An aqueous solution freezes on `0.36^@C K_f and K_b` for water are 1.8 and 0.52 k kg `mol^-1` respectively then value of boiling point of solution as 1 atm pressure is

Options

A. `101.04^@C`
B. `100.104^@C`
C. `0.104^@C`
D. `100^@C`

Answer

To solve the problem, we need to find the boiling point of the solution given its freezing point depression. We will use the formulas for freezing point depression and boiling point elevation.

### Step-by-Step Solution:

1. **Identify the Given Data:**
- Freezing point depression (ΔTf) = 0.36°C
- Kf (freezing point depression constant for water) = 1.8 °C kg/mol
- Kb (boiling point elevation constant for water) = 0.52 °C kg/mol

2. **Determine the Freezing Point of Pure Water:**
- The freezing point of pure water is 0°C.
- Since the solution freezes at -0.36°C, the depression in freezing point is ΔTf = 0°C - (-0.36°C) = 0.36°C.

3. **Use the Freezing Point Depression Formula:**
\[ \Delta T_f = i \cdot m \cdot K_f \]
Here, \(i\) is the van 't Hoff factor (which we will assume to be 1 for a non-dissociating solute), and \(m\) is the molality of the solution.

4. **Rearranging to Find Molality (m):**
\[ m = \frac{\Delta T_f}{i \cdot K_f} \]
Substituting the known values:
\[ m = \frac{0.36}{1 \cdot 1.8} = \frac{0.36}{1.8} = 0.2 \, \text{mol/kg} \]

5. **Use the Boiling Point Elevation Formula:**
\[ \Delta T_b = i \cdot m \cdot K_b \]
Again, assuming \(i = 1\):
\[ \Delta T_b = 1 \cdot 0.2 \cdot 0.52 = 0.104°C \]

6. **Calculate the Boiling Point of the Solution:**
- The boiling point of pure water is 100°C.
- Therefore, the boiling point of the solution is:
\[ \text{Boiling Point} = 100°C + \Delta T_b = 100°C + 0.104°C = 100.104°C \]

### Final Answer:
The boiling point of the solution at 1 atm pressure is **100.104°C**.