An aqueous solution containing an ionic salt having molality equal to `0.19` freezes at `-0.704^(@)C`. The Van't Hoff factor of the ionic salt is (`K_(f)` for water=`1.86 K m^(-1)`)

Options

A. 3
B. 2
C. 4
D. 5

Explanation

To solve the problem, we will use the formula for depression in freezing point, which relates the change in freezing point to the molality of the solution and the Van't Hoff factor.

### Step-by-step Solution:

1. **Identify the given values:**
- Molality (m) = 0.19 mol/kg
- Freezing point of the solution (Tf) = -0.704 °C
- Freezing point of pure water (T0f) = 0 °C
- Freezing point depression constant (Kf) for water = 1.86 K kg/mol

2. **Calculate the depression in freezing point (ΔTf):**
\[ \Delta T_f = T_0^f - T_f \]
Substituting the values:
\[ \Delta T_f = 0 - (-0.704) = 0.704 \, \text{°C} \]

3. **Use the formula for freezing point depression:**
\[ \Delta T_f = i \cdot K_f \cdot m \]
Rearranging to solve for the Van't Hoff factor (i):
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

4. **Substituting the known values into the equation:**
\[ i = \frac{0.704}{1.86 \cdot 0.19} \]

5. **Calculate the denominator:**
\[ 1.86 \cdot 0.19 = 0.3534 \]

6. **Now calculate the Van't Hoff factor (i):**
\[ i = \frac{0.704}{0.3534} \approx 1.992 \]

7. **Round the result to the nearest whole number:**
\[ i \approx 2 \]

### Conclusion:
The Van't Hoff factor (i) of the ionic salt is approximately 2.

### Final Answer:
The correct option is **B) 2**.

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