Question

Calculate the enthalpy change of vaporisation of benzene if 13 gram of benzene vaporised by supplying 5.1 kJ of heat.

• A. \( 43.5 \text{ kJ mol}^{-1} \)
• B. \( 35.3 \text{ kJ mol}^{-1} \)
• C. \( 30.6 \text{ kJ mol}^{-1} \)
• D. \( 40.7 \text{ kJ mol}^{-1} \)

Hint:

Always convert the given mass to moles first to find molar enthalpy values.

Answer 1

The Correct Option is C

Step 1: Concept
Enthalpy of vaporization (\(\Delta H_{vap}\)) is the heat required to vaporize one mole of a substance at constant pressure.

Step 2: Meaning
Molar mass of benzene (\(C_6H_6\)) is \(78 \text{ g/mol}\).

Step 3: Analysis
Number of moles (\(n\)) = \(\frac{\text{mass}}{\text{molar mass}} = \frac{13}{78} = \frac{1}{6} \text{ mol}\).
\(\Delta H_{vap} = \frac{\text{Heat supplied}}{\text{moles}} = \frac{5.1 \text{ kJ}}{1/6 \text{ mol}}\).

Step 4: Conclusion
\(\Delta H_{vap} = 5.1 \times 6 = 30.6 \text{ kJ/mol}\). Final Answer: (C)