0.15 molal solution of NaCI has freezing point `-0.52 ^(@)C`
Calculate van't Hoff factor .`(K_(f) = 1.86 K kg mol^(-1) )`

Explanation

To calculate the van't Hoff factor (i) for a 0.15 molal solution of NaCl with a freezing point of -0.52 °C, we can follow these steps:

### Step 1: Identify the given values
- Molality (m) = 0.15 mol/kg
- Freezing point of the solution (Tf) = -0.52 °C
- Freezing point of pure solvent (T0) = 0 °C (for water)
- Freezing point depression constant (Kf) = 1.86 K kg/mol

### Step 2: Convert the freezing point to Kelvin
To use the freezing point in calculations, we need to convert it to Kelvin:
- T0 = 0 °C = 273 K
- Tf = -0.52 °C = 273 K - 0.52 °C = 272.48 K

### Step 3: Calculate the change in freezing point (ΔTf)
ΔTf is calculated as:
\[ \Delta T_f = T_0 - T_f = 273 K - 272.48 K = 0.52 K \]

### Step 4: Use the formula for freezing point depression
The formula for freezing point depression is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- ΔTf = change in freezing point
- i = van't Hoff factor
- Kf = freezing point depression constant
- m = molality

### Step 5: Rearrange the formula to solve for i
Rearranging the formula gives us:
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

### Step 6: Substitute the known values into the formula
Substituting the known values:
\[ i = \frac{0.52 K}{1.86 K \cdot 0.15 \, \text{mol/kg}} \]

### Step 7: Calculate the denominator
Calculating the denominator:
\[ 1.86 K \cdot 0.15 \, \text{mol/kg} = 0.279 K \, \text{mol/kg} \]

### Step 8: Calculate the van't Hoff factor (i)
Now substituting back into the equation:
\[ i = \frac{0.52 K}{0.279 K \, \text{mol/kg}} \approx 1.864 \]

### Final Answer
The van't Hoff factor (i) for the NaCl solution is approximately **1.864**.

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