# 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.

Aug 23, 2016

The energy required is 90.1 kJ.

#### Explanation:

The mass of ethanol is

$\text{Mass" = 125 color(red)(cancel(color(black)("mL"))) × "0.7849 g"/(1 color(red)(cancel(color(black)("mL")))) = "98.11 g}$

The moles of ethanol are

$\text{Moles" = 98.11 color(red)(cancel(color(black)("g"))) × "1 mol"/(46.07 color(red)(cancel(color(black)("g")))) = "2.130 mol}$

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

$n = \text{2.130 mol}$
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"