Question

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.)

Answer 1

Answer:

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