Molal elevation constant of chloroform is `3.88`, `0.3g` of camphor added to `25.2g` of chloroform raised the boiling point of the solvent by `0.299^(@)C`. Calculate the molecular weight of camphor.

Answer

To calculate the molecular weight of camphor using the given data, we can follow these steps:

### Step 1: Understand the formula for boiling point elevation
The formula for boiling point elevation is given by:
\[ \Delta T_b = K_b \cdot m \]
where:
- \(\Delta T_b\) = change in boiling point (°C)
- \(K_b\) = molal elevation constant (°C kg/mol)
- \(m\) = molality of the solution (mol/kg)

### Step 2: Identify the given values
From the question, we have:
- \(K_b\) for chloroform = 3.88 °C kg/mol
- Mass of camphor (solute) = 0.3 g
- Mass of chloroform (solvent) = 25.2 g
- Change in boiling point (\(\Delta T_b\)) = 0.299 °C

### Step 3: Convert the mass of chloroform to kg
To use the molality formula, we need the mass of the solvent in kilograms:
\[ \text{Mass of chloroform in kg} = \frac{25.2 \, \text{g}}{1000} = 0.0252 \, \text{kg} \]

### Step 4: Calculate the molality of the solution
Rearranging the boiling point elevation formula to find molality:
\[ m = \frac{\Delta T_b}{K_b} \]
Substituting the values:
\[ m = \frac{0.299}{3.88} \approx 0.0770 \, \text{mol/kg} \]

### Step 5: Calculate the number of moles of camphor
Using the definition of molality:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Rearranging gives us:
\[ \text{moles of camphor} = m \cdot \text{mass of solvent in kg} = 0.0770 \cdot 0.0252 \approx 0.00194 \, \text{moles} \]

### Step 6: Calculate the molecular weight of camphor
The molecular weight (M) can be calculated using the formula:
\[ M = \frac{\text{mass of solute}}{\text{moles of solute}} \]
Substituting the values:
\[ M = \frac{0.3 \, \text{g}}{0.00194 \, \text{moles}} \approx 154.64 \, \text{g/mol} \]

### Conclusion
The molecular weight of camphor is approximately **154.64 g/mol**.

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