Calculate the `K_(b)` for chloroform from the following data
(a) Boiling point of pure `CHCl_(3)=61.3^(@)C`
(b) The solutionn containing `5.02g` of naphthalene `(C_(10)H_(8))` in 18g of `CHCl_(3)` boils at `69.5^(@)C`

Answer

To calculate the `K_b` (boiling point elevation constant) for chloroform (CHCl₃), we can follow these steps:

### Step 1: Determine the boiling point elevation (\( \Delta T_b \))
The boiling point elevation can be calculated using the formula:
\[ \Delta T_b = T_{solution} - T_{pure} \]
Where:
- \( T_{solution} = 69.5^\circ C \) (boiling point of the solution)
- \( T_{pure} = 61.3^\circ C \) (boiling point of pure chloroform)

Substituting the values:
\[ \Delta T_b = 69.5^\circ C - 61.3^\circ C = 8.2^\circ C \]

### Step 2: Calculate the moles of solute (naphthalene)
To find the moles of naphthalene, we use the formula:
\[ \text{moles} = \frac{\text{mass}}{\text{molar mass}} \]
Given:
- Mass of naphthalene = 5.02 g
- Molar mass of naphthalene (C₁₀H₈) = 128 g/mol

Calculating the moles:
\[ \text{moles of naphthalene} = \frac{5.02 \, g}{128 \, g/mol} \approx 0.0392 \, mol \]

### Step 3: Calculate the molality of the solution
Molality (m) is defined as the moles of solute per kilogram of solvent. The mass of chloroform is given as 18 g, which is 0.018 kg.

Using the formula:
\[ m = \frac{\text{moles of solute}}{\text{mass of solvent in kg}} \]
Substituting the values:
\[ m = \frac{0.0392 \, mol}{0.018 \, kg} \approx 2.1778 \, mol/kg \]

### Step 4: Use the boiling point elevation formula to find \( K_b \)
The boiling point elevation can also be expressed as:
\[ \Delta T_b = K_b \times m \]
Rearranging this gives us:
\[ K_b = \frac{\Delta T_b}{m} \]
Substituting the values:
\[ K_b = \frac{8.2^\circ C}{2.1778 \, mol/kg} \approx 3.76 \, °C \, kg/mol \]

### Final Answer
The value of \( K_b \) for chloroform is approximately **3.76 °C kg/mol**.

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