# 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?

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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.

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.

The energy required is 90.1 kJ.

The mass of ethanol is

The moles of ethanol are

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

where

In your problem,

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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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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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