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 data provided. Here are the steps to arrive at the solution:

### Step 1: Calculate the number of moles of naphthalene
To find the number of moles of naphthalene, we use the formula:
\[ \text{Number of moles} = \frac{\text{mass}}{\text{molar mass}} \]
Given:
- Mass of naphthalene = 0.65 g
- Molar mass of naphthalene \((C_{10}H_8)\) = \(12 \times 10 + 1 \times 8 = 128 \, \text{g/mol}\)

Calculating the number of moles:
\[ \text{Number of moles} = \frac{0.65 \, \text{g}}{128 \, \text{g/mol}} = 0.005078 \, \text{mol} \]

### 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 equal to 0.1 kg.
\[ \text{Molality} (m) = \frac{\text{Number of moles of solute}}{\text{mass of solvent (kg)}} \]
Calculating molality:
\[ m = \frac{0.005078 \, \text{mol}}{0.1 \, \text{kg}} = 0.05078 \, \text{mol/kg} \]

### Step 3: Use the elevation in boiling point to find \(K_b\)
The elevation in boiling point (\(\Delta T_b\)) is given as 0.103°C. The relationship between the elevation in boiling point and molality is given by:
\[ \Delta T_b = K_b \cdot m \]
Rearranging to find \(K_b\):
\[ K_b = \frac{\Delta T_b}{m} \]
Substituting the values:
\[ K_b = \frac{0.103 \, \text{°C}}{0.05078 \, \text{mol/kg}} = 2.028 \, \text{°C kg/mol} \]

### Step 4: Calculate the molar heat of vaporization (\(\Delta H_{vap}\))
Using the formula:
\[ K_b = \frac{R \cdot M \cdot T_b^2}{1000 \cdot \Delta H_{vap}} \]
Where:
- \(R = 2 \, \text{cal/(K mol)}\) (gas constant)
- \(M\) (molar mass of methyl acetate) = 74 g/mol
- \(T_b\) (boiling point of methyl acetate in Kelvin) = \(57 + 273 = 330 \, \text{K}\)

Rearranging to find \(\Delta H_{vap}\):
\[ \Delta H_{vap} = \frac{R \cdot M \cdot T_b^2}{1000 \cdot K_b} \]
Substituting the values:
\[ \Delta H_{vap} = \frac{2 \cdot 74 \cdot (330)^2}{1000 \cdot 2.028} \]
Calculating:
\[ \Delta H_{vap} = \frac{2 \cdot 74 \cdot 108900}{1000 \cdot 2.028} = \frac{16056 \, \text{cal}}{2.028} \approx 7916.6 \, \text{cal/mol} \]
Converting to kilocalories:
\[ \Delta H_{vap} \approx 7.9166 \, \text{kcal/mol} \]

### Final Answer
The molar heat of vaporization of methyl acetate is approximately:
\[ \Delta H_{vap} \approx 7.917 \, \text{kcal/mol} \]