Number of HCl molecules present in 10 mL of 0.1 N HCl solution is

Options

A. `6.022 xx 10^(23)`
B. `6.022 xx 10^(22)`
C. `6.022 xx 10^(21)`
D. `6.022 xx 10^(20)`

Answer

To find the number of HCl molecules present in 10 mL of a 0.1 N HCl solution, we will follow these steps:

### Step 1: Understand the Normality (N) Concept
Normality (N) is defined as the number of equivalents of solute per liter of solution. For HCl, which is a strong acid, the number of equivalents is equal to the number of moles since it dissociates completely.

### Step 2: Calculate the Number of Moles
We can use the formula for normality:
\[ N = \frac{\text{Number of moles}}{\text{Volume (L)}} \]
Rearranging this gives:
\[ \text{Number of moles} = N \times \text{Volume (L)} \]
Given:
- Normality (N) = 0.1 N
- Volume = 10 mL = 0.01 L (since 1 L = 1000 mL)

Now substituting the values:
\[ \text{Number of moles} = 0.1 \times 0.01 = 0.001 \text{ moles} \]

### Step 3: Convert Moles to Molecules
To find the number of molecules, we use Avogadro's number, which is approximately \(6.022 \times 10^{23}\) molecules/mole. The formula to convert moles to molecules is:
\[ \text{Number of molecules} = \text{Number of moles} \times N_A \]
Where \(N_A\) is Avogadro's number.

Substituting the values:
\[ \text{Number of molecules} = 0.001 \text{ moles} \times 6.022 \times 10^{23} \text{ molecules/mole} \]
Calculating this gives:
\[ \text{Number of molecules} = 6.022 \times 10^{20} \text{ molecules} \]

### Final Answer
The number of HCl molecules present in 10 mL of 0.1 N HCl solution is \(6.022 \times 10^{20}\) molecules.

---