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Question 3

The molar enthalpy of vaporisation (ΔHvap\Delta H_{\text{vap}}ΔHvap​) of a liquid is the enthalpy change when one mole of liquid is converted to vapour at its boiling point.

A chemist performs an experiment to determine ΔHvap\Delta H_{\text{vap}}ΔHvap​ for tetrachloromethane (CCl4\text{CCl}_4CCl4​).

The chemist:

  • places a flask containing tetrachloromethane on a continuous-recording balance
  • uses an immersed electrical heating element rated at 0.85 kW0.85\text{ kW}0.85 kW to bring the liquid to its boiling point (76.7 ∘C76.7\text{ }^{\circ}\text{C}76.7 ∘C)
  • records the initial mass once steady boiling is established
  • maintains the heating at 0.85 kW0.85\text{ kW}0.85 kW for exactly 5.0 minutes5.0\text{ minutes}5.0 minutes (300 s300\text{ s}300 s)
  • records the final mass of the flask and remaining liquid.

The mass of tetrachloromethane vaporised and lost from the flask during this period is 1.31 kg1.31\text{ kg}1.31 kg.

Calculate the molar enthalpy of vaporisation, ΔHvap\Delta H_{\text{vap}}ΔHvap​, for tetrachloromethane in kJ mol−1\text{kJ mol}^{-1}kJ mol−1.

(Molar mass of CCl4=154.0 g mol−1\text{CCl}_4 = 154.0\text{ g mol}^{-1}CCl4​=154.0 g mol−1)

[1 kW=1 kJ s−1][1\text{ kW} = 1\text{ kJ s}^{-1}][1 kW=1 kJ s−1]

[3]

Energetics Questions

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