In a normal solution of `BaCl_(2)`, normalities of `Ba^(+2)` and `Cl^(-)` are in the ratio

Options

A. `2:1`
B. `1:2`
C. `1:1`
D. `2:3`

Explanation

To find the ratio of normalities of \( \text{Ba}^{2+} \) and \( \text{Cl}^- \) in a normal solution of \( \text{BaCl}_2 \), we can follow these steps:

### Step 1: Understand the Dissociation of \( \text{BaCl}_2 \)
When \( \text{BaCl}_2 \) dissolves in water, it dissociates into its ions:
\[ \text{BaCl}_2 \rightarrow \text{Ba}^{2+} + 2\text{Cl}^- \]
This means that for every 1 mole of \( \text{BaCl}_2 \), we get 1 mole of \( \text{Ba}^{2+} \) and 2 moles of \( \text{Cl}^- \).

### Step 2: Define Normality
Normality (N) is defined as the number of equivalents of solute per liter of solution. For ionic compounds, the equivalents can be determined based on the charge of the ions.

### Step 3: Calculate Normality of \( \text{Ba}^{2+} \)
For \( \text{Ba}^{2+} \):
- The valency factor (n) is 2 (since it has a +2 charge).
- If we have 1 normal solution of \( \text{BaCl}_2 \), it means we have 1 equivalent of \( \text{Ba}^{2+} \) per liter of solution.
- Therefore, the normality of \( \text{Ba}^{2+} \) is:
\[ N_{\text{Ba}^{2+}} = \frac{2 \times \text{moles of } \text{Ba}^{2+}}{1 \text{ L}} = 2 \text{ N} \]

### Step 4: Calculate Normality of \( \text{Cl}^- \)
For \( \text{Cl}^- \):
- The valency factor (n) is 1 (since it has a -1 charge).
- From the dissociation, for every mole of \( \text{BaCl}_2 \), we get 2 moles of \( \text{Cl}^- \).
- Therefore, the normality of \( \text{Cl}^- \) is:
\[ N_{\text{Cl}^-} = \frac{1 \times \text{moles of } \text{Cl}^-}{1 \text{ L}} = 2 \text{ N} \]

### Step 5: Find the Ratio of Normalities
Now, we can find the ratio of the normalities of \( \text{Ba}^{2+} \) to \( \text{Cl}^- \):
\[ \text{Ratio} = \frac{N_{\text{Ba}^{2+}}}{N_{\text{Cl}^-}} = \frac{2}{2} = 1 \]

### Conclusion
Thus, the normalities of \( \text{Ba}^{2+} \) and \( \text{Cl}^- \) are in the ratio of:
\[ \text{Ratio} = 1 : 1 \]