The mass of a mixtutre contining HCl and `H_(2)SO_(4)` is 0.1g. On treatment withan excess of an `AgNO_(3)` solution, reacted with this acid mixture gives 0.1435g of AgCl. Mass % of the `H_(2)SO_(3)` mixture is:

Options

A. 36.5
B. 63.5
C. 50
D. none of these

Answer

To solve the problem, we need to find the mass percentage of H₂SO₄ in a mixture of HCl and H₂SO₄ that weighs 0.1 g. The mixture reacts with AgNO₃ to produce AgCl, and we know the mass of AgCl produced is 0.1435 g.

### Step-by-Step Solution:

1. **Identify the Reaction**:
The reaction of HCl with AgNO₃ produces AgCl and HNO₃. The reaction can be represented as:
\[ \text{HCl} + \text{AgNO}_3 \rightarrow \text{AgCl} + \text{HNO}_3 \]

2. **Define Variables**:
Let \( x \) be the mass of HCl in the mixture. Therefore, the mass of H₂SO₄ in the mixture will be \( 0.1 - x \) grams.

3. **Calculate Moles of HCl**:
The molar mass of HCl is approximately 36.5 g/mol. The number of moles of HCl can be calculated as:
\[ \text{Moles of HCl} = \frac{x}{36.5} \]

4. **Relate Moles of HCl to Moles of AgCl**:
According to the stoichiometry of the reaction, 1 mole of HCl produces 1 mole of AgCl. Therefore, the moles of AgCl produced will also be:
\[ \text{Moles of AgCl} = \frac{0.1435}{143.5} = 0.001 \]

5. **Set Up the Equation**:
Since the moles of HCl are equal to the moles of AgCl produced:
\[ \frac{x}{36.5} = 0.001 \]

6. **Solve for \( x \)**:
Rearranging the equation gives:
\[ x = 0.001 \times 36.5 = 0.0365 \text{ g} \]

7. **Calculate Mass of H₂SO₄**:
The mass of H₂SO₄ in the mixture is:
\[ \text{Mass of H₂SO₄} = 0.1 - x = 0.1 - 0.0365 = 0.0635 \text{ g} \]

8. **Calculate Mass Percentage of H₂SO₄**:
The mass percentage of H₂SO₄ in the mixture is calculated as:
\[ \text{Mass \% of H₂SO₄} = \left( \frac{0.0635}{0.1} \right) \times 100 = 63.5\% \]

### Final Answer:
The mass percentage of H₂SO₄ in the mixture is **63.5%**.