Calculate the van't Hoff factor of `CdSO_(4)` (molecular mass 208.4)
if the dissolution of 5.21 g of `CdSO_(4)` in half litre water gives a depression in freezing point of `0.
168^(@)C` (`K_(f)` of water is `"1.86 K kg mol"^(-1)`)

Explanation

To calculate the van't Hoff factor (i) of `CdSO_(4)`, we can follow these steps:

### Step 1: Identify the given data
- Mass of `CdSO_(4)` (solute) = 5.21 g
- Volume of water (solvent) = 0.5 L = 500 g (since the density of water is approximately 1 g/mL)
- Depression in freezing point (ΔTf) = 0.168 °C
- Freezing point depression constant (Kf) of water = 1.86 K kg mol⁻¹
- Molecular mass of `CdSO_(4)` = 208.4 g/mol

### Step 2: Calculate the molality (m) of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent.

1. Calculate the number of moles of `CdSO_(4)`:
\[ \text{Moles of } CdSO_4 = \frac{\text{mass of } CdSO_4}{\text{molar mass of } CdSO_4} = \frac{5.21 \text{ g}}{208.4 \text{ g/mol}} \approx 0.0250 \text{ mol} \]

2. Convert the mass of the solvent (water) to kilograms:
\[ \text{Mass of solvent in kg} = \frac{500 \text{ g}}{1000} = 0.5 \text{ kg} \]

3. Calculate the molality (m):
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.0250 \text{ mol}}{0.5 \text{ kg}} = 0.0500 \text{ mol/kg} \]

### Step 3: Use the freezing point depression formula
The freezing point depression is given by the formula:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- ΔTf = depression in freezing point
- i = van't Hoff factor
- Kf = freezing point depression constant
- m = molality

### Step 4: Rearrange the formula to solve for i
Rearranging the formula gives:
\[ i = \frac{\Delta T_f}{K_f \cdot m} \]

### Step 5: Substitute the known values into the equation
Substituting the values we have:
\[ i = \frac{0.168 \text{ °C}}{1.86 \text{ K kg mol}^{-1} \cdot 0.0500 \text{ mol/kg}} \]

### Step 6: Calculate the van't Hoff factor (i)
Calculating the denominator:
\[ 1.86 \cdot 0.0500 = 0.093 \]
Now, substituting this back into the equation for i:
\[ i = \frac{0.168}{0.093} \approx 1.806 \]

### Conclusion
The van't Hoff factor (i) for `CdSO_(4)` is approximately **1.806**.

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