0.80g of impure `(NH_(4))_(2) SO_(4)` was boiled with 100mL of a 0.2N NaOH solution till all the `NH_3` (g) evolved. the remaining solution was diluted to `250 mL`. `25 mL`of this solution was neutralized using `5mL` of a `0.2N H_(2)SO_(4)` solution. The percentage purity of the `(NH_(4))_(2)SO_(4)` sample is:

Options

A. 82.5
B. 72.5
C. 62.5
D. 17.5

Answer

To solve the problem of determining the percentage purity of the impure ammonium sulfate `(NH₄)₂SO₄`, we will follow these steps:

### Step 1: Calculate the total equivalents of NaOH used
Given:
- Volume of NaOH solution = 100 mL
- Normality of NaOH solution = 0.2 N

To find the total equivalents of NaOH, we use the formula:
\[ \text{Number of equivalents} = \text{Normality} \times \text{Volume (in L)} \]
Converting 100 mL to liters:
\[ \text{Volume} = \frac{100}{1000} = 0.1 \text{ L} \]
Now, substituting the values:
\[ \text{Number of equivalents of NaOH} = 0.2 \times 0.1 = 0.02 \text{ equivalents} = 20 \text{ mEq} \]

### Step 2: Calculate the equivalents of NaOH used for neutralization
Given:
- Volume of H₂SO₄ solution = 5 mL
- Normality of H₂SO₄ solution = 0.2 N
- Volume of diluted solution = 25 mL

Using the same formula for H₂SO₄:
\[ \text{Number of equivalents of H₂SO₄} = \text{Normality} \times \text{Volume (in L)} \]
Converting 5 mL to liters:
\[ \text{Volume} = \frac{5}{1000} = 0.005 \text{ L} \]
Now, substituting the values:
\[ \text{Number of equivalents of H₂SO₄} = 0.2 \times 0.005 = 0.001 \text{ equivalents} = 1 \text{ mEq} \]

### Step 3: Calculate the equivalents of NaOH that reacted with `(NH₄)₂SO₄`
Since the reaction between NaOH and H₂SO₄ is a neutralization reaction, the equivalents of NaOH that reacted with ammonium sulfate can be calculated as:
\[ \text{Equivalents of NaOH reacting with (NH₄)₂SO₄} = \text{Total equivalents of NaOH} - \text{Equivalents of NaOH used for neutralization} \]
\[ = 20 \text{ mEq} - 10 \text{ mEq} = 10 \text{ mEq} \]

### Step 4: Relate equivalents of NaOH to moles of `(NH₄)₂SO₄`
From the balanced reaction:
\[ (NH₄)₂SO₄ + 2 NaOH \rightarrow Na₂SO₄ + 2 NH₃ + 2 H₂O \]
This shows that 2 moles of NaOH react with 1 mole of `(NH₄)₂SO₄`. Therefore, the moles of `(NH₄)₂SO₄` can be calculated as:
\[ \text{Moles of } (NH₄)₂SO₄ = \frac{\text{Equivalents of NaOH reacting}}{2} = \frac{10 \text{ mEq}}{2} = 5 \text{ mEq} = 5 \times 10^{-3} \text{ moles} \]

### Step 5: Calculate the mass of `(NH₄)₂SO₄`
Using the molar mass of `(NH₄)₂SO₄`, which is approximately 132 g/mol:
\[ \text{Mass of } (NH₄)₂SO₄ = \text{Moles} \times \text{Molar Mass} = 5 \times 10^{-3} \text{ moles} \times 132 \text{ g/mol} = 0.66 \text{ g} \]

### Step 6: Calculate the percentage purity of `(NH₄)₂SO₄`
The percentage purity can be calculated using the formula:
\[ \text{Percentage Purity} = \left( \frac{\text{Mass of pure } (NH₄)₂SO₄}{\text{Mass of impure sample}} \right) \times 100 \]
Substituting the values:
\[ \text{Percentage Purity} = \left( \frac{0.66 \text{ g}}{0.8 \text{ g}} \right) \times 100 = 82.5\% \]

Thus, the percentage purity of the `(NH₄)₂SO₄` sample is **82.5%**.