10 ml of concentrated HCl were diluted to 1 litre. 20 ml of this diluted solution required 25 ml of 0.1 N sodium hydroxide solution for complete neutralization, the normality of the concentrated acid will be

Options

A. 8
B. 9.5
C. 12.5
D. 15

Explanation

To find the normality of the concentrated HCl solution, we can follow these steps:

### Step 1: Determine the normality of the diluted HCl solution.

Given that 20 ml of the diluted HCl solution requires 25 ml of 0.1 N NaOH for complete neutralization, we can use the formula for neutralization:

\[ N_1V_1 = N_2V_2 \]

Where:
- \(N_1\) = Normality of HCl (unknown)
- \(V_1\) = Volume of HCl solution (20 ml)
- \(N_2\) = Normality of NaOH (0.1 N)
- \(V_2\) = Volume of NaOH solution (25 ml)

Substituting the known values into the equation:

\[ N_1 \times 20 \, \text{ml} = 0.1 \, \text{N} \times 25 \, \text{ml} \]

Calculating the right side:

\[ N_1 \times 20 = 0.1 \times 25 \]
\[ N_1 \times 20 = 2.5 \]

Now, solving for \(N_1\):

\[ N_1 = \frac{2.5}{20} = 0.125 \, \text{N} \]

### Step 2: Calculate the normality of the concentrated HCl solution.

The diluted HCl solution was prepared by diluting 10 ml of concentrated HCl to a total volume of 1 litre (1000 ml). The dilution factor can be calculated as:

\[ \text{Dilution Factor} = \frac{\text{Final Volume}}{\text{Initial Volume}} = \frac{1000 \, \text{ml}}{10 \, \text{ml}} = 100 \]

The normality of the concentrated HCl (\(N_c\)) can be calculated using the dilution relationship:

\[ N_c = N_1 \times \text{Dilution Factor} \]

Substituting the values we have:

\[ N_c = 0.125 \, \text{N} \times 100 = 12.5 \, \text{N} \]

### Final Answer:
The normality of the concentrated HCl solution is **12.5 N**.

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