The total number of ions present in 1 ml of 0.1M barium nitrate solution is :-

Options

A. `6.02xx10^(18)`
B. `6.02xx10^(19)`
C. `3.0xx6.02xx10^(19)`
D. `3.0xx6.02xx10^(18)`

Explanation

To find the total number of ions present in 1 ml of a 0.1 M barium nitrate solution, we can follow these steps:

### Step 1: Understand the Composition of Barium Nitrate
Barium nitrate is represented by the formula \( \text{Ba(NO}_3\text{)}_2 \). Each formula unit of barium nitrate dissociates into one barium ion (\( \text{Ba}^{2+} \)) and two nitrate ions (\( \text{NO}_3^{-} \)). Therefore, one formula unit of barium nitrate produces a total of 3 ions:
- 1 \( \text{Ba}^{2+} \)
- 2 \( \text{NO}_3^{-} \)

### Step 2: Calculate the Number of Moles in 1 ml of 0.1 M Solution
The concentration of the solution is 0.1 M, which means there are 0.1 moles of barium nitrate in 1 liter (1000 ml) of solution. To find the number of moles in 1 ml, we can use the following calculation:

\[ \text{Moles of Ba(NO}_3\text{)}_2 \text{ in 1 ml} = 0.1 \, \text{mol/L} \times \frac{1 \, \text{ml}}{1000 \, \text{ml}} = 0.1 \times 10^{-3} \, \text{mol} = 10^{-4} \, \text{mol} \]

### Step 3: Calculate the Total Number of Ions
Since each mole of barium nitrate produces 3 moles of ions, we can calculate the total moles of ions produced from \( 10^{-4} \) moles of barium nitrate:

\[ \text{Total moles of ions} = 10^{-4} \, \text{mol} \times 3 = 3 \times 10^{-4} \, \text{mol of ions} \]

### Step 4: Convert Moles of Ions to Number of Ions
Using Avogadro's number (\( 6.022 \times 10^{23} \) ions/mol), we can find the total number of ions:

\[ \text{Total number of ions} = 3 \times 10^{-4} \, \text{mol} \times 6.022 \times 10^{23} \, \text{ions/mol} \]

Calculating this gives:

\[ \text{Total number of ions} = 3 \times 6.022 \times 10^{19} \approx 1.8066 \times 10^{20} \text{ ions} \]

### Final Answer
The total number of ions present in 1 ml of 0.1 M barium nitrate solution is approximately \( 1.81 \times 10^{20} \) ions.

---