   Chapter 11, Problem 12PS

Chapter
Section
Textbook Problem

The enthalpy of vaporization of liquid mercury is 59.11 kJ/mol. What quantity of energy as heat is required to vaporize 0.500 mL of mercury at 357 °C, its normal boiling point? The density of mercury is 13.6 g/mL.

Interpretation Introduction

Interpretation: The energy required to evaporate the given volume of mercury has to be determined.

Concept Introduction:

The energy required to overcome the intermolecular forces of attraction for 1mol of a substance is called standard enthalpy of vaporization,ΔvapH0

Equation for density is,

Density=MassVolume

From its given mass is,

Number of moles=GivenmassMolecularmass

Explanation

Given,

Density=13.6g/mLVolume=0.500mLMolarmass=200.59g/molEnthaplyofvaporization,ΔvapH0=59.11kJ/mol

Equation for density is,

Density=MassVolume

Hence,

The mass of ethanol is,

Mass=Density×Volume=(13.6g1mL)×0.500mL=6.8g =2

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