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 an aqueous solution given its freezing point depression and the constants \( K_f \) and \( K_b \) for water. Here’s a step-by-step solution:

### Step 1: Understand the given data
- Freezing point depression (\( \Delta T_f \)) = 0.36°C
- \( K_f \) for water = 1.8 °C kg/mol
- \( K_b \) for water = 0.52 °C kg/mol

### Step 2: Calculate the molality of the solution
Using the formula for freezing point depression:
\[ \Delta T_f = K_f \times m \]
Where \( m \) is the molality of the solution.

Rearranging the formula to find molality:
\[ m = \frac{\Delta T_f}{K_f} \]
Substituting the values:
\[ m = \frac{0.36}{1.8} = 0.2 \text{ mol/kg} \]

### Step 3: Calculate the elevation in boiling point (\( \Delta T_b \))
Using the formula for boiling point elevation:
\[ \Delta T_b = K_b \times m \]
Substituting the values:
\[ \Delta T_b = 0.52 \times 0.2 = 0.104 \text{ °C} \]

### Step 4: Calculate the boiling point of the solution
The boiling point of the solution (\( T_b \)) can be calculated using the formula:
\[ T_b = T_{b,\text{solvent}} + \Delta T_b \]
Where \( T_{b,\text{solvent}} \) is the boiling point of the solvent (water), which is 100°C.

Substituting the values:
\[ T_b = 100 + 0.104 = 100.104 \text{ °C} \]

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

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