`x` mole of `KCl` and `y` mole of `BaCl_(2)` are both dissolved in `1kg` of water. Given that `x+y=0.1` and `K_(f)` for water is `1.85K//"molal`, what is the observed range of `DeltaT_(f,)` if the ratio of `x` to `y` is varied ?

Options

A. `0.37^(0)` to `0.555^(0)`
B. `0.185^(0)` to `0.93^(0)`
C. `0.56^(0)` to `0.93^(0)`
D. `0.37^(0)` to `093^(0)`

Answer

To solve the problem, we need to calculate the freezing point depression (\( \Delta T_f \)) for a solution containing \( x \) moles of \( KCl \) and \( y \) moles of \( BaCl_2 \) dissolved in 1 kg of water. We know that \( x + y = 0.1 \) and the van 't Hoff factors (\( i \)) for the solutes are as follows:

- For \( KCl \): \( KCl \rightarrow K^+ + Cl^- \) (i = 2)
- For \( BaCl_2 \): \( BaCl_2 \rightarrow Ba^{2+} + 2Cl^- \) (i = 3)

### Step-by-Step Solution:

1. **Identify the Variables**:
- Let \( x \) be the moles of \( KCl \).
- Let \( y \) be the moles of \( BaCl_2 \).
- Given: \( x + y = 0.1 \).

2. **Write the Expression for Freezing Point Depression**:
The formula for freezing point depression is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( i \) is the van 't Hoff factor,
- \( K_f \) is the cryoscopic constant of the solvent (water in this case, \( K_f = 1.85 \, \text{K kg/mol} \)),
- \( m \) is the molality of the solution.

3. **Calculate the Contribution of Each Solute**:
- For \( KCl \):
\[ \Delta T_{f,KCl} = i_{KCl} \cdot K_f \cdot x = 2 \cdot 1.85 \cdot x \]
- For \( BaCl_2 \):
\[ \Delta T_{f,BaCl_2} = i_{BaCl_2} \cdot K_f \cdot y = 3 \cdot 1.85 \cdot y \]

4. **Total Freezing Point Depression**:
Combine the contributions from both solutes:
\[ \Delta T_f = (2 \cdot 1.85 \cdot x) + (3 \cdot 1.85 \cdot y) \]
Simplifying gives:
\[ \Delta T_f = 1.85 \cdot (2x + 3y) \]

5. **Substitute \( y \) in Terms of \( x \)**:
Since \( y = 0.1 - x \):
\[ \Delta T_f = 1.85 \cdot (2x + 3(0.1 - x)) \]
Expanding this:
\[ \Delta T_f = 1.85 \cdot (2x + 0.3 - 3x) = 1.85 \cdot (0.3 - x) \]

6. **Determine the Range of \( \Delta T_f \)**:
- When \( x = 0.1 \) (and \( y = 0 \)):
\[ \Delta T_f = 1.85 \cdot (0.3 - 0.1) = 1.85 \cdot 0.2 = 0.37 \, \text{°C} \]
- When \( x = 0 \) (and \( y = 0.1 \)):
\[ \Delta T_f = 1.85 \cdot (0.3 - 0) = 1.85 \cdot 0.3 = 0.555 \, \text{°C} \]

### Conclusion:
The observed range of \( \Delta T_f \) is from \( 0.37 \, \text{°C} \) to \( 0.555 \, \text{°C} \).