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:
\[ \text{Molar mass of Na}_2\text{SO}_4 = 2 \times \text{(mass of Na)} + \text{(mass of S)} + 4 \times \text{(mass of O)} \]
\[ = 2 \times 23 + 32 + 4 \times 16 = 46 + 32 + 64 = 142 \text{ g/mol} \]

### Step 2: Determine the degree of ionization (α)
Given that Na₂SO₄ is 81.5% ionized, we can express this as:
\[ \alpha = 0.815 \]

### Step 3: Calculate the van 't Hoff factor (i)
The dissociation of Na₂SO₄ can be represented as:
\[ \text{Na}_2\text{SO}_4 \rightarrow 2 \text{Na}^+ + \text{SO}_4^{2-} \]
This means that one formula unit of Na₂SO₄ produces 3 ions (2 Na⁺ and 1 SO₄²⁻). Therefore, the total number of ions (n) is 3.

Using the formula for the van 't Hoff factor:
\[ i = 1 + (n - 1) \alpha \]
Substituting the values:
\[ i = 1 + (3 - 1) \times 0.815 = 1 + 2 \times 0.815 = 1 + 1.63 = 2.63 \]

### Step 4: Use the freezing point depression formula
The freezing point depression formula is given by:
\[ \Delta T_f = i \cdot K_f \cdot m \]
Where:
- \(\Delta T_f = 3.82^\circ C\)
- \(K_f = 1.86^\circ C \cdot \text{kg/mol}\)
- \(m\) is the molality, defined as:
\[ m = \frac{\text{moles of solute}}{\text{kg of solvent}} \]

### Step 5: Calculate the moles of Na₂SO₄
The number of moles of Na₂SO₄ can be calculated using the mass and molar mass:
\[ \text{Moles of Na}_2\text{SO}_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{5 \text{ g}}{142 \text{ g/mol}} \approx 0.0352 \text{ mol} \]

### Step 6: Calculate the molality (m)
The molality can be expressed as:
\[ m = \frac{0.0352 \text{ mol}}{\frac{x}{1000}} = \frac{0.0352 \times 1000}{x} \]

### Step 7: Substitute into the freezing point depression equation
Now substituting the values into the freezing point depression formula:
\[ 3.82 = 2.63 \cdot 1.86 \cdot \frac{0.0352 \times 1000}{x} \]

### Step 8: Solve for x
Rearranging the equation to solve for x:
\[ x = \frac{2.63 \cdot 1.86 \cdot 0.0352 \cdot 1000}{3.82} \]
Calculating the right-hand side:
\[ x \approx \frac{2.63 \cdot 1.86 \cdot 0.0352 \cdot 1000}{3.82} \approx \frac{244.59}{3.82} \approx 64.1 \text{ g} \]

### Step 9: Final calculation
After calculating, we find:
\[ x \approx 45.09 \text{ g} \text{ (rounded to 45 g)} \]

### Conclusion
Thus, the value of x (the mass of water) is approximately **45 g**.