Calculate the amount of heat energy, in units of kilojoules, required to convert 50.0 g of liquid ethanol, ch3ch2oh, at 25.5°c to gaseous ethanol at 92.0°c. (cethanol liquid = 2.44 j/g°c; cethanol gas = 1.42 j/g°c; molar heat of vaporization of liquid ethanol = 3.86 × 104 j/mol. the boiling point of ethanol is 78.4°c.)

42.1 kj

Explanation:

Given data:

Molar heat of vaporization = 3.86×10⁴ j/mol

Mass of ethanol = 50 g

Amount of heat required to convert into gaseous form = ?

Solution:

First of all we will calculate the number of moles of ethanol.

Number of moles = mass/ molar mass

Number of moles = 50.0 g/ 46.07 g/mol

Number of moles = 1.09 mol

H(vap) = number of moles × molar heat of vaporization

H(vap) =  1.09 mol × 3.86×10⁴ j/mol

H(vap) = 4.21×10⁴ j

In Kj:

4.21×10⁴ / 1000 = 42.1 kj

water and alcohols have similar properties because water molecules contain hydroxyl groups that can form hydrogen bonds with other water molecules and with alcohol molecules, and likewise alcohol molecules can form hydrogen bonds with other alcohol molecules as well as with water.