5 g of `Na_(2)SO_(4)` was dissolved in x g of `H_(2)O` . The change in freezing point was found to be `3.82^(@)C` . If `Na_(2)SO_(4)` is `81.5%` ionised , the value of x
(`k_(f)` for water =`1.86^(@)C` kg `"mol"^(-1)`) is apporximately :
(molar mass of S=32 g `"mol"^(-1)` and that of Na=23 g `"mol"^(-1)`)

Options

A. 25 g
B. 65 g
C. 15 g
D. 45 g

Answer

To solve the problem, we will follow these steps:

### Step 1: Calculate the molar mass of Na₂SO₄
The molar mass of Na₂SO₄ can be calculated using the atomic masses provided:
- Sodium (Na) = 23 g/mol
- Sulfur (S) = 32 g/mol
- Oxygen (O) = 16 g/mol

The molar mass of Na₂SO₄ is calculated as follows:
\[ \text{Molar mass of Na₂SO₄} = 2 \times 23 + 32 + 4 \times 16 = 46 + 32 + 64 = 142 \text{ g/mol} \]

### Step 2: Determine the degree of ionization (α) and van 't Hoff factor (i)
Given that Na₂SO₄ is 81.5% ionized, we can express this as:
\[ \alpha = \frac{81.5}{100} = 0.815 \]

The van 't Hoff factor (i) for Na₂SO₄, which dissociates into 2 Na⁺ and 1 SO₄²⁻, can be calculated using:
\[ i = 1 + (n - 1) \cdot \alpha \]
where \( n \) is the number of particles the solute dissociates into. For Na₂SO₄, \( n = 3 \) (2 Na⁺ + 1 SO₄²⁻):
\[ i = 1 + (3 - 1) \cdot 0.815 = 1 + 2 \cdot 0.815 = 1 + 1.63 = 2.63 \]

### Step 3: Use the freezing point depression formula
The freezing point depression (\( \Delta T_f \)) is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \( \Delta T_f = 3.82^\circ C \)
- \( K_f = 1.86 \, ^\circ C \, \text{kg/mol} \)
- \( m \) is the molality of the solution.

### Step 4: Calculate molality (m)
We can rearrange the formula to find molality:
\[ m = \frac{\Delta T_f}{i \cdot K_f} \]
Substituting the known values:
\[ m = \frac{3.82}{2.63 \cdot 1.86} = \frac{3.82}{4.8958} \approx 0.780 \, \text{mol/kg} \]

### Step 5: Relate molality to the mass of the solvent (x)
Molality is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent (kg)}} \]
The number of moles of Na₂SO₄ in 5 g is:
\[ \text{moles of Na₂SO₄} = \frac{5 \, \text{g}}{142 \, \text{g/mol}} \approx 0.0352 \, \text{mol} \]

Let \( x \) be the mass of water in grams. To convert to kg, we have:
\[ \text{mass of solvent} = \frac{x}{1000} \, \text{kg} \]

Setting up the equation:
\[ 0.780 = \frac{0.0352}{\frac{x}{1000}} \]
Rearranging gives:
\[ x = \frac{0.0352 \times 1000}{0.780} \approx 45.05 \, \text{g} \]

### Final Answer
Thus, the value of \( x \) is approximately **45 g**.

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