How much water is to be added to dilute `10 mL` of `10 N HCl` to make it decinormal?

Options

A. `990 mL`
B. `1010 mL`
C. `100 mL`
D. `1000 mL`

Explanation

To solve the problem of how much water is to be added to dilute 10 mL of 10 N HCl to make it decinormal (0.1 N), we can follow these steps:

### Step 1: Understand the relationship between normality and volume
Normality (N) is a measure of concentration equivalent to the number of equivalents of solute per liter of solution. In this case, we have a 10 N solution of HCl, and we want to dilute it to 0.1 N.

### Step 2: Use the dilution formula
The dilution formula can be expressed as:
\[ N_1 \times V_1 = N_2 \times V_2 \]
Where:
- \( N_1 \) = initial normality (10 N)
- \( V_1 \) = initial volume (10 mL)
- \( N_2 \) = final normality (0.1 N)
- \( V_2 \) = final volume (unknown)

### Step 3: Plug in the values
Substituting the known values into the equation:
\[ 10 \, \text{N} \times 10 \, \text{mL} = 0.1 \, \text{N} \times V_2 \]

### Step 4: Solve for \( V_2 \)
\[ 100 \, \text{N mL} = 0.1 \, \text{N} \times V_2 \]
To find \( V_2 \):
\[ V_2 = \frac{100 \, \text{N mL}}{0.1 \, \text{N}} = 1000 \, \text{mL} \]

### Step 5: Calculate the amount of water to be added
Now that we know the final volume \( V_2 \) is 1000 mL, we can find out how much water needs to be added:
\[ \text{Volume of water to be added} = V_2 - V_1 \]
\[ \text{Volume of water to be added} = 1000 \, \text{mL} - 10 \, \text{mL} = 990 \, \text{mL} \]

### Final Answer
To dilute 10 mL of 10 N HCl to make it decinormal (0.1 N), you need to add **990 mL** of water.

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