Chapter Contents Understandin Calculate the amount of energy necessary to raise

Question

Chapter Contents Understandin Calculate the amount of energy necessary to raise the temperature of 1.00 L of ethanol (d 0.7849 g/cm*) from 250°C to its boiling point (78.3°C) and then to vaporize the liquid. (Cethanol 2.44 J/g K; heat of vaporization at 78.3 °C = 38.56 kJ/mol.) Answer 1.00 L (1000 mL/1 L) (0.7849 g/cm3) 785 g Heat liquid from 25.0 °C to 783 oC. T 78.3°C-25.0 °C = 53.3 °C-53.3 K q = (2.44 J/g-K) (785 g) (53.3 K) = 1.02 × 10 Boil the liquid. q = 38.56 kJ/mol (785 g) (1 mol/46.08 g) 657 kJ Total 102 kJ + 657 kJ = 759 kJ 102 kJ

Explanation / Answer

volume of ethanol = 1 litre

weight of 1 litre of ethanool = 785 gram

now for the last line : heat required to vaporise for 1 mol of ethanol = 38.56 Kj

so moles of 785 gram = 785 / molecular weight of ethanol

= 785 / 46.08

so heat required to vaporise 785 / 46.08 mole of ethanol = 38.56 * 785 / 46.08 = 657 KJ

so this 1/46.08 gram is used to convert the weight of ethanol (785 gram ) into mols of ethanol because heat of vaporisation is given in unit of KJ/mol.

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