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.

Answer 1

The energy required is 90.1 kJ.

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"#
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Answer 2

To calculate the heat required, use the formula: ( \text{heat} = \text{mass} \times \text{heat of vaporization} ). The heat of vaporization for ethanol is approximately 38.6 kJ/mol. Convert moles to grams for the mass term. Ethanol molar mass is approximately 46.07 g/mol. ( \text{Heat} \approx 125 , \text{mL} \times \frac{1 , \text{g}}{1 , \text{mL}} \times \frac{1 , \text{mol}}{46.07 , \text{g}} \times 38.6 , \text{kJ/mol} ).

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Answer from HIX Tutor

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

When evaluating a one-sided limit, you need to be careful when a quantity is approaching zero since its sign is different depending on which way it is approaching zero from. Let us look at some examples.

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