0.7g of `(NH_(4))_(2)SO_(4)` sample was boiled with 100mL of 0.2 N NaOH solution was diluted to 250 ml. 25mL of this solution was neutralised using 10mL of a 0.1 N `H_(2)SO_(4)` solution. The percentage purity of the `(NH_(4))_(2)SO_(4)`sample is :

Options

A. 94.3
B. 50.8
C. 47.4
D. 79.8

Explanation

To solve the problem step by step, we will follow the procedure outlined in the video transcript while providing clear calculations.

### Step 1: Calculate the milliequivalents of H₂SO₄ used
We know that:
- Normality (N) of H₂SO₄ = 0.1 N
- Volume of H₂SO₄ used = 10 mL

Using the formula for milliequivalents:
\[ \text{Milliequivalents of H₂SO₄} = \text{Normality} \times \text{Volume (in L)} = 0.1 \, \text{N} \times 0.010 \, \text{L} = 0.001 \, \text{N} \times 10 = 1 \, \text{mEq} \]

### Step 2: Calculate the total milliequivalents of H₂SO₄ in the diluted solution
Since we took 25 mL of the diluted solution, we can find the total milliequivalents in 250 mL:
\[ \text{Total mEq of H₂SO₄ in 250 mL} = \text{mEq in 25 mL} \times \frac{250 \, \text{mL}}{25 \, \text{mL}} = 1 \, \text{mEq} \times 10 = 10 \, \text{mEq} \]

### Step 3: Relate milliequivalents of H₂SO₄ to (NH₄)₂SO₄
From the neutralization reaction, we know that:
\[ \text{mEq of H₂SO₄} = \text{mEq of } (NH₄)_2SO₄ \]
Thus, we have:
\[ \text{mEq of } (NH₄)_2SO₄ = 10 \, \text{mEq} \]

### Step 4: Calculate the weight of (NH₄)₂SO₄ corresponding to the milliequivalents
The molar mass of (NH₄)₂SO₄ is calculated as follows:
- Molar mass of N = 14 g/mol
- Molar mass of H = 1 g/mol
- Molar mass of S = 32 g/mol
- Molar mass of O = 16 g/mol

Molar mass of (NH₄)₂SO₄:
\[ = 2 \times (14 + 4 \times 1) + 32 + 4 \times 16 = 2 \times 18 + 32 + 64 = 36 + 32 + 64 = 132 \, \text{g/mol} \]

The n-factor for (NH₄)₂SO₄ is 2 because it can donate 2 equivalents of NH₄⁺.

Now, using the formula:
\[ \text{Weight} = \text{mEq} \times \frac{\text{Molar mass}}{\text{n-factor}} \]
Substituting the values:
\[ \text{Weight} = 10 \, \text{mEq} \times \frac{132 \, \text{g/mol}}{2} = 10 \times 66 = 660 \, \text{mg} = 0.66 \, \text{g} \]

### Step 5: Calculate the percentage purity of (NH₄)₂SO₄
Given the initial weight of the sample is 0.7 g, the percentage purity is calculated as:
\[ \text{Percentage Purity} = \left( \frac{\text{Weight of pure (NH₄)₂SO₄}}{\text{Weight of sample}} \right) \times 100 = \left( \frac{0.66 \, \text{g}}{0.7 \, \text{g}} \right) \times 100 \approx 94.29\% \]

### Final Answer
The percentage purity of the (NH₄)₂SO₄ sample is approximately **94.3%**.

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