Energetics

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Question 5
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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.

To determine ΔHvap\Delta H_{\text{vap}}ΔHvap​ for propanone (CH3COCH3\text{CH}_3\text{COCH}_3CH3​COCH3​), a student uses an electrical immersion heater to boil a sample of propanone in a flask. The system is maintained at the boiling point of propanone (56 ∘C56\text{ }^{\circ}\text{C}56 ∘C).

The student:

  • records the mass of the flask and boiling propanone
  • switches on the 1.25 kW1.25\text{ kW}1.25 kW heater and allows the liquid to boil for 3.50 minutes3.50\text{ minutes}3.50 minutes
  • switches off the heater and records the mass of the flask and remaining propanone.

The loss in mass of propanone during this time is 515 g515\text{ g}515 g.

Calculate the molar enthalpy of vaporisation, ΔHvap\Delta H_{\text{vap}}ΔHvap​, for propanone.

(Molar mass of propanone = 58.0 g mol−158.0\text{ g mol}^{-1}58.0 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

  1. A Level
  2. /Chemistry
  3. /Energetics