Ethanol, #CH_3CH_2OH#, has a vapor pressure of 59 mm Hg at 25 °C. What quantity of energy as heat is required to evaporate 125 mL of the alcohol at 25 °C?

The enthalpy of vaporization of the alcohol at 25 °C is 42.32 kJ/mol. The density of the liquid is 0.7849 g/mL.

1 Answer
Aug 23, 2016

Answer:

The energy required is 90.1 kJ.

Explanation:

The mass of ethanol is

#"Mass" = 125 color(red)(cancel(color(black)("mL"))) × "0.7849 g"/(1 color(red)(cancel(color(black)("mL")))) = "98.11 g"#

The moles of ethanol are

#"Moles" = 98.11 color(red)(cancel(color(black)("g"))) × "1 mol"/(46.07 color(red)(cancel(color(black)("g")))) = "2.130 mol"#

The energy required to evaporate a given amount of a liquid is given by the formula

#color(blue)(|bar(ul(color(white)(a/a) q = nΔ_"vap"Hcolor(white)(a/a)|)))" "#

where

#q# is the energy required
#n# is the number of moles
#Δ_"vap"H# is the molar enthalpy of vaporization

In your problem,

#n = "2.130 mol"#
#Δ_"vap"H = "42.32 kJ/mol"#

#q = 2.130 color(red)(cancel(color(black)("mol"))) × "42.32 kJ"/(1 color(red)(cancel(color(black)("mol")))) = "90.1 kJ"#