Q22 / 150 · MHT CET 2025 (22 Apr Shift 1) · Chemistry

1.8 g water is vapourised by supplying 4 kJ heat at $100^{\circ} \mathrm{C}$. What is the heat of vapourisation of water at same temperature?

  • A
    $8 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • Correct answer
    B
    $40 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • C
    $18 \mathrm{~kJ} \mathrm{~mol}^{-1}$
  • D
    $32 \mathrm{~kJ} \mathrm{~mol}^{-1}$

Solution

Heat of vaporization is the heat required to vaporize one mole of a substance at its boiling point. Given 4 kJ vaporizes 1.8 g of water, and the molar mass of water is 18 g/mol, the number of moles is $n = \frac{1.8}{18} = 0.1$ mol.

The heat of vaporization is calculated as $\Delta H_{\text{vap}} = \frac{\text{total heat}}{n} = \frac{4}{0.1} = 40$ kJ/mol.

The result corresponds to option B.

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

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