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, we can use the concept of normality and the dilution equation, which is based on the law of conservation of mass. Here’s a step-by-step solution:

### Step 1: Understand the given values
- We have:
- Initial normality (n1) = 10 N
- Initial volume (V1) = 10 mL
- Final normality (n2) = 0.1 N (decinormal)

### Step 2: Set up the dilution equation
Using the dilution equation:
\[ n_1 \times V_1 = n_2 \times V_2 \]
Where:
- \( V_2 \) is the final volume after dilution, which will be the sum of the initial volume and the volume of water added (V).

### Step 3: Express the final volume
Let \( V \) be the volume of water to be added. Then:
\[ V_2 = V + 10 \text{ mL} \]

### Step 4: Substitute the known values into the equation
Substituting the known values into the dilution equation:
\[ 10 \, \text{N} \times 10 \, \text{mL} = 0.1 \, \text{N} \times (V + 10) \]

### Step 5: Simplify the equation
This simplifies to:
\[ 100 = 0.1 \times (V + 10) \]

### Step 6: Solve for V
Now, multiply both sides by 10 to eliminate the decimal:
\[ 1000 = V + 10 \]

Now, isolate \( V \):
\[ V = 1000 - 10 \]
\[ V = 990 \, \text{mL} \]

### Conclusion
Therefore, to dilute 10 mL of 10 N HCl to make it decinormal, you need to add **990 mL of water**.

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