Q11 / 150 · MHT CET 2025 (19 Apr Shift 2) · Chemistry

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

  • A
    $43 \cdot 5 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • B
    $35 \cdot 3 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • Correct answer
    C
    $30 \cdot 6 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • D
    $40 \cdot 7 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

Enthalpy change of vaporisation calculation for benzene

The molar mass of benzene $(C_6H_6)$ is $6 \times 12 + 6 \times 1 = 78$ g/mol. With $13$ g vaporized, the number of moles is $\frac{13}{78} = \frac{1}{6}$ mol.

Given that $5.1$ kJ of heat is supplied, the enthalpy change per mole is calculated as $\Delta H_{\text{vap}} = \frac{5.1}{1/6} = 5.1 \times 6 = 30.6$ kJ/mol.

The calculated value of $30.6$ kJ/mol corresponds to option C.

Final answer: $\boxed{C}$

Hence, the correct answer is option: 'C'.

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