If sodium sulphate is considered to be completely dissociated into cations and anions in aqueous solution,the change in freezing point of water`(DeltaT_(f))`,when `0.01`mole of sodium sulphate is dissoved in `1` kg of water,is `(K_(f)=1.86Kkg mol^(-1))`

Options

A. `0.0372K`
B. `0.0558K`
C. `0.0744K`
D. `0.0186K`

Explanation

To solve the problem of determining the change in freezing point of water when 0.01 moles of sodium sulfate (Na₂SO₄) is dissolved in 1 kg of water, we will follow these steps:

### Step 1: Determine the van 't Hoff factor (i)
Sodium sulfate dissociates in water as follows:
\[ \text{Na}_2\text{SO}_4 \rightarrow 2 \text{Na}^+ + \text{SO}_4^{2-} \]

From this dissociation, we can see that one formula unit of sodium sulfate produces 3 ions (2 sodium ions and 1 sulfate ion). Therefore, the van 't Hoff factor (i) is:
\[ i = 3 \]

### Step 2: Calculate the molality (m)
Molality (m) is defined as the number of moles of solute per kilogram of solvent. Here, we have:
- Moles of solute (Na₂SO₄) = 0.01 moles
- Mass of solvent (water) = 1 kg

Using the formula for molality:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
\[ m = \frac{0.01 \text{ moles}}{1 \text{ kg}} = 0.01 \text{ mol/kg} \]

### Step 3: Use the freezing point depression formula
The change in freezing point (\(\Delta T_f\)) can be calculated using the formula:
\[ \Delta T_f = K_f \cdot m \cdot i \]

Where:
- \( K_f = 1.86 \, \text{K kg mol}^{-1} \) (freezing point depression constant for water)
- \( m = 0.01 \, \text{mol/kg} \)
- \( i = 3 \)

Substituting the values into the formula:
\[ \Delta T_f = 1.86 \, \text{K kg mol}^{-1} \cdot 0.01 \, \text{mol/kg} \cdot 3 \]
\[ \Delta T_f = 1.86 \cdot 0.01 \cdot 3 \]
\[ \Delta T_f = 0.0558 \, \text{K} \]

### Final Answer
The change in freezing point of water (\(\Delta T_f\)) when 0.01 moles of sodium sulfate is dissolved in 1 kg of water is:
\[ \Delta T_f = 0.0558 \, \text{K} \]

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