calculate the mass of NH4Cl:
\[ \text{Mass of NH4Cl} = \text{Moles} \times \text{Molar Mass} = 0.046 \times 53.5 = 2.461 \, \text{g} \]

### Step 8: Calculate the percent purity of the sample
The percent purity of the ammonium chloride sample is calculated as:
\[ \text{Percent Purity} = \left( \frac{\text{Mass of pure NH4Cl}}{\text{Mass of sample}} \right) \times 100 \]
\[ \text{Percent Purity} = \left( \frac{2.461}{3.0} \right) \times 100 = 82.03\% \]

### Final Answer
The percent purity of the given sample of ammonium chloride is approximately **82.03%**.

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