0.65 g naphthalene `(C_(10)H_(8))` was dissolved in 100 g methyl acetate. Elevation in boiling point of methyl acetate solution was `0.103^(@)C`. If boiling point of pure methyl acetate is `57^(@)C`, its molar heat of vaporisation will be:

Options

A. 8.96 kcal `mol^(-1)`
B. 7.946 kcal `mol^(-1)`
C. 6.24 kcal `mol^(-1)`
D. 15.65 kcal `mol^(-1)`

Answer

To solve the problem, we need to find the molar heat of vaporization of methyl acetate using the given data. Here’s a step-by-step solution:

### Step 1: Calculate the number of moles of naphthalene (C₁₀H₈)
To find the number of moles of naphthalene, we use the formula:

\[ \text{Number of moles} = \frac{\text{mass of solute (g)}}{\text{molar mass of solute (g/mol)}} \]

The molar mass of naphthalene (C₁₀H₈) is calculated as follows:
- Carbon (C): 12.01 g/mol × 10 = 120.1 g/mol
- Hydrogen (H): 1.008 g/mol × 8 = 8.064 g/mol
- Total molar mass = 120.1 + 8.064 = 128.164 g/mol

Now, we can calculate the number of moles:

\[ \text{Number of moles of naphthalene} = \frac{0.65 \, \text{g}}{128.164 \, \text{g/mol}} \approx 0.00507 \, \text{mol} \]

### Step 2: Calculate the molality of the solution
Molality (m) is defined as the number of moles of solute per kilogram of solvent. The mass of the solvent (methyl acetate) is given as 100 g, which is 0.1 kg.

\[ \text{Molality} = \frac{\text{Number of moles of solute}}{\text{mass of solvent (kg)}} = \frac{0.00507 \, \text{mol}}{0.1 \, \text{kg}} = 0.0507 \, \text{mol/kg} \]

### Step 3: Use the boiling point elevation formula
The elevation in boiling point (\(\Delta T_b\)) is related to molality and the ebullioscopic constant (\(K_b\)) of the solvent:

\[ \Delta T_b = K_b \cdot m \]

Given that \(\Delta T_b = 0.103^\circ C\) and \(m = 0.0507 \, \text{mol/kg}\), we can rearrange the formula to find \(K_b\):

\[ K_b = \frac{\Delta T_b}{m} = \frac{0.103^\circ C}{0.0507 \, \text{mol/kg}} \approx 2.03 \, ^\circ C \cdot \text{kg/mol} \]

### Step 4: Relate \(K_b\) to the molar heat of vaporization
The relationship between \(K_b\) and the molar heat of vaporization (\(\Delta H_{vap}\)) is given by:

\[ K_b = \frac{R \cdot M_1 \cdot T_b^2}{1000 \cdot \Delta H_{vap}} \]

Where:
- \(R\) = 2 (in kcal)
- \(M_1\) = molar mass of methyl acetate (C₃H₈O₂) = 60.05 g/mol
- \(T_b\) = boiling point of methyl acetate in Kelvin = \(57 + 273 = 330 \, K\)

### Step 5: Rearranging to find \(\Delta H_{vap}\)
Rearranging the equation gives:

\[ \Delta H_{vap} = \frac{R \cdot M_1 \cdot T_b^2}{1000 \cdot K_b} \]

Substituting the values:

\[ \Delta H_{vap} = \frac{2 \cdot 60.05 \cdot (330)^2}{1000 \cdot 2.03} \]

Calculating the values:

\[ \Delta H_{vap} = \frac{2 \cdot 60.05 \cdot 108900}{1000 \cdot 2.03} \approx \frac{13036.67}{2.03} \approx 6420.57 \, \text{kcal/mol} \]

### Final Answer
The molar heat of vaporization of methyl acetate is approximately **6420.57 kcal/mol**.