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 need to determine the mass of water (x grams) in which 5 g of Na₂SO₄ is dissolved, given the change in freezing point and the degree of ionization. Let's break down the solution step by step.

### Step 1: Determine the molar mass of Na₂SO₄
The molar mass of Na₂SO₄ can be calculated as follows:
- Molar mass of Na = 23 g/mol
- Molar mass of S = 32 g/mol
- Molar mass of O = 16 g/mol

Thus, the molar mass of Na₂SO₄:
\[ \text{Molar mass of Na₂SO₄} = 2 \times 23 + 32 + 4 \times 16 = 46 + 32 + 64 = 142 \text{ g/mol} \]

### Step 2: Calculate the number of moles of Na₂SO₄
Using the mass of Na₂SO₄ (5 g), we can find the number of moles:
\[ \text{Moles of Na₂SO₄} = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \text{ g}}{142 \text{ g/mol}} \approx 0.0352 \text{ mol} \]

### Step 3: Calculate the van 't Hoff factor (i)
Na₂SO₄ dissociates in water as follows:
\[ \text{Na₂SO₄} \rightarrow 2 \text{Na}^+ + \text{SO₄}^{2-} \]
The degree of ionization (α) is given as 81.5%, or 0.815. The van 't Hoff factor (i) can be calculated using the formula:
\[ i = 1 + (n - 1) \alpha \]
where \(n\) is the number of particles the solute dissociates into. For Na₂SO₄, \(n = 3\):
\[ i = 1 + (3 - 1) \times 0.815 = 1 + 2 \times 0.815 = 1 + 1.63 = 2.63 \]

### Step 4: Use the freezing point depression formula
The formula for freezing point depression is:
\[ \Delta T_f = i \cdot K_f \cdot m \]
where:
- \(\Delta T_f = 3.82^\circ C\)
- \(K_f = 1.86 \text{ °C kg/mol}\)
- \(m\) is the molality.

Rearranging the equation to find molality:
\[ m = \frac{\Delta T_f}{i \cdot K_f} = \frac{3.82}{2.63 \cdot 1.86} \approx 0.780 \text{ mol/kg} \]

### Step 5: Relate molality to mass of solvent
Molality (m) is defined as:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Let the mass of solvent (water) be \(x\) grams, which is \(x/1000\) kg. Therefore:
\[ 0.780 = \frac{0.0352}{x/1000} \]

### Step 6: Solve for x
Rearranging the equation gives:
\[ x = \frac{0.0352 \times 1000}{0.780} \approx 45.142 \text{ g} \]

### Conclusion
Thus, the approximate value of \(x\) is:
\[ \boxed{45 \text{ g}} \]