The normality of a mixture of HCI and `H_2 SO_4` solution is N/5 Now, 0.287 g of AgCI is obtained when 20 ml of the solution is treated with excess of `AgNO_3`. The percentage of both acids in the mixture will be?

Options

A. `HCI = 32.44%, H_2 SO_4 = 67.56%`
B. `HCI = 42.69%, H_2 SO_4 = 57.31%`
C. `HCI= 48.75%, H_2 SO_4 = 51.25%`
D. `HCI=92.51%, H_2 SO_4 = 7.49%`

Answer

To solve the problem step-by-step, we will follow the outlined procedure based on the information given in the question.

### Step 1: Calculate the number of equivalents of the mixture
The normality of the mixture is given as \( \frac{N}{5} \). The volume of the solution used is 20 mL, which is equivalent to 0.02 L.

\[ \text{Number of equivalents of the mixture} = \text{Normality} \times \text{Volume in L} = \frac{1}{5} \times 0.02 = 0.004 \text{ equivalents} \]

### Step 2: Calculate the number of equivalents of AgCl
We know that 0.287 g of AgCl is formed. To find the number of equivalents of AgCl, we need to use its equivalent weight.

The molar mass of AgCl is:
- Ag: 107.87 g/mol
- Cl: 35.45 g/mol
- Molar mass of AgCl = 107.87 + 35.45 = 143.32 g/mol

The equivalent weight of AgCl (since it has an n-factor of 1) is the same as its molar mass:
\[ \text{Equivalent weight of AgCl} = 143.32 \text{ g} \]

Now, we can calculate the number of equivalents of AgCl:
\[ \text{Number of equivalents of AgCl} = \frac{\text{mass}}{\text{equivalent weight}} = \frac{0.287 \text{ g}}{143.32 \text{ g/equiv}} \approx 0.002 \text{ equivalents} \]

### Step 3: Determine the number of equivalents of HCl
Since the reaction between HCl and AgNO3 produces AgCl, the number of equivalents of HCl will be equal to the number of equivalents of AgCl:
\[ \text{Number of equivalents of HCl} = 0.002 \text{ equivalents} \]

### Step 4: Calculate the number of equivalents of H2SO4
The total number of equivalents in the mixture is 0.004. Thus, the number of equivalents of H2SO4 can be calculated as:
\[ \text{Number of equivalents of H2SO4} = \text{Total equivalents} - \text{Equivalents of HCl} = 0.004 - 0.002 = 0.002 \text{ equivalents} \]

### Step 5: Calculate the equivalent weights of the acids
- For HCl:
- Molar mass of HCl = 36.5 g/mol
- Equivalent weight of HCl = 36.5 g (n-factor = 1)

- For H2SO4:
- Molar mass of H2SO4 = 98 g/mol
- Equivalent weight of H2SO4 = \(\frac{98}{2} = 49 \text{ g}\) (n-factor = 2)

### Step 6: Calculate the mass of HCl and H2SO4 in the mixture
- Mass of HCl:
\[ \text{Mass of HCl} = \text{Number of equivalents of HCl} \times \text{Equivalent weight of HCl} = 0.002 \times 36.5 = 0.073 \text{ g} \]

- Mass of H2SO4:
\[ \text{Mass of H2SO4} = \text{Number of equivalents of H2SO4} \times \text{Equivalent weight of H2SO4} = 0.002 \times 49 = 0.098 \text{ g} \]

### Step 7: Calculate the total mass of the mixture
\[ \text{Total mass} = \text{Mass of HCl} + \text{Mass of H2SO4} = 0.073 + 0.098 = 0.171 \text{ g} \]

### Step 8: Calculate the percentage of each acid in the mixture
- Percentage of HCl:
\[ \text{Percentage of HCl} = \left( \frac{\text{Mass of HCl}}{\text{Total mass}} \right) \times 100 = \left( \frac{0.073}{0.171} \right) \times 100 \approx 42.69\% \]

- Percentage of H2SO4:
\[ \text{Percentage of H2SO4} = 100 - \text{Percentage of HCl} = 100 - 42.69 \approx 57.31\% \]

### Final Result
- Percentage of HCl in the mixture: **42.69%**
- Percentage of H2SO4 in the mixture: **57.31%**

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