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

Options

A. `6.02 xx 10^(8)`
B. `6.02 xx 10^(10)`
C. `3.0 xx 6.03 xx 10^(19)`
D. `3.0 xx 6.02 xx 10^(8)`

Answer

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: Write the formula for barium nitrate
Barium nitrate is represented as \( \text{Ba(NO}_3\text{)}_2 \).

### Step 2: Determine the dissociation of barium nitrate in solution
When barium nitrate dissolves in water, it dissociates into ions:
\[ \text{Ba(NO}_3\text{)}_2 \rightarrow \text{Ba}^{2+} + 2 \text{NO}_3^{-} \]
From this dissociation, we can see that 1 mole of barium nitrate produces 1 mole of barium ions and 2 moles of nitrate ions.

### Step 3: Calculate the total number of moles of ions produced from 1 mole of barium nitrate
From the dissociation:
- 1 mole of \( \text{Ba}^{2+} \) ions
- 2 moles of \( \text{NO}_3^{-} \) ions

Thus, the total number of moles of ions produced from 1 mole of barium nitrate is:
\[ 1 + 2 = 3 \text{ moles of ions} \]

### Step 4: Calculate the number of moles of barium nitrate in 1 ml of 0.1 M solution
To find the number of moles in 1 ml of a 0.1 M solution, we use the formula:
\[ \text{Number of moles} = \text{Molarity} \times \text{Volume (in liters)} \]
Converting 1 ml to liters:
\[ 1 \text{ ml} = 0.001 \text{ L} \]
Now, substituting the values:
\[ \text{Number of moles} = 0.1 \, \text{M} \times 0.001 \, \text{L} = 0.0001 \, \text{moles} \]

### Step 5: Calculate the total number of ions in 0.0001 moles of barium nitrate
Since each mole of barium nitrate produces 3 moles of ions, the total number of moles of ions from 0.0001 moles of barium nitrate is:
\[ \text{Total moles of ions} = 0.0001 \, \text{moles} \times 3 = 0.0003 \, \text{moles of ions} \]

### Step 6: Convert moles of ions to number of ions
Using Avogadro's number (\(6.022 \times 10^{23}\) ions/mole), we can find the total number of ions:
\[ \text{Total number of ions} = 0.0003 \, \text{moles} \times 6.022 \times 10^{23} \, \text{ions/mole} \]
Calculating this gives:
\[ \text{Total number of ions} = 0.0003 \times 6.022 \times 10^{23} \approx 1.807 \times 10^{20} \, \text{ions} \]

### Final Answer
The total number of ions present in 1 ml of 0.1 M barium nitrate solution is approximately \(1.807 \times 10^{20}\) ions.

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