Colligative property measurement is one of the techniques used in the measurement of chemical quantities with reasonable accuracy.
If a 40.65 gm sample of `K_2SO_4` and `BaSO_4` is dissolved in 900 gm of pure water to form a solution 'A' at `57^@C`.Its vapours pressure is found to be 39.6 torr while vapour pressure of pure water at `57^@C` is 40 torr. Density of solution A is 1.24g/ml.
In a different experiment when small amount of pure `BaSO_4` is mixed with water at `57^@C` it gives the osomotic rise at `4.05xx10^(-5)` atm.
`(R=0.082 Lt.-atm//"mol"-K,K=39,Ba=137,S=32,O=16)`
Elevation in boiling point of solution A is `(K_b "water"=0.54 K-kg//mol)`:

Options

A. 0.3 K
B. 0.1 K
C. 0.04 K
D. 0.05 K

Explanation

To solve the problem, we need to calculate the elevation in boiling point (\( \Delta T_b \)) of the solution A using the formula for colligative properties. Here are the steps:

### Step 1: Calculate the molality of the solution
We need to find the number of moles of solute (\( K_2SO_4 \) and \( BaSO_4 \)) and then calculate the molality.

1. **Calculate the molar mass of \( K_2SO_4 \)**:
- Potassium (K): 39 g/mol × 2 = 78 g/mol
- Sulfur (S): 32 g/mol × 1 = 32 g/mol
- Oxygen (O): 16 g/mol × 4 = 64 g/mol
- Total = 78 + 32 + 64 = 174 g/mol

2. **Calculate the number of moles of \( K_2SO_4 \)**:
\[ \text{Moles of } K_2SO_4 = \frac{\text{mass}}{\text{molar mass}} = \frac{40.65 \text{ g}}{174 \text{ g/mol}} \approx 0.233 \text{ moles} \]

3. **Calculate the mass of water in kg**:
\[ \text{Mass of water} = 900 \text{ g} = 0.9 \text{ kg} \]

4. **Calculate molality (m)**:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} = \frac{0.233 \text{ moles}}{0.9 \text{ kg}} \approx 0.259 \text{ mol/kg} \]

### Step 2: Determine the van 't Hoff factor (i)
For \( K_2SO_4 \), it dissociates into 3 ions:
\[ K_2SO_4 \rightarrow 2K^+ + SO_4^{2-} \]
Thus, \( i = 3 \).

### Step 3: Use the boiling point elevation formula
The formula for boiling point elevation is:
\[ \Delta T_b = i \cdot K_b \cdot m \]
Where:
- \( K_b \) for water = 0.54 K·kg/mol
- \( m \) = 0.259 mol/kg
- \( i = 3 \)

### Step 4: Calculate \( \Delta T_b \)
Substituting the values into the formula:
\[ \Delta T_b = 3 \cdot 0.54 \cdot 0.259 \approx 0.419 \text{ K} \]

### Step 5: Round off the value
Rounding off the calculated value gives:
\[ \Delta T_b \approx 0.42 \text{ K} \]

### Conclusion
The elevation in boiling point of solution A is approximately \( 0.42 \text{ K} \).

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