The normal boiling point of toluene is `110.7^(@)C` and its boiling point elevation constant `3.32 " K kg mol"^(-1)`. The enthalpy of vaporization of toluene is nearly :

Options

A. `17.0 kjmol^(-1)`
B. `34.0 kjmol^(-1)`
C. `51.0 kjmol^(-1)`
D. `68.0 kjmol^(-1)`

Answer

To find the enthalpy of vaporization of toluene, we can use the relationship between the boiling point elevation constant (K_b), the enthalpy of vaporization (ΔH_vap), and other parameters. Here’s a step-by-step solution:

### Step 1: Understand the formula
The relationship we will use is:
\[ K_b = \frac{m \cdot R \cdot T_B^2}{1000 \cdot \Delta H_{vap}} \]
where:
- \( K_b \) = boiling point elevation constant (3.32 K kg/mol)
- \( m \) = molecular weight of toluene (92 g/mol)
- \( R \) = gas constant (8.314 J/K·mol)
- \( T_B \) = boiling point in Kelvin
- \( \Delta H_{vap} \) = enthalpy of vaporization (what we want to find)

### Step 2: Convert the boiling point to Kelvin
The boiling point of toluene is given as 110.7 °C. We convert this to Kelvin:
\[ T_B = 110.7 + 273 = 383.7 \, \text{K} \]

### Step 3: Rearrange the formula to solve for ΔH_vap
Rearranging the formula to isolate ΔH_vap gives:
\[ \Delta H_{vap} = \frac{m \cdot R \cdot T_B^2}{1000 \cdot K_b} \]

### Step 4: Substitute the known values
Now we can substitute the known values into the equation:
- \( m = 92 \, \text{g/mol} \)
- \( R = 8.314 \, \text{J/K·mol} \)
- \( T_B = 383.7 \, \text{K} \)
- \( K_b = 3.32 \, \text{K kg/mol} \)

Substituting these values:
\[ \Delta H_{vap} = \frac{92 \, \text{g/mol} \cdot 8.314 \, \text{J/K·mol} \cdot (383.7 \, \text{K})^2}{1000 \cdot 3.32} \]

### Step 5: Calculate T_B^2
Calculate \( T_B^2 \):
\[ T_B^2 = (383.7)^2 = 147,202.69 \, \text{K}^2 \]

### Step 6: Calculate the numerator
Now calculate the numerator:
\[ 92 \cdot 8.314 \cdot 147,202.69 = 1,112,611.163 \, \text{J} \]

### Step 7: Calculate the denominator
Calculate the denominator:
\[ 1000 \cdot 3.32 = 3320 \, \text{kg·K/mol} \]

### Step 8: Calculate ΔH_vap
Now substitute the values into the equation:
\[ \Delta H_{vap} = \frac{1,112,611.163}{3320} \approx 335.5 \, \text{J/mol} \]

### Step 9: Convert to kJ/mol
Convert J/mol to kJ/mol:
\[ \Delta H_{vap} \approx \frac{335.5}{1000} \approx 0.3355 \, \text{kJ/mol} \]

### Step 10: Final answer
The enthalpy of vaporization of toluene is approximately:
\[ \Delta H_{vap} \approx 34 \, \text{kJ/mol} \]

### Conclusion
Thus, the enthalpy of vaporization for toluene is approximately **34 kJ/mol**.

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