10 mL of an HCl solution gave `0.1435` g of AgCl when treated with excess of `AgNO_(3)` .The normality of the resulting solution is

Options

A. `0.1`
B. 3
C. `0.3`
D. `0.2`

Answer

To find the normality of the HCl solution that produced 0.1435 g of AgCl when treated with excess AgNO3, we can follow these steps:

### Step 1: Calculate the moles of AgCl produced
To find the moles of AgCl, we use the formula:

\[ \text{Moles of AgCl} = \frac{\text{mass of AgCl}}{\text{molar mass of AgCl}} \]

The molar mass of AgCl is approximately 143.5 g/mol.

\[ \text{Moles of AgCl} = \frac{0.1435 \, \text{g}}{143.5 \, \text{g/mol}} = 0.001 \, \text{mol} \]

### Step 2: Calculate the molarity of HCl
Since the reaction between HCl and AgNO3 produces AgCl in a 1:1 molar ratio, the moles of HCl will also be 0.001 mol.

Now, we need to convert the volume of HCl from mL to L:

\[ \text{Volume of HCl} = 10 \, \text{mL} = 0.010 \, \text{L} \]

Now we can calculate the molarity (M) of HCl:

\[ \text{Molarity (M)} = \frac{\text{moles of HCl}}{\text{volume of solution in L}} = \frac{0.001 \, \text{mol}}{0.010 \, \text{L}} = 0.1 \, \text{M} \]

### Step 3: Calculate the normality of HCl
For HCl, which is a strong acid, the normality (N) is equal to the molarity (M) because it dissociates completely to give one H+ ion per molecule of HCl.

\[ \text{Normality (N)} = \text{Molarity (M)} = 0.1 \, \text{N} \]

### Final Answer
The normality of the resulting HCl solution is **0.1 N**.

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