How many kilojoules of heat are required to vaporize 50.0 g of ethanol (MM = 46.07 g). Its molar heat of vaporization is 43.5 kJ/mol.

ΔH = 41.7 kj

Given data:

Mass of ethanol = 50 g

Heat required to vaporize = ?

Boiling point of ethanol = 78.3°C

Molar heat of vaporization = 38.6 kj/mol

Solution:

First of all we will calculate the number of moles.

Number of moles = mass / molar mass

Number of moles = 50 g/ 46.07 g/mol

Number of moles = 1.08 mol

ΔH = 38.6 kj/mol ×1.08 mol

ΔH = 41.7 kj

Explanation:

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