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12: Intermolecular Forces

These are homework exercises to accompany the Textmap created for "General Chemistry: Principles and Modern Applications " by Petrucci et al. Complementary General Chemistry question banks can be found for other Textmaps and can be accessed here. In addition to these publicly available questions, access to private problems bank for use in exams and homework is available to faculty only on an individual basis; please contact Delmar Larsen for an account with access permission.


Determine which of the following have either Hydrogen Bonding, Dipole, or London Forces?

  1. \(H_2O\)
  2. \(H_2S\)
  3. \(BF_3\)
  4. \(CHCl_3\)


  1. H-bonding
  2. Dipole
  3. London
  4. Dipole


For each of the following substances, describe which intermolecular forces are present:

  1. \(LiBr\)
  2. \(GeH_4\)
  3. \(SO_2\)
  4. \(CH_4\)


  1. Ion-Ion
  2. London Forces
  3. Dipole-Dipole
  4. Hydrogen bonding


Describe the types of forces for the following substances: (a) \(HF\) (b) \(ICl\) (c) \(HCl\) (d) \(Br_2\) (e) \(CH_4\)                    


  1. In \(HF\), there are  weak London forces and molecule hydrogen bonding is the strongest force present
  2. Because of the polarity in the \(I-Cl\) bond, it contains dipole interactions
  3. Since \(HCl\) is not a diatomic molecule, London forces are weak. Hydrogen bonding is weak because it is not an (\(F\),\(O\),\(N\)) atom.  However, \(Cl\) is electronegative and hydrogen only a little electronegative, so \(HCl\) has a dipole moment.
  4. \(Br_2\) clearly doesn't have a hydrogen bond, neither does it have a dipole moment.  Therefore the strongest force is the London force because \(Br_2\) had a larger atomic mass
  5. \(CH_4\), not an (\(F\),\(O\),\(N\)) atom therefore it has no H-bond, it has no  dipole dipole interactions, plus London forces are weak which is why it has a a very low critical temperature.


Describe the importance of London forces, dipole-dipole interactions and hydrogen bonding in the following substances:


There are typically three main forces affecting liquid and solids: hydrogen bonding, dipole-dipole interactions and London forces

  • \(CHCl_3\): Since N,O or F is not present, there is no hydrogen bonding. It’s a polar molecule, and has a strong dipole-dipole interaction.  
  • \(F_2\): In the molecule \(F_2\), there are no \(H\) atoms, and thus can have neither dipole moments (dipole-dipole interactions) nor hydrogen bonding. Thus, London forces are the most important. 
  • \(CO\): Since there is no Hydrogen, there is no hydrogen bonding. \(CO\) has a very small dipole moment, but they do affect the molecule. Overall, London forces are the strongest force.
  • \(OH\): Since this molecule is small, London forces are not very strong. Here, hydrogen bonding is the strongest force.
  • \(CH_3CH_3\): The only significant force here is London forces because of the molecules lack of a great difference in electronegativity and shape.


Which substance has a higher viscosity?

  1. \(H_2O\) vs. \(H_2S\)
  2. \(CH_4\) vs. \(CHCl_3\)
  3. \(CH_3OH\) vs. \(CH_3CH_3\)
  4. \(Ar\) vs. \(He\)


The greater the forces that hold the liquid together (molecular forces) the greater its resistance to flow (viscosity) 

  1. \(H_2O\)
  2. \(CHCl_3\)
  3. \(CH_3OH\)
  4. \(Ar\)


List the following molecules in order of increasing viscosity.

  1. ethanol
  2. acetic acid
  3. butanol 
  4. ethanethiol 


The greater the forces that hold the liquid together (molecular forces) the greater its resistance to flow (viscosity) 



One of the following is liquid at room temperature and the others are gaseous: \(BF_3\), \(C_3H_8O\), \(CF_2Cl_2\), \(SF_6\). Which do you think is the liquid?




All of the following substances are gaseous at room temperature except one that is liquid: \(C_2H_6\), \(NH_3\), \(NO_2\), \(H_2S\), and \(H_2O\). Which one is the liquid and explain why you think so?


\(H_2O\): The attraction of the hydrogen bonds keeps water a liquid over a wide temperature range and the energy required to break these multiple hydrogen bonds causes water to stay intact (ie, the bonds don’t break until a high enough temperature is reached for it to vaporize).


Which is a gas and which is a liquid at room temperature:  \(CH_3OH\), \(C_2H_6\), \(O_2\), \(N_2O\)


  1. \(CH_3OH\) is the liquid. \(C_2H_6\) has more electrons so it has a stronger London force.
  2. \(CH_3OH\) also has hydrogen bonding.


One of the following substances, \(CO_2\); \(O_2\), \(C_5H_{12}\), \(NaCl\). is liquid at room temperature and the rest are gaseous. Which is the liquid? Why


\(NaCl\) is a liquid at room temperature. This is because out of all of the listed elements, it has the strongest intermolecular forces and therefore the highest boiling point.


Using your knowledge of intermolecular forces, guess which of these substances is most likely to be a liquid at room temperature

  • \(C_2H_6\)
  • \(O_2\)
  • \(C_2H_5 OH\)
  • \(CH_4\)


The answer here is C, \(C_2H_5OH\). Without knowing the specific properties of this substance, we expect this to be in liquid form because of high potential to form hydrogen bonds (the OH suffix), and so its intermolecular attractions should be much stronger than usually seen in gasses. 


Explain how detergent can help lower the energy required to spread drops of water into a film.


The detergent lowers the surface tension of the water, thereby lowering the energy and allowing water to spread more easily. Detergent is therefore known as a wetting agent.


List the following molecules in order of increasing viscosity.

  1. \(C_4H_{10}O\) 
  2. \(CH_4O\)
  3. \(CCl_4\)


(c) < (a) < (b)


Why doesn’t oil and water mix?


Water is a polar molecule meaning that there is unequal sharing of electrons between the hydrogen and oxygen atoms. Oils are made of hydrocarbon chains that are bonded by hydrogen. The elements and structure of oil cause them to become non-polar which means they won’t combine with water.


Why does vapor pressure increase with temperature, while surface tension and viscosity decrease with temperature?


Increasing the temperature of the liquid causes the molecules to move faster. And the more energy the more molecules are able to break free at the surface causing the vapor pressure to increase.


To keep wooden decks and fences from becoming too wet and moldy, most water repellents, use chemicals such as 1,2,4-Trimethylbenzene (\(C_9H_{12}\)), a hydrocarbon (non-polar oil), to repel to treat wood. Explain how non-polar oils such as (\(C_9H_{12}\)) repel water.


The most important part of the answer is given in the question; hydrocarbons are non-polar. Wood is carbon based, and is non-polar, but is very porous so it does allow water to enter it. Since both the deck and (\(C_9H_{12}\)) are non-polar, they naturally adhere to each other. Water, a polar substance, does not adhere very well to non-polar materials. Once the deck is coated with the non-polar oil, water is naturally repelled.   


Fluorine is one of the most common, natural elements to be found on Earth. Fluorinated compounds are found in non- stick frying pans, heat resistant cables, colorfast paints, and similar items. Explain why fluorine is used in these items.


Fluorine has high electronegativity and small dimensional size. Also, Fluorine has a strong chemical bond with carbon, creating a stable organic chemical, which has good heat, chemical, and weather resistant properties.


How do intermolecular forces affect surface tension and vapor pressures differently as temperature increases?


As the temperature increases, there is additional movement between the molecules forces, leading to easily broken bonds and weak intermolecular forces. Thus, as the temperature increases, the forces controlling surface tension decrease, decreasing the surface tension. Conversely, vapor pressure change is indicated by the Clausius-Clapeyron equation. Usually, vapor pressure is always moving towards its boiling point, or when the vapor pressure is equal to the atmospheric pressure. As you increase the vapor pressure, you reach boiling point faster, which takes less change in heat (or temperature). Thus, as temperature increases, so does the vapor pressure. 


Does honey have high viscosity and Why?


Honey has viscosity because of the strong intermolecular attractions, and lots of internal friction. These properties make the substance thick and sluggish.


Is there any scientific basis for the colloquial expression “Thicker then cold syrup”? Explain.


Syrup has a high viscosity. The coldest temperature will produce the thickest liquid because the higher the viscosity, the colder the liquid all due to intermolecular forces.


Why does detergent lower the surface tension of water?


Detergent is a wetting agent which lowers the energy required to spread out a water molecule throughout a surface.


What does it mean when something “wets” the surface? What are wetting agents?


When something “wets” the surface it means that a drop of liquid is spreading into a film of liquid across a surface. This depends on the intermolecular forces and their strengths. “Wetting agents” reduce the surface tension of water so it is able to glide along surfaces more easily.


Would it be easier or harder to evaporate water at high altitudes?


At higher altitudes, air pressure is lower and the reduced air pressure lowers the temperature at which water boils in an open container making it easier to evaporate.


Ethanol has a vapor pressure of 34.2 mmHg at 24.2 °C and 89.2 mmHg at 69.2 °C. Calculate its enthalpy of vaporization.


First conver temperature to absolution temperature

  • 24.2 °C + 273.15 = 297.35 K
  • 69.2 °C + 273.15 = 342.35 K

\[ln \left(\dfrac{89.2}{34.2} \right) = \dfrac{∆H}{ 8.3145} \left[ \dfrac{1}{297.35} – \dfrac{1}{342.35} \right] \]

\[∆H = 18.0\; kJ\]


Why does ethyl alcohol feel cool against the skin?


Ethyl alcohol feels cool because of heat of evaporation. While the alcohol is evaporating, it absorbs heat from both the skin underneath it and also from the air around it therefore we feel it is cold when we rub it against the skin.


List and describe the three main forces that occur readily with vaporization.


  • Increased temperature: more molecules have sufficient kinetic energy to overcome intermolecular forces
  • Increased surface area: there are more liquid molecules at the surface
  • Decreased strength of intermolecular forces: the energy needed to overcome intermolecular forces is less, and more molecules have the energy to escape.


At 100 C° what part of \(H_2O\) is in which phase?


Because this is the boiling point, the liquid at the top will be becoming vapor while the bottom remains liquid.


What is the difference in vapor pressure between a liquid heated to the point where its surface is evaporating and a liquid that is boiling? How does that difference affect vaporization?


Vapor pressure is lower during evaporation than when the whole liquid is boiling. Because the vapor pressure is not as high when the temperature of the liquid is lower, atmospheric pressure is greater which causes just the top of the liquid to vaporize. When boiling, the temperature of the liquid is higher and its vapor pressure is equal to the atmospheric pressure. This causes vapor to form throughout the liquid and be released. 


The enthalpy of vaporization of diethyl ether, \((C_2H_5)_2O\) is 27.25 kJ/mol. How many liters of diethyl ether gas is produced at 23 C and 1.1 atm (conditions remain constant), when 2.56 kJ of heat is absorbed diethyl ether liquid?


Vaporization: liquid to gas

  • Pressure= 1.1 atm
  • Temperature= 23C+ 273= 300K


\[n=\dfrac{27.25\ \frac{kJ}{mol}}{ 2.56\ kJ} =10.64\ mol\ (C_2H_5)_2O\]

\[V= \dfrac{(10.64\ mol) (300\ K) (0.0821\ L\ \frac{atm}{mol\ K})}{1.1\ atm}\]

\[V= 238.2\ L\]


What volume of \(H_2O_{(g)}\), measured at 100℃ and standard pressure, is formed when 1.25 kJ of heat is absorbed by \(H_2O_{(l)}\) at a constant temperature of 100℃?


Given: \(\Delta H_{vap}=40.67\ \frac{kJ}{mol}\)  at 100℃

\(100^oC + 273.15 = 373.15\ K\)

\(V= \dfrac{nRT}{P} = 1.25\ kJ\times \frac{(1\ mol)}{(40.67\ kJ)} \times 0.08206\frac{(L\ atm)}{(mol\ K)}\times 373.15 \frac{K}{1\ atm} = 0.941L\ H_2O_{(l)}\)


The enthalpy of vaporization of Cyclohexane, \(C_6H_{12(l)}\), is \(-33\ \frac{kJ}{mol}\) at 298 K.  At 298 K and 105.3 mmHg how many liters of \(C_6H_{12(g)}\) is formed when 1.69 kJ of heat is absorbed by \(C_6H_{12}\)?


\[V=\dfrac{nRT}{P} =\dfrac{1.69kJ\times \frac{1mol}{33kJ}\times 0.08206\frac{L\ atm}{mol\ K}\times 298K}{105.3mmHg\times \frac{1atm}{760mmHg}}=9.04L\ C_6H_{12(l)}\]


Explain the process of vaporization?


Vaporization occurs when a water molecule gains enough energy to escape from the waters surface.


The enthalpy of vaporization of Ethyl alcohol, \(CH_3CH_3OH_{(l)}\), is \(42.6\frac{kJ}{mol}\) at 298 K. How many liters of \(CH_3CH_2OH_{(g)}\), measured at 298 K and 92.4 mmHg, are formed when 1.69 kJ of heat is absorbed by \(CH_3CH_2OH_{(l)}\) at a constant temperature of 298 K?



\(V= \frac{(1.69 kJ\times \frac{1mol}{42.6\ kJ}) \times(0.08206 \frac{L\ atm}{mol\ K}) (298K)} {(92.4mmHg\times \frac{1\ atm}{760 \ mmHg})}\)

\(V = 7.98L\ CH_3CH_2OH_{(l)}\)


How many liters of \(C_2H_6\) measured at 24.5 °C and 756 mmHg, must be burned to provide the heat needed to vaporize 4.23 L of \(CH_3OH\) at 65°C?

\(\Delta {H}_c^°: CH_3OH = -1.56\times 10^3 \frac{kJ}{mol}\)

\(\Delta H_{vap}: C_2H_6= 38.0 \frac{kJ}{mol}\); 

\(CH_3OH\ B.P.= 64.7^oC\)


Heat needed:

\(4.23L\ CH_3OH\times \frac{1000cm^3}{1L}\times \frac{0.798\ g}{1\ cm^3}\times \frac{1mol}{34.042}\times \frac{38kJ}{1\ mol}= 3768\ kJ\)

Amount of \(C_2H_6\) needed:

\(3768\ kJ\times \frac{1\ mol}{1.56\times 10^3\ kJ}= 2.42\ moles\ of\ C_2H_6\)


 What is the vapor pressure of Water at 50 ºC, in atm?


Even though it is unstated in the problem, we are asked about the vapor pressure of water, which is a liquid at two different temperatures and pressures. We know that at water’s BP (100 ºC) pressure is 1 atm. When asked about 2 states and 1 compound of interest, we use Clausius-Clapeyron equation. We also know that enthalpy of vaporization for water is \(40.7\frac{kJ}{mol}\)

  • \(T_1=100 + 273.15 = 373.15\ K\)
  • \(P_1=1\ atm\)
  • \(T_2= 50+ 273.15= 323.15\ K\)  
  • \(P_2=?\)

$$\ln \left(\dfrac{P_2}{P_1}\right)= \dfrac{ΔH_{vap}}{R} \left( \dfrac{1}{T_1} - \dfrac{1}{T_2} \right)$$

$$ln(\frac{P_2}{1atm})=[\frac{40,700J}{8.3145\frac{J}{mol\ K}}]\times (\frac{1}{373.15} - \frac{1}{323.15})=2.0297$$

$$\ln P_2 - \ln (1\ atm) = 2.0297$$

$$P_2=7.612\ atm$$


Benzene has a vapor pressure of 12.0 mmHg at 61.0 ˚C and 120 mmHg at 108.7 ˚C.  Find the enthalpy of vaporization.


\(T_1 = (61 + 273.2)K = 334.2\ K\)

\(T_2 = (108.7 + 273.2)K = 381.9\ K\)

\(ln(\frac{P1}{P2}) = \frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1} - \frac{1}{T_2})\)

\(\Delta H_{vap}= \frac{[-R\times ln (\frac{P1}{P2})]}{(\frac{1}{T_1} - \frac{1}{T_2})}\)

\(ln(\frac{12.0mmHg}{120 mmHg})=\frac{\Delta H_{vap}}{8.3145\frac{J}{mol\ K}}\times (\frac{1}{381.9}- \frac{1}{334.2})\)

\(\Delta H_{vap}=5.122\times 10^4 \frac{J}{mol}\)


Cyclohexane has a vapor pressure of 16 mmHg @ 63oC and 136 mmHg @ 115oC. Calculate its \(\Delta H_{vap}\).


\(T_1=(63+ 273.2)= 336.2K\)    

\(T_2=(115+ 273.2)= 388.2K\)

\(ln(\frac{16}{136})=\frac{\Delta H_{vap}}{8.3145\frac{J}{mol\ K}}\times (\frac{1}{388.2}-\frac{1}{336.2})\)

\(\Delta H_{vap}= 44.7\ \frac{kJ}{mol}\)


Hexane, \(C_6H_{14}\), has a vapor pressure of 25.0 mmHg at 311K and 84 mmHg at 365K. Calculate the enthalpy of vaporization, \(\Delta H_{vap}\).


\(25\ mmHg=25\ mmHg\times \frac{1atm}{760\ mmHg} = 0.03289atm\)  

\(84\ mmHg=84\ mmHg \times \frac{1atm}{760\ mmHg} = 0.1105atm\)

\(ln(\frac{0.1105}{0.03289})= \frac{\Delta H_{vap}}{8.314 \frac{J}{(K\ mol)}}\times (\frac{1}{311}-\frac{1}{365})\)

\(1.2118= 5.722\times 10^{-5}\Delta H_{vap}\)

\(\Delta H_{vap}=21.2\ \frac{kJ}{mol}\)


Propane has a vapor pressure of 40.0 mmHg at -92.4°C and 760 mmHg at -42.1°C. Calculate the enthalpy of vaporization, \(\Delta H_{vap}\).


\(T_i=(-92.5+273) K= 180.9K\)                         

\(T_f=(-42.1+273.2) K= 231.1 K\)

\(ln(\frac{40.0mmHg}{760mmHg}) = \frac{\Delta H_{vap}}{8.3145\frac{J}{mol\ K}}(\frac{1}{231.1K} – \frac{1}{180.9 K})\)

\(-2.944=-1.44\times 10^{-4} \Delta H_{vap}\)

\(\Delta H_{vap}= 2.04\times 10^4 \frac{J}{mol}= 20.4 \frac{kJ}{mol}\)


Butane (\(C_4H_10\)) has a vapor pressure of 2854.66 mmHg at 49.0℃ and 155.14 mmHg at 4.5℃. Calculate its enthalpy of vaporization, \(\Delta H_{vap}\).


  • \(T_1=(49℃+273.15\;K)= 322.15\; K\)
  • \(T_2=(4.5℃+273.15\;K)= 277.65 \;K\)

\(ln (\frac{2854.66 mmHg}{155.14 mmHg}) =  \dfrac{∆H_{vap}}{(8.3145 \frac{J}{mol\ K})} \times (\frac{1}{277.65 K} - \frac{1}{322.15 K})\)

\(2.91=5.98\times 10^{-5}\Delta H_{vap}\)

\(\Delta H_{vap} = 4.866 \times 10^4\ \frac{J}{mol} = 48.66\ \frac{kJ}{mol}\)


 Cyclohexanol has a vapor pressure of 80.0 mmHg at 56.0oC and 10.0 mmHg at 25.0oC.  Calculate its enthalpy of vaporization, \(\Delta H_{vap}\).


\(ln(\frac{P_2}{P_1})= \frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1}-\frac{1}{T_2})\)

\(\Delta H_{vap}=ln(\frac{10.0mmHg}{80.0mmHg})\times \frac{(8.3145\frac{J}{mol\ K})}{(\frac{1}{56.0+273K}- \frac{1}{25.0+273K})}\)

\(\Delta H_{vap}=54.7\ kJ\)



The normal boiling point of ethanol (\(C_2H_6O\)) is 78.37℃ and its \(∆H_{vap}\) is \(38.6\frac{kJ}{mol}\). At what temperature does ethanol have a vapor pressure of 335 mmHg?


\(T_1=(78.37^oC+273.15)= 400.52 K\)

\(ln(\frac{760 mmHg}{335 mmHg})=  \frac{38.6\times 10^3 \frac{J}{mol}}{(8.3145\frac{J}{mol\ K})}\times (\frac{1}{T}-\frac{1}{400.52 K})=0.819\)

\(\frac{1}{T}- \frac{1}{400.52 K}= \frac{0.819\times 8.3145 \frac{J}{mol\ K}}{38.6\times 10^3 \frac{J}{mol}}\)

\(\frac{1}{T}-\frac{1}{400.52 K}=1.764\times 10^{-4}\)

\(\frac{1}{T}=\frac{1}{400.52 K}+1.764\times 10^{-4}\)

\(T=374.09\ K\)


Methanol, \(CH_3OH\), has a temperature at 240°C and pressure at 77.5 atm. Estimate its vapor pressure at 105°C.


\(ln(\frac{77.5atm}{P_1})=\frac{38\times 10^3 \frac{J}{mol}}{8.314\frac{J}{mol\ }}\times (\frac{1}{378}-\frac{1}{513})\)





What is the enthalpy of vaporization of carbon tetrachloride if it has a vapor pressure of 0.28 atm at 313.2 K and 1.1 atm at 353.2 K?


\(ln(\frac{P_1}{P_2}) = \frac{\Delta H_{vap}}{R}\times (\frac{1}{T_1} - \frac{1}{T_2})\)

\(ln(\frac{0.28atm}{1.1atm}) =\frac{\Delta H_{vap}}{8.3145\frac{J}{mol\ K}}\times (\frac{1}{353.2}- \frac{1}{313.2})\)

\(\Delta H_{vap}= \frac{R \times ln(\frac{P_1}{P_2})}{(\frac{1}{T_1} - \frac{1}{T_2})}\)

\(\Delta H_{vap} = 31.4\ \frac{kJ}{mol}\)


Hydrazine, \(N_2H_4\) has a normal boiling point of 113.5°C and a critical point of 380°C and 145.4 atm. Estimate vapor pressure at 100°C.


\(T_1=(113.5+273.2) K = 386.7 K\)                    

\(T_2=(380+273.2) K = 653.2 K\)

\(ln(\frac{P_2}{P_1})= \frac{\Delta H_{vap}}{R} \times (\frac{1}{T_1}-\frac{1}{T_2})\)

\(ln(\frac{145.4}{1})=\frac{\Delta H_{vap}}{8.3145\frac{J}{K\ mol}}\times (\frac{1}{386.7K} – \frac{1}{653.2K})\)

\(\Delta H_{vap}= 39.2 \frac{kJ}{mol}\)

\(ln(\frac{1}{P})= \frac{39,200\frac{J}{mol}}{8.3145\frac{J}{mol\ K}}\times (\frac{1}{373.2K} – \frac{1}{386.7K})\)

\(P= 0.44103\ atm\)


The normal melting point of Aluminum is 933.5K and \(\Delta H_{fus}\) of Aluminum is \(10.7 \frac{kJ}{mol}\).

  1. How much heat is evolved when 3.78 kg of Aluminum freezes?
  2. How much heat must be absorbed at 933.5 K to melt a can of aluminum that is \(64\times 12\times 10\ cm^3\)?


\(Melting\ point\ of\ Al=933.5K\) 

\(\Delta H_{fus}\  Al=10.7 \frac{kJ}{mol}\) 


2.37kg of molten aluminum freezes can 64x12x10

  1. heat evolved = \( (2.37kg\ Al )(\frac{1000g}{1kg})(\frac{1mol\ Al}{26.98g})(\frac{10.7kJ\ Al}{1mol\ Al})= 939.92\ kJ\)
  2. heat absorbed= \( (64\times 12\times 10)(\frac{2.7g}{1cm^3})(\frac{1mol\ Cu}{26.98g})(\frac{10.7kJ\ Al}{1mol\ Al})= 8.2\times 10^{23} kJ\)


Water is held in a 5L evacuated flask at various temperatures. What is the temperature of the vapor in ˚C if the pressure is

a) 0.037atm

b) 1 atm  

c) 0.77atm.


  1. \(0.037atm\times (\frac{760mmHg}{1atm})= 28.3mmHg\)
  2. \(1atm\times (\frac{760mmHg}{1atm}) = 760mmHg\)
  3. \(0.77atm\times (\frac{760mmHg}{1atm}) =588.6mmHg\)

Utilizing the table of Vapor pressure of water at various temperatures in the text book

a) 28.0˚C


c) 93.0˚C


In a 4.20L evacuated flask there is 0.360g of liquid water. Calculate the pressure of the water vapor in the flask at

(a) 40 ˚C

(b) 60 ˚C

(c) 80 ˚C


\(0.360\ g\ of\ H_2O = 0.0199\ mol\)


40 ˚C, vapor pressure of \(H_2O = 52.8\ mmHg = 0.0694\ atm\)

\(n=\frac{PV}{RT}= \frac{0.0694atm\times 4.20L}{0.0820\frac{L\ atm}{mol\ K} \times 313.2 K}\)

\(n= 0.0113\ mol\)

0.0113 mol of water vapor is less than 0.0199 mol of water there for not all of the water evaporates.


60 ˚C, vapor pressure of \(H_2O = 122.6\ mmHg = 0.1613\ atm\)

\(n=\frac{PV}{RT}=\frac{0.1613atm\times 4.20L}{0.08206\frac{L\ atm}{mol\ K}\times 333.2K}\)

\(n= 0.0248\ mol\ H_2O\ vapor\)

Since 0.0248 mol > 0.0199 mol,

\(P=\frac{nRT}{V} =\frac{0.0199\ mol\times 0.08206\frac{L\ atm}{mol\ K}\times 333.2K}{4.20L}= 0.130\ atm\)

80˚C, vapor pressure of \(H_2O = 150.4\ mmHg = 0.198\ atm\)

All water evaporates as seen in part (b) so

\(P=\frac{nRT}{V} =\frac{0.0199\ mol\times 0.08206\frac{L\ atm}{mol\ K}\times 353.2K}{4.20L}= 0.137\ atm\)


A sample of liquid \(Cl_2\) is sealed into an evacuated flask. What is the pressure of the vapor in the flask if the temperature is:

  1. 32.0°C;
  2. 49.0°C;
  3. 76.3°C?

\(\Delta H_{vap}: (Cl_2)\) is \(20.41\frac{kJ}{mol}\) and the vapor pressure of \(Cl_2\) @ 21°C is 6.86 atm.



\(ln (\frac{P_2}{P_1}) = \frac{\Delta H_{vap}}{R}\times (\frac{1}{T_2} – \frac{1}{T_1})\)

\(P_1 = 6.86\ atm\)

\(T_1 = 21^oC = 294.15\ K\)

\(T_2 = 32.0^oC = 305.15\ K\)
\(ln (\frac{P_2}{6.86}) = \frac{20410}{8.3145}\times (\frac{1}{305.15 K} – \frac{1}{294.15 K})\)

\(P_2= 9.27\ atm\)


\(ln (\frac{P_2}{P_1}) = \frac{\Delta H_{vap}}{R}\times (\frac{1}{T_2} – \frac{1}{T_1})\)

\(P_1 = 6.86\ atm\)

\(T_1 = 21^oC = 294.15\ K\)

\(T_2 = 49.0^oC = 322.15\ K\)

\(ln (\frac{P_2}{6.86}) = \frac{20410}{8.3145}\times (\frac{1}{322.15 K} – \frac{1}{294.15 K})\)

\(P_2= 14.2\ atm\)


\(ln (\frac{P_2}{P_1}) = \frac{\Delta H_{vap}}{R}\times (\frac{1}{T_2} – \frac{1}{T_1})\)

\(P_1 = 6.86\ atm\)

\(T_1 = 21^oC = 294.15\ K\)

\(T_2 = 76.3^oC = 349.45\ K\)

\(ln (\frac{P_2}{6.86}) = \frac{20410}{8.3145}\times (\frac{1}{349.45 K} – \frac{1}{294.15 K})\)

\(P_2= 25.7\ atm\)


A 0.350 g sample of \(H_2O_{(l)}\) is sealed into an evacuated 4.10 L flask. What is the pressure of the vapor in the flask if the temperature is a) 40°C; b) 60°C; and c) 90°C. Given the vapor pressure of \(H_2O_{(l)}\) at 40°C is 55.3 mmHg and at 60°C is 149.4 mmHg.



\(0.350g\ H_2O_{(l)}\times \frac{1\ mol\ H_2O}{18.0g H_2O} = 0.0195\ mol\ H_2O\)

\(55.3mmHg\times \frac{1\ atm}{760mmHg} = 0.0728\ atm\)


\(n = \frac{(0.0728\ atm)(4.10\ L)}{0.08206 \frac{L\ atm}{mol\ K}\times (313.15 K)}\) 

\(n = 0.0116\ mol\ H_2O\)

→ 0.0116 mol < 0.0195 mol 

→ P =0.0728 atm 


\(149.4mmHg\times \frac{1 atm}{760mmHg} = 0.1966\ atm\) 


\(n = \frac{(0.1966 atm)(4.10 L)}{0.08206 \frac{L\ atm}{mol\ K}\times (333.15 K)}\) 

\(n = 0.0295\ mol\ H_2O\)

→ 0.0295 mol > 0.0195 mol


\(P =  \frac{nRT}{V} = \frac{0.0195 mol\times 0.08206\frac{L\ atm}{mol\ K} \times 333.15 K}{4.10\ L}\)  

\(P = 1.300\ atm\)


\(P =  \frac{nRT}{V} = \frac{0.0195 mol\times 0.08206\frac{L\ atm}{mol\ K} \times 363.15 K}{4.10\ L}\)

\(P = 1.417\ atm\)


A 0.696 g sample of \(H_2O_{(l)}\) is sealed into an evacuated 7.8 L flask. What is the pressure of the vapor in the flask if the temperature is 92°C?


0.696 g sample corresponds to 0.0386mol of \(H_2O\).

\(P=\frac{nRT}{V} = \frac{0.0386mol\times 0.08206\frac{L\ atm}{mol\  K}\times 365.2K}{7.8\ L} = 0.148 atm\)


To an isolated container with 120g of \(H_2O_{(l)}\) at 22, 158g steam at \(100^oC\) and 2.73 kg of ice at \(0^oC\) are added.

  1. What mass ice remains unmelted after equilibrium is reached?
  2. what additional mass of steam should be introduced to melt all of the ice.?



heat lost by water:

 \(Q_w = m\times c\times \Delta T\)

\(Q_w=(120)(4.185\frac{J}{g^oC})(0-22^0C)(\frac{1kK}{1000J})= -11.04kJ\)

\(\Delta H_{condensation} = -\Delta H_{vaporization}\): heat lost= heat loss of condensation and cooling

\(Q_{steam}=(158)(4.185\frac{J}{g^oC})(0-100^0C)(\frac{1kK}{1000J}) +  (158)(\frac{1mol}{18.015g})(-40.7\frac{kJ}{1mol})- 356.96= -423kJ\)

\(Total\ energy\ to\ melt\ ice= Q_w+Q_{steam} =-11.04+-423= -434.04kJ\)

\(moles\ of\ ice\ melted= 434.04\times \frac{1}{6.01}=72.22\ mol\)  

\(mass\ of\ ice\ melted= (72.22)(\frac{18.015}{1})(\frac{1}{1000})= 1.3 kg\)

\(mass\ unmelted\ ice= 1.73-1.3= 0.43 kg\)


\(heat\ required\ to\ melt\ ice = n \Delta H_{fus}\)

\(heat\ required =(0.43kg)(\frac{1000g}{1kg})(\frac{1mol}{18.015g})(\frac{6.01kJ}{1mol})= 143.45\ kJ\)
\(heat\ per\ 1\ mol\ H_2O= (-4.07\frac{kJ}{1mol})+(18.015)(4.18)(0-100^oC) (\frac{1kJ}{1000J}) = -40.7kj- 7.53kj= -48.2\frac{kJ}{mol}\)

\(mass\ steam\ required =(143.45)(\frac{1}{48.2})(\frac{18.015g}{1mol})= 53.62g\ steam\)


Describe the structure of a diamond using the properties in network covalent solids.


By exposing graphite to high pressure and a comparatively low temperature, the carbon atoms can bond to form a crystalline solid structure, where each atom is bonded to four others. The covalent bonds become extremely strong and thus make diamond one of the hardest solids.


It is winter time and the temperatures at night drop well below freezing. You have tropical plants outside on your lawn, and know that they will not be able to withstand the freezing temperatures. When looking for advice on how on how to possibly save your plants, you come across an article that states leaving your sprinklers on during the night will allow your plants to keep. After turning on your sprinklers you notice that the water is freezing on the plants. Should you still keep the sprinklers running?


Yes. \(-\Delta H_{fus}\) (freezing) is an exothermic reaction, therefore, the freezing water on the plants means that heat is being released. Thus, keeping your sprinklers on will keep the plants warm.


You put a beer in the freezer to cool it. You would think it would freeze, but instead when you take it out it is still liquid. When you open the bottle, the beer freezes (turns into slush). Why does the beer not freeze until it is opened?


The reason for this is because it contains the solute carbon dioxide, along with the solvent water. When carbon dioxide is dissolved into water the freezing point gets lowered. When you open the bottle there is a pressure difference between the inside of the can and the outside and the pressure lowers and this leads to carbon dioxide to leave. This then creates the freezing temperature to increase because the carbon dioxide is leaving. The temperature is getting colder so the beer gradually freezes from the top of the beer to the bottom.


You decide to heat up soup in a clean container in the microwave. You purposely don’t fill it to the top of the container because you don’t want it to spill out. When you take it out of the microwave, the container is burning hot and you notice the soup spilled out of the container in the microwave. Explain.


The liquid in the container is superheated, which is when a liquid is heated to a temperature higher than it’s boiling point. The larger the bubbles in the liquid, the lower the surface tension, causing the liquid to expand explosively and boil over. 


Calculate the normal boiling point of heptane, if its vapor pressure at 294 K is 0.29 atm and its enthalpy of vaporization is 4.18 kJ.


\(ln(\frac{1}{0.29}) = \frac{4.18\times 1000}{8.3145}(\frac{1}{194} – \frac{1}{T_{boiling}})\)

\(T_{boiling} = 371\ K = 98.4\ ^oC\)


The saying “diamonds are forever” is technically not true. Instead, diamond turns into graphite over a large amount of time. Explain why this phenomenon occurs. 


If the free energy of the product is less than the free energy of the reactant, then the product is considered stable. Converting diamond to graphite, the free energy is positive meaning diamonds are higher in energy than graphite. Therefore diamond is unstable. Because diamond is unstable, it eventually turns into graphite over millions of years. 


Based on your knowledge of network covalent solids, would you expect a diamond or graphite to have a higher density?


You would expect a diamond to have a higher density when compared to graphite, because unlike graphite that has \(sp^2\) sheets held together by p orbitals, diamonds form \(sp^3\) bonds resulting high in its strength. P bonds does not form strong bonds, which causes these sheets in graphite to be easily broken by sheer force.


Other structures, such as silicon carbide, \(SiC\), crystallize in the same fashion as a diamond. What can you infer about the structure of silicon carbide? Unlike silicon carbide, boron nitride crystallizes in  similar fashion to graphite. Knowing this, what could you suggest about the bonding scheme for \(BN\)?


If silicon atoms are substituted for half the carbon atoms in a diamond crystal structure, the resulting structure is that of silicon carbide. Carbon atoms can bond together to produce a solid with properties different than that of diamonds. Like graphite, boron nitride, involves the orbital set \(sp^2 + p\).


Name the crystalline solid type for each.

  1. \(NaCl\)
  2. \(solid\ HCl\)
  3. \(quartz\)
  4. \(Fe\)


in order, ionic, molecular, covalent network, metallic.


Boron nitrate, \(BN\), crystallizes in a from similar to graphite, whereas silicon carbide, \(SiC\), crystallizes in a form similar to form similar diamond. Propose a bonding scheme for \(BN\).


B and N must have \(sp^2\) hybridization


The hardness of crystals is rated based on Mohs hardness values.  The higher the Mohs value, the harder the material is to scratch.  Which crystal will have the LOWEST Mohs value: \(NaF\), \(NaCl\), or \(KCl\)?


\(KCl\) because it is the easiest to scratch.


There are two arrangements for the closest packing of spheres. Name the two arrangements and explain what they are.


One arrangement is hexagonal closest packed (HCP) in which the third layer has the same arrangement of spheres as the first layer and covers all the tetrahedral holes. Thus, the stacking for HCP is “ababab” because the structure repeats itself after every two layers. The other arrangement is cubic closest packed (CCP) which is similar to the HCP, since the second layer of spheres is placed on to half of the depressions of the first layer. However, the spheres in the third layer do not line up with those of layer A. Thus, the arrangement of layers is “abcabc”.


Explain why there can’t be a single arrangements for the closest packing of spheres.


There can be 2 because in one layer has 6 spheres touching one spheres. 2nd layer is placed on top of 1st layer by placing spheres in the hole of 1st. 3rd layer is placed on top of those two. The closest packing arrangement abab results, when one see a sphere of the bottom first layer. No first layer is visible through the indentation; the closest packing arrangement abcabc results.


Explain how both Cubic Close Packed and Hexagonal Closed Packed sphere arrangements result in  maximized density.


Cubic Close Packed and Hexagonal Closed Packed maximize the density of packing  spheres because they are both include sheets of spheres arranged as a triangle at the vertices. In both cases, each sphere has 12 neighbors. Each sphere will have 3 gaps that are empty and decrease the density of the packed spheres. For one gap, surrounded by spheres in an octahedral configuration, the distance to the center of the gap from the center of the sphere is \(\sqrt2\) times the radius of the sphere. For the other two gaps, surrounded by spheres in tetrahedral configurations, the distance to the center of the gap from the center of the sphere is \(\sqrt{\frac{3}{2}}\) times the radius of the sphere. By minimizing the empty space, 74% of the space is filled, thus maximizing the density.


What is a unit cell? Name the five different unit cells along with the characteristics for each (coordination number and atoms per unit cell).


Crystal structures occur whenever there is a regular pattern of atoms, ions, or molecules. A unit cell is the smallest representation of an entire crystal, and all crystal lattices are built of repeating unit cells. The simple cubic has a coordination number of 6 and contains 1 atom per unit. The body centered cubic (BCC) has a coordination number of 8 and contains 2 atoms per unit cell. The face centered cubic (FCC) has a coordination number of 12 and contains 4 atoms per unit cell. The hexagonal closest packed (HCP) has a coordination number of 12 and contains 2 atoms per unit cell. Lastly, the cubic closest packed (CCP) has a coordination number of 12 and contains 4 atoms per unit cell.


Consider the two-dimensional lattice shown at page 555.

  1. What is the unit cell?
  2. Is the amount of each element in the unit cell?
  3. Identify some simpler units than the unit cell, and why they can’t function as a unit cell.


  1. Smallest repeated unit of crystal lattice. Grey squares, open circles, and colored diamonds
  2. 1 light colored sq, 2 circles, 1 dark colored diamond


What are the coordination numbers for the following Atoms per unit cell?

  1. simple cubic
  2. Body-centered cubic
  3. hexagonal closet
  4. face-centered cubic


(a)6        (b)8        (c)12      (d)12


Calculate the density of iron \(Fe\) given that the metallic radius is 150 pm for one atom and that it displays a body centered cubic shape.


\(molar\ mass\ of\ Fe = 55.85 \frac{g}{mol}\)

\( \frac{(55.85\frac{g}{mol})}{(6.02\times 10^{23})} = 9.277\times 10^{-23} \frac{g}{unit cell}\)

\(L = \frac{4r}{\sqrt3}\)              

(substitute radius (in cm) into equation to get the length of one side)

\(L = \frac{4(1.5\times 10^{-8})}{\sqrt3}\)

\(L = 3.46\times 10^{-8}\)

\(Volume = L^3\)

So, \(Volume = 4.16\times 10^{-23}\ cm^3\)

Now that you have mass and volume, you are able to calculate density using \(D =\frac{M}{V}\)

\(D = \frac{(9.277\times 10^{-23})}{(4.16\times 10^{-23})}\)

\(D = 2.23\frac{g}{cm^3}\)


Iron has a body-centered cubic crystal structure. Using a metallic radius of 98 meter for the Fe atom, calculate the density of iron.



\(S=226.3213055\ meters\)


\(D= \frac{55.845g}{(226.3213055\ m)^3}\)

\(D= 4.8\times 10^{-6} \frac{g}{m^3}\)


Tungsten has a body-centered cubic crystal structure. Using a metallic radius of 150 pm for the W atom, calculate the density of tungsten.


\(V = (\frac{4r}{\sqrt3})^3\) 

\(V= (\frac{4\times 150\ pm \times \frac{100cm}{10^{12}\ pm}}{ {\sqrt3}})^3 = 4.157\times 10^{-23}\ cm^3\)

\(M=(\frac{2W\ atoms}{1unit cell})(\frac{1mol\ W}{6.022\times 10^{23}\ W\ atoms})(\frac{183.85g\ W}{1mol\ W})\)

\(M=6.106\times 10^{-22}\)


\(D= \frac{6.106\times 10^{-22}}{4.157\times 10^{-23}\ cm^3} =14.69\frac{g}{cm^3}\)


What is the density of Po which has a packing structure of simple cubic? The radius of Po is 135 pm.


\(M = (\frac{1\ atom}{unit\ cell}) (\frac{209g}{mol}) (\frac{mol}{6.022\times 10^{23} atoms}) = 3.47\times 10^{-22}\ g\)

\(135pm (\frac{1m}{10^{12}pm}) (\frac{100cm}{1m}) = 1.35\times 10^{-8}cm\)

\(V=L^3 = (2r)^3= (2\times 1.35\times 10^{-8} cm)^3 = 1.97\times 10^{-23} cm^3\)

\(D = \frac{M}{V} = \frac{3.47\times 10^{-22}\ g}{1.97\times 10^{-23} cm^3} = 17.6 \frac{g}{cm^3}\)


Calculate the density of Iron using a metallic radius of 124 pm. Assume Iron has a body-centered cubic packing structure.


In order to calculate the density, calculate the volume of Iron. We know that the length of one side of this structure is equal to \(\frac{4r}{\sqrt3}\)

So, \(\frac{4(124)}{\sqrt3}= 286.3657 pm\).  

Before calculating the volume, convert 286.3657 pm into cm.

\(286.3657\times \frac{(1 m)}{10^{12} pm}\times \frac{(100 cm)}{(1 m)}= 2.863657\times 10^{-8}\)

\(Volume= (2.863657\times 10^{-8})^3= 2.348351\times 10^{-23}\)

Now, calculate the mass in grams of Iron per unit cell. Since BCC packing structure implies to atoms per unit cell, this can be done be calculating the mass of one iron atom and multiplying by 2.

\(\frac{Mass\ of\ iron}{unit\ cell}= 2\times \frac{(1\ mol\ Fe)}{(6.022X10^{23})}\times \frac{(55.845 g)}{(1\ mol\ Fe)}=9.2735\times 10^{-23}\)

Finally, to find the density of iron, just divide the mass per unit cell by volume per unit cell.

\(Density= \frac{9.2735\times 10^{-23}}{2.348351\times 10^{-23}}= 3.9489 \frac{g}{cm^3}\)


Sodium is one of several elements known to have a body-centered cubic crystal system. The distance between neighboring \(Na\) atoms in this structure is 395 pm.

  1. What is the radius of one \(Na\) atom?;
  2. What is the density of \(Na\) metal?;
  3. At what angle (in radians) to the parallel faces of the \(Na\) unit cells would first-order diffraction be observed when using x-rays of wavelength \(3.43\times 10^{-9}\ m\)? (if the reflective index of sodium = 0.23)


            a)         length of a body-centered cubic unit cell = \(\frac{4r}{\sqrt3}\)

                        distance between neighbors = length of unit cell = \(395 pm = 3.95\times 10^{-10}\ m\)

                        \(3.95\times 10^{-10} =  \frac{4r}{\sqrt3}\)                            

                        \(r = 1.71\times 10^{-10}\ m = 1.71\times 10^{-8}\ cm\)

            b)        \(M\ of\ unit\ cell = (22.99\frac{g}{mol})(\frac{1\ mol}{6.022\times 10^{23}\ atoms})(\frac{2\ atoms}{unit\ cell})\)

                        \(= 7.64\times 10^{-23}\frac{g}{unit\ cell}\)

                        \(V\ unit\ cell = \frac{64r^3}{3^{3/2}} = \frac{64(1.71\times 10^{-8}\ cm)^3}{3^{3/2}} = 6.16\times 10^{-23}\ cm^3\)
                        \(D = \frac{M}{V}= \frac{7.64\times 10^{-23}}{6.16\times 10^{-23}} = 1.24\frac{g}{cm^3}\)

            c)         \(n\times L = 2\times d\times sin(x)\)

                        \((0.23)(3.43\times 10^{-9}) = 2(3.95\times 10^{-10} m)sin(x)\)

                        \(x = 1.52 radians\)


Polonium \((Po)\) is the largest member of the group 16 and is the only element known to take on the simple cubic crystal system. The distance between nearest neighbor Po atoms in this structure is 400 pm.

  1. What is the density of \(Po\) atom?
  2. What is the density of \(Po\) metal?
  3. At what angle (in degree) to the parallel faces of the \(Po\) unit cells would first-order diffraction be observed when using X-rays of wavelength \(1.8 \times 10^{-10}\;m\)?


  1. Diameter of Po=400 pm
  2. \(Unit\ cell=(400pm)^3=6.4\times 10^7 pm^3=6.4\times 10^{-23} cm^3\ per\ unit\ cell\) \[\rho=\dfrac{M}{V}=\dfrac{3.47 \times 10^{-22}\;g}{6.4 \times 10^{-23}\; cm^3}=5.42\; g/cm^3\]
  3. \(n=1\), \(d=400pm\) and \(wavelength=180.0pm\)

       \(sin(\Theta)=\frac{n\times gamma}{2d}=\frac{(1)(1.8\times 10^{-10})}{2(400\times 10^{-12})}=0.225\)



One unit cell of silver has an edge of 4.09 Angstroms and occupies a CCP.

  1. What is the density of the silver \(Ag\) atom?
  2. At what angle (in degrees) to the parallel faces of the Silver unit cells would first-order diffraction be observed when using X-rays of wavelength \(1.5\times 10^{-10}\ m\)? (not verified)




\(V\ per unit\ cell= L^3=(4.09\ Å)^3= 68.418\ Å = 6.84\times 10^{-7}\ cm^3\)

\(M\ per unit\ cell= \frac{(107.868 g)}{(1\ mol)}\times \frac{(1\ mol)}{(6.022\times 10^{23}\ atoms})= 1.791\times 10^{-22} \frac{g}{atom} \times 2=3.582\times 10^{-22}\frac{g}{unit cell}\)

\(Density= \frac{3.582\times 10^{-22}}{6.84\times 10^{-7}}= 5.5285\times 10^{-16}\frac{g}{cm^3}\)


\(n= 1\), \(d=4.09\), \(\pi=1.5\times 10^{-10}\)

\(sin\Theta = \frac{1.5\times 10^{-10}}{2(4.09\times 10^{-10}\ Å)}\)

\(\Theta= 10.566^o\)


Gold is given as a face centered cubit unit cell. A cold atom has a radius of 144 pm. What is the volume per unit cell of gold (in \(cm ^3\))? What is the density of gold (in \(\frac{g}{cm^3})\)? What is the distance between crystal planes if the incident angle of the x-ray is 20 and the wavelength is 5.0 nm? (use n=1 and put units in pm)



\(L= 144\sqrt8\)

\(L= 407\ pm\)

\(V=(407)^3 = 6.7\times 10^7\ pm^3\)

\(6.7\times 10^7\ pm^3 \times (\frac{1m}{1\times 10^{12}pm})^3\times (\frac{100cm}{1m})^3 = 6.7\times 10^{-23} cm^3\)

\(Mass\ of\ the\ unit\ cell\ (uc) = (\frac{4\ atoms}{uc})(\frac{196.67g}{mol})(\frac{mol}{6.022\times 10^{23}}) = 1.31\times 10^{-21}\ g\)

\(D = \frac{M}{V} = \frac{(1.31\times 10^{-21}g)}{(6.7\times 10^{-23}cm^3)}= 19.4\frac{g}{cm^3}\)

\(n\lambda= d(sin 20°)\)

\((1)(5 nm) = d(sin 20°)\)

\(d = 0.068 nm= 68 pm\)


Refer back to Figure 12-49 for reference on the unit cell coordination structure for cesium chloride. What can we expect from the radius ratio inequalities of the \(Cs^+\) and \(Cl^-\) ions \((\frac{rCs^+}{RCl^-} = 0.934)\)?


We expect from the radius ratio inequalities that the \(Cs^+\) ion will occupy a cubic hole in a simple cubic lattice of \(Cl^-\) ions. This unit cell is in accord with one to one ratio of \(Cs^+\) to \(Cl^-\) ions since there is one \(Cs^+\) in the center of the unit cell and \(8\times (\frac{1}{8})\ Cl^-\)ions at the corners.


Prove that the unit cells for \(TiO_2\) in Figure 12-50 are consistent with its formulas.


\(TiO_2\): 8 \(Ti^{4+}\) ions on the corners, shared among 8 unit cells, 1 \(Ti^{4+}\) corner ion per unit cell, There is 1 in the center contained within the unit cell. There is also 2 \(O^{2-}\) on the faces, shared between two unit cells, 2 face atoms per unit cells. 2 \(O^{2-}\) ions within the unit cell. It totals to 4 \(O^{2-}\) ions per unit cell. The ratio of Ti and O is 2\(Ti\) per 4\(O^{2-}: Ti_2O_4\).


Show that the unit cells for \(MgCl_2\) and \(MnO_2\) in the following pictures are consistent with their formulas. 


  • \(MgCl_2\): In each of the eight corners there are \(Mg^{2+}\) ions, which are shared among eight unit cells, for a total of one \((8\times \frac{1}{8})\) ion per unit cell. There are also six \(Mg^{2+}\) ions on each face of the unit cell, for a total of three \((6\times \frac{1}{2})\) ions per unit cell. This altogether gives a total on 4 \(Mg^{2+}\) per unit cell. There are eight \(Cl^-\) ions contained within the unit cell. This gives a ratio of 4 \(Mg^{2+}\) ions to 8 \(Cl^-\) ions, or \(MgCl_2\).
  • \(MnO_2\): In each of the eight corners there are \(Mn^{2+}\) ions, which are shared among eight unit cells, for a total of one \((8\times \frac{1}{8})\) ion per unit cell. There is also one \(Mn^{2+}\) ion in the center of the unit cell. Therefore, there are a grand total of two \(Mn^{2+}\) ions per unit cell. There are four \(O^{2-}\) ions on four faces of the unit cell, each of which is shared between two unit cells, for a total of two \((4\times \frac{1}{2})\) atoms per unit cell. There are an additional two \(O^{2-}\) ions completely contained within the unit cell. This gives a grand total of four \(Mn^{2+}\) ions per unit cell. Thus, the ratio is two \(Mn^{2+}\) ions to four \(O^{2-}\) ions, or \(MnO_2\).


The crystal structure of lithium fluoride, \(LiF\), is a face-centered cubic, like that of \(NaCl\). Given: the radius of \(Li^+ = 59 \;pm\) and the radius of \(F^- = 133\; pm\), identify:

  1. the coordination numbers of \(Li^+\) and \(F^-\)
  2. the number of formula units in the unit cell
  3. the volume and length of a unit cell (in pm)
  4. the density of \(LiF\)


a) In a face-centered cubic unit cell, there are six anions around each cation, and six cations surrounding each anion. Therefore, the coordination number of \(Li^+\) and \(F^-\) is 6.

b) There is a fluoride ion at each of the eight quarters, in the unit cell; each of which is shared with the eight surrounding unit cells. There is also one fluoride ion on each of the six faces; each is shared between two unit cells.  Thus, the total number of fluoride ions

\(= (8\times \frac{1}{8})+(6\times \frac{1}{2})=3\ F^-\ ions\)

There are twelve edges on the unit cell, each of which has a lithium ion on it, and each of these is shared between four unit cells. There is also a lithium ion completely contained within the unit cell. Thus, the total number of lithium ions

\(= (12\times \frac{1}{4}) + 1 = 4\ Li^+\ ions\)

Thus, there are four formula units per unit cell of \(LiF\).

c) Along the edge of the unit cell, combined, there is one \(Li^+\) and one \(F^-\) ion. Therefore the length of the unit cell is \(2\times Li^+\  radius+2\times F^-\  radius=(2\times 59)+(2\times 133)=384pm\)

Since the unit cell is a cube, the volume of the unit cell is the length cubed, 

\((384\ pm)^3 = 5.66\times 10^7\ pm\)

d) \(5.66\times 10^7pm \times (\frac{1m}{10^12pm})^3  × (\frac{100 cm}{1m})^3= 5.66\times 10^{-23}\ cm^3\)

\(D=\frac{M}{V}= \frac{4\ LiF\ f.u.}{5.66\times 10^{-23}\ cm^3}\times \frac{1\ mol\ LiF}{6.022\times 10^{23}\ f.u.}\times \frac{25.94g\  LiF}{1\ mol\ LiF}=3.04\frac{g}{cm^3}\) 


The crystal structure of magnesium oxide, \(Mg_O\), is of the \(NaCl\) type (Fig. 12-48). Use this fact, together with ionic radii from Figure 8-8, to prove the following:

  1. the distance and volume of a unit cell
  2. the density of \(MgO\)
  3. the coordination numbers of \(Mg\) and \(O\)
  4. the number of formula units in the unit cell


  1. The length of the edge is equal to the radius of 1 \(O\), plus the diameter of \(Mg\), the radius of another 0 edge length\(=2\times O radius+2 mg radius-2\times 140pm+2\times 72 pm=424 pm\)                                                                                                     \(V=(424\ pm)^3= 7.62 \times 10^7\ pm^3\times (\frac{1m}{10^{12}\ pm})^3 \times (\frac{100\ cm}{1\ m})^3=7.62 \times 10^{-23}\ cm^3\)
  2. \(\rho = \frac{m}{V}=\frac{4MgO\ formula\ unit}{7.62\times 10^{-23}\ cm^3}\times \frac{1mol\ MgO}{6.022\times 10^{23}\ formula\ units}\times \frac{40.3g\ MgO}{1mol\ MgO}=3.51\frac{g}{cm^3}\)
  3. \(Mg\) is 6 and \(O\) is 6
  4. \(Total\ Oxide\ ion=8\ corners\times \frac{1\ oxide\ ion}{8\ unit\ cell}+6\ faces \times \frac{1\ oxide\ ion}{2\ unit\ cells}= 4\ O\ ions\)

\(Total\ Mg\ ions=12\ adjoining\ cell\times \frac{1\ magnesiumion}{4\ unit\ cells}+central\ Mg\ ion=4\ Mg\ ions\)


Determine the density (in \(\frac{g}{cm^3})\) of tungsten if the atomic radius is 140.8 pm and the crystal packing structure is body centered cubic.


\(BCC = 2\frac{atoms}{unit\ cells}\)

\(m = (\frac{2\ atoms}{unit\ cells})(183.8\frac{g}{mol})(\frac{1\ mol}{6.022\times 10^{23}\ atoms}) = 6.104\times 10^{-22}\ g\)

\(V= L^3 = (\frac{4r}{3^0.5})^3 = (\frac{(4)(140.8\times 10^{-10}cm)}{(3^0.5)})^3 = 3.43\times 10^{-23}\ cm^3\)

\(r = 140.8pm = 140.8\times 10^{-12}m = 140.8\times 10^{-10}\ cm\)

\(d = \frac{m}{V} = \frac{6.104\times 10^{-22}\ g}{3.43\times 10^{-23}\ cm^3} = 17.8 \frac{g}{cm^3}\)


The ionic radii of \(Na^+\) and \(Cl^-\) in \(NaCl\) are 99pm and 181pm, respectively. What is the length of the unit cell of \(NaCl\)?


\(Length = (rCl^-) + (rNa^+) + (rNa^+) + (rCl^-)= 2(rNa^+) + 2(rCl^-) = (2\times 99) +(2\times 181) = 560pm\)


Using the data from figure 8-9, predict the type of cubic unit cell adopted by a)  \(NaCl\) b) \(MgS\) and c) \(LiF\)


  1. Radius ratio= \(\frac{99pm}{181pm}= 0.547\) = Cations occupy octahedral holes of an fcc array of anions.
  2. Radius ratio= \(\frac{72pm}{184pm}= 0.3913\) = cations occupy tetrahedral holes of an fcc array of anions.
  3. Radius ratio= \(\frac{59pm}{133pm}= 0.44361\) = cations occupy octahedral holes of an fcc array of anions.


Determine if the following molecules are face-centered cubic (fcc), or body-centered cubic (bcc). Use a table comparing some atomic and ionic radii. 

  1. \(RbI\)
  2. \(CsCl\)
  3. \(MgO\)
  4. \(AgCl\)


  1. \(RbI\) has a \(Rb\) radius =149pm and \(I\) radius = 220pm therefore is face centered cubic.
  2. \(CsCI\) has a \(Cs\) radius = 77pm and \(Cl\) radius = 220pm therefore is body centered cubic. 
  3. \(MgO\)  has a \(Mg\) radius = 72pm and \(O\) radius = 140pm therefore is face centered cubic.
  4. \(AgCl\) has an \(Ag\) radius = 115pm and \(Cl\) radius = 181pm therefore is face centered cubic. 


Use data from Figure 8-9 to predict the type of cubic unit cell adopted by (a) \(LiO_2\) (b) \(CaO\) (c) \(CuCl\)


  1. \(\frac{rLi}{rO2}=\frac{59pm}{128pm}=0.461\) cations are in the octahedral holes of a face centered cubic
  2. \(\frac{rCa}{rO}=\frac{100pm}{140pm}=0.714\) cations are in octahedral holes of a face centered cubic
  3. \(\frac{rCu}{rCl}=\frac{96pm}{181pm}=0.53\) cations are in octahedral holes of a face centered cubic


Explain the concept of lattice energies and describe how they are related to Hess’s Law in determining enthalpies of formation. Include the steps required to find lattice energy.


The lattice energy of a compound is found indirectly by using Hess’s Law to cancel out ions and ultimately calculating the desired enthalpy. Step 1: \(\Delta H\) of sublimation of initial solid. Step 2: Dissociation of \(0.5mol\) of 2nd gas. Step 3: Ionization of \(1mol\) of 1st gas (previously solid) to proton-donated ion. Step 4: convert \(1mol\) of 2nd gas to its proton-gained ion. Step 5: Allow the 1st gas and 2nd gas to form one mole of newly formed solid.


Without doing calculations, indicate how you would except the lattice energies of \(CsCl_{(s)}\), \(LiCl_{(s)}\), \(RbCl_{(s)}\), \(KCl_{(s)}\) to compare with the value of \(-987\frac{kJ}{mol}\) determined for \(NaCl_{(s)}\) on page 547.


Those lattice energies vary approximately with cation size. More exothermic lattice energy will produce in smaller cation. \(LiCl_{(s)}\) is the most exothermic and \(CsCl_{(s)}\) is the least.


Determine \(\Delta H^o\) for the reaction by using the following data.\(N_2H_{4\ (l)} + O_{2\ (g)}\rightarrow N_{2\ (g)} + H_2O_{(l)}\)       \(\Delta H^o= -622.2 kJ\)

\(N_2H_{4(l)} +2H_2O_{2(l)} \rightarrow N_{2(g)} + 4H_2O_{(l)}\)
\(H_{2\ (g)} + \frac{1}{2}O_{2\ (g)} \rightarrow H_2O_{(l)}\)                \(\Delta H^o= -285.8 kJ\)

\(H_{2\ (g)} + O_{2\ (g)} \rightarrow H_2O_{2\ (l)}\)                \(\Delta H^o= -187.8 kJ\)


\(Rxn1 + 2Rxn2 – 2Rxn3\)

\(\Delta H^o_{rxn} = -622.2 kJ + (2\times -285.8 kJ) + (2\times 187.8 kJ) = -818.2 kJ\)


Without doing calculations, indicate how the lattice energies of \(MgS_{(s)}\), \(CaS_{(s)}\), \(SrS_{(s)}\), and \(BaS_{(s)}\) compare to one another. Which is the most endothermic/exothermic?


When the anion is kept constant, one can compare the lattice energy by investigating the size of the cation. The smaller the cation, the more energy is capable of being released, causing a more exothermic lattice energy. Therefore, it is evident that \(MgS_{(s)}\) has the most exothermic lattice energy of the series, and \(BaS_{(s)}\) has the most endothermic lattice energy of the series.


What relation do you see between the Lattice energies of alkali metals and their cation sizes?


A smaller cation will produce a more exothermic lattice energy. 


Calculate the reaction enthalpy for the hydrogenation of graphite:

\[C_{(graphite)} + 2H_{2 (g)} \rightarrow CH_{4 (g)}\]

from the following information:

  • \(H_{2 (g)} + ½O_{2 (g)} \rightarrow H_2O_{ (l)}\)  \(\Delta H^o= -285.8 \frac{kJ}{mol}\)
  • \(C_{(graphite)} + O_{2 (g)} \rightarrow  CO_{2 (g)}\)  \(\Delta H^o= -293.5 \frac{kJ}{mol}\)
  • \(CH_{4 (g)} + 2O_{2 (g)} \rightarrow CO_{2 (g)} + 2H_2O_{ (l)}\)  \(\Delta H^o= -890.4 \frac{kJ}{mol}\)


\(\Delta H^o_{rxn}=(2\times -258.8)+(1\times -293.5)+(-1\times -890.4)=25.3kJ\)


Refer to Example 12-12. Together with the data given there, use the data here to calculate \(\Delta H_f\) for 1 mol \(MgCl_{2\ (s)}\). Explain why you would except \(2MgCl_2\) to be a much more stable compound than \(MgCl\). (Second ionization energy of \(Mg\), \(I_2=1500 \frac{kJ}{mol}\); lattice energy of \(2MgCl_{2(s)}=-2526\frac{kJ}{mol})\).


\(Mg^+_{(g)} \rightarrow Mg^{2+}_{(g)}+e^-\)   \(I_2=1500\frac{kJ}{mol}\)

\(2Mg^{2+}_{(g)}\rightarrow 2Cl^-_{(g)}+2MgCl_{2(s)}\)      \(LE=-2526\frac{kJ}{mol}\)

\(Mg_{(s)} \rightarrow Mg_{(g)}\)  \(\Delta H_{sub}=146\frac{kJ}{mol}\)

\(Mg_{(s)} \rightarrow Mg^+_{(g)}+e^-\)   \(I_1=750\frac{kJ}{mol}\)

\(Cl_{2(g)} \rightarrow 2Cl_{(g)}\)    \(DE=(2\times 122)\ \frac{kJ}{mol}\)

\(2Cl_{(g)}+2e^- \rightarrow 2Cl^-_{(g)}\)     \(2\times EA=2(-349)\ \frac{kJ}{mol}\)

\(\Delta H_f=\Delta H_{sub}+I_1+I_2+DE+(2\times EA)+LE=146+750+1500+244-698-2526-645\)

The example of 12-12, \(\Delta H_f\) for \(MgCl\) is \(19\frac{kJ}{mol}\). \(MgCl_2\) is much more stable than \(MgCl\), since it has more energy release when it form \(MgCl_{2\ (s)}\) is more stable than \(MgCl_{(s)}\).


One type of phase change is the direct passage of moles from the solid to the vapor state called sublimation. Explain why snow may disappear from the ground even though temperature may fail to rise above 0°C.


Under these conditions snow does not melt but sublimes. The solid has a vapor pressure of ice at 0°C is 4.58 mmHg. If the air is not already saturated with water vapor, the ice will sublime.


When a wax candle is burned, the fuel has gaseous hydrocarbons showing at the tip of the candle wick. Describe the phase changes and steps by which the solid wax is consumed ultimately.


The candle has cotton thread surrounded by paraffin which is a heavy chain hydrocarbon and hence initially in the solid phase. When the cotton thread is lighted, the tip starts to burn and the heat melts the paraffin into a liquid phase. The liquid of hydrocarbons enter into the fibers of the cotton thread. The hydrogen carbons end up vaporing and finally burn out giving out carbon dioxide and water vapor.

$$C_nH_{2n(l)} + O_{2(g)} \rightarrow nCO_{2(g)} + H_2O_{(g)}$$


When wood is burned, describe the phase changes and processes that occur.


When wood is burned, oxygen from the air reacts with various hydrocarbons in the wood, forming carbon dioxide, steam and some other chemicals. The overall process is exothermic and involves no liquid stage.


When ice is heated up in a pot, describe the phase changes in between in order for it to end up into a gas.


When the ice reaches \(0^oC\) it phase changes from solid ice to liquid water. As the water continues to boil, the liquid water will transform into vapor gas once \(100^oC\) is reached. At \(0^oC\) this is enthalpy of condensation and at \(100^oC\) this is called enthalpy of vaporization.


Using the follow data determine the quantity of heat needed to convert 13.0 g of ice at \(-11.0°C\) to water vapor at \(105°C\). Heat of fusion =\(334\frac{J}{g}\), heat of vaporization of water = \(2257\frac{J}{g}\), specific heat of ice =\(2.09\frac{J}{g°C}\), specific heat of water =\(4.18\frac{J}{g°C}\), specific heat of vapor =\(2.09\frac{J}{g°C}\). 


Heat required to raise the temperature of ice

=\((13.0g)\times \frac{(2.09\ J)}{(g°C)}\times (0--10)°C=271.7\ J\)

Heat required to convert ice to liquid water

=\((13.0g)\times \frac{334\ J}{g}= 4342\ J\)

Heat required to raise the temperature of water

=\((13.0g)\times \frac{4.18\ J}{g°C}\times (100- -0)°C=5434\ J\)

Heat required to convert water to steam

=\((13.0g)\times \frac{2257J}{g°C}= 29341\ J\)

Heat required to raise the temperature of water

=\((13.0g)\times \frac{(2.09\ J)}{(g°C)}\times (105- -100)°C=135.85\ J\)

Sum together all the heats



Determine the quantity of heat need to convert 15.0 g of solid mercury at \(-50.0°C\) to mercury vapor at \(25°C\).


Specific heat : \(Mercury_{(s)}= 0.121\frac{J}{g°C}\) 

\(Melting (mp= 38.87°C) = 2.33\ \Delta H^o\ \frac{kJ}{mol}\)
\(Mercury_{(l)} = 0.140\frac{J}{g°C}\)

\(Vaporization = 61.3 \Delta H°\ \frac{kJ}{mol}\)

\(q=mCp\Delta T = (15.0g) (0.121\frac{J}{g°C})(-38.87-(50))= 20.2\ J\)

\(q = n\Delta H_{fus} = (15.0 g) (\frac{mol}{200.59g}) (2330\frac{J}{mol}) = 174\ J\)

\(q=mCp\Delta T = (15.0 g) (0.140\frac{J}{g°C}) (25- (-38.87)) = 134\ J\)

\(q = n\Delta H_{vap} = (15.0 g) (\frac{mol}{200.59g}) (61300\frac{J}{mol}) = 4580\ J\)

\(q\ total = 20.2 + 174 + 134+ 4580 = 4910\ J\)


Determine the amount of heat required to heat 4.2 g of liquid Cl2 at \(-54.3°C\) to vapor at \(20°C.\) (specific heats \(Cl_{2(l)}\), \(19.4\frac{J}{g°C}\) ; \(Cl_{2\ (g)}\), \(14.7 \frac{J}{g°C}\) (enthalpy of vaporization of \(Cl_2 = 4.90\frac{kJ}{mol})\) (boiling point \(Cl_2 = -34.0 °C)\)


            \(q= mC\Delta T\)                     \(q = n\Delta H\)

            \(q_1 = (4.2 g)(19.4\frac{J}{g°C})(-34.0 + 54.3°C) = 1654\ J\)

            \(moles = (4.2g\ Cl_2)(\frac{1\ mol}{70.9\ g}) = 0.0592\ mol\ Cl_2\)

            \(q_2= (0.592 mol)(4.90\frac{kJ}{mol}) = 0.290\ kJ = 290\ J\)

            \(q_3 = (4.2 g)(14.7\frac{J}{g°C})(20 + 34.0°C) = 3334\ J\)

            \(q_{total} = 1654 + 290 + 3334 = 5278\ J\)


Using the following data and data from Appendix D to determine the quantity of heat needed to convert 24.0g of solid mercury at \(-47^oC\) to mercury vapor at \(33^oC\). Specific heats: \(Hg_{(s)}:\ 24.3\frac{J}{mol\ K}\), \(Hg_{(l)}:\ 28.0\frac{J}{mol\ K}\). Melting point of \(Hg_{(s)}:\ -38.87^oC\). Heat of fusion: \(2.33 \frac{kJ}{mol}\).


\(Mol\ Hg_{(s)}=\frac{Mass}{atomic\ mass}=\frac{24.0g}{200.59\frac{g}{mol}}=0.120mol\)

\(Heat\ required,\ q_l=mol\times molar\ specific\ heat \times temp\ differ=0.12mol\times 24.3\frac{J.}{mol\ K}\times (-38.7+50^oC)=32.85J\)

\(Heat\ of\ formation= 61.32\frac{kJ}{mol}\)

\(\Delta H_{sub}=\Delta H_{fus}+\Delta H_{vap}=2.33+61.32=3.65kJ\)

\(Heat\ required, q_2=mol\times \Delta H_{sub}=0.12\times 63.65\frac{kJ}{mol} \times \frac{10^3J}{1kJ}=7638\ J\)

\(Heat\ required, q_3=mass\times specific\ heat\times temp\ diff=24.0g\times 0.104\frac{J}{g^oC} \times (33+38.7^oC)=178.96J\)

\(Total\ heat\ required=q_1+q_2+q_3=32.85+7638+178.96=7849.81J\)


Estimate how much heat is absorbed when 2.5 g of Instant Car Kooler vaporizes. Comment on the effectiveness of this spray in cooling the interior of a car. Assume the spray is 10% \(C_2H_5OH_{(aq)}\) by mass, the temperature is \(75^oC\), the heat capacity of air is \(36\frac{J}{mol\ K}\), and use \(\Delta H_{vap}\) data from Table 12.3


\(Mass\ of\ C_2H_5OH\ when\ there\ is\ 2.5g\ of\ Instant\ Car\ Kooler=\frac{10}{100}\times 2.5g=0.25g\)

\(Heat\ absorbed\ by\ 0.25g\ ethanol=0.25g\times \frac{1mol}{46.07g}\times 42.6\frac{kJ}{mol}\times \frac{10^3J}{1kJ}=231.17J\)

\(Average\ volume\ of\ a\ car=100ft^3\times (\frac{12in}{1ft})^3\times (\frac{2.54cm}{1in})^3=2.83\times 10^6\ cm^3\)

\(Approximate\ density\ of\ air\ at\ this\ temperature=1.067\frac{kg}{m^3}\times \frac{10^3g}{1kg}\times (\frac{1m}{10^2cm})^3=0.00107\frac{g}{cm^3}\)

\(Approximate\ mass\ of\ air\ in\ a\ car=D\times V= 0.00107\frac{g}{cm^3}times (2.83\times 10^6cm^3)=3028.1g\)

\(Mol\ of\ air=\frac{M}{average\ molar\ mass}=\frac{3028.1g}{28.97\frac{g}{mol}}=104.52\ mol\)

\(Heat\ capacity\ of\ air=36\frac{J}{mol\ K}\)

\(Heat=mol\ of\ air\times specific\ heat\ capacity\times temp\ difference\)

\(231.17J=104.52mol\times 36\frac{J}{mol\ K}\times (75-T_f^oC)\)



Determine the quantity of heat needed to warm \(50.0g\) of water from \(25°C\) to steam at \(110°C\). \((\Delta H_{vap}= 44\times 10^3 \frac{J}{mol})\).


\(q=mCp\Delta T= (50.0 g) (4.184 \frac{J}{g\ °C}) (100 – 25) = 15690\ J\)

\(q = n\Delta H_{vap}= (50.0 g) (\frac{mol}{18.02\ g}) (44\times 10^3 \frac{J}{mol}) = 122086.57\ J\)

\(q=mCp\Delta T= (50.0 g) (2.08\frac{J}{g\ °C}) (110 – 100) = 1040\ J\)

\(q\ total = 15690 + 122086.57 + 104 J= 1.39\times 10^5\ J\)


If temperature of \(H_2O\) is increased when it had already reached the critical point, what phase is \(H_2O\) in?


Past the critical point, no phase boundaries exist.


Using information on page 538, explain how the total number of atoms in a bcc unit cell is two when it takes nine atoms to draw.


Only the center atoms belong entirely to the bcc unit cell. Only \(\frac{1}{8}^{th}\) of each corner atoms on the cell belongs entirely to the unit cell. So the 8 corner atoms only equal 1 atom. By that, the total number of atoms is then only 2. \([1 + (8\times (\frac{1}{8}))]\)


There is one \(Mg\) atoms in the simple cubic unit cell. To calculate the mass and density of the unit cell, use the length of each unit cell, \(L= 409pm\). Also, using this information, calculate the density of magnesium.


\(1\ mol\ Mg = 6.022\times 10^{23}\ atoms = 24.32\frac{g}{mol}\ Mg\)

\(m= 1\ atom\ Mg\ atoms \times \frac{24.32g\ Mg}{6.022\times 10^{23}\ Mg\ atoms} = 4.04\times 10^{-23}\ g\ Mg\)

\(V= L^3= (409 pm\times \frac{10^3}{10^9})^3 cm^3= 6.84\times 10^{-11}\)

\(Density\ of\ Mg=\frac{mass}{volume} = \frac{4.04\times 10^{-23}\ g\ Mg}{6.84\times 10^{-11}\ cm^3}\)

\( = 5.9\times 10^{-13}\frac{g}{cm^3}\)


Calculate the atomic radius of vanadium (V) in pm, if the density is \(6.11\frac{g}{cm^3}\) and its packing structure is body centered cubic.


\(\rho = \dfrac{m}{V}\)

\(m = (2 \frac{atoms}{unit cell}) (\frac{1\ mol}{6.02\times 10^{23}\ atoms})(50.94 \frac{g}{mol})\)

\(V = L^3 = (4r\sqrt3)^3\)

\(6.11\frac{g}{cm^3} = (1.61918\times 10^{-22}\ g)(12.32r^3)\)

\(r = 131\ pm\)


Using the analyses of a bcc structure on page 538 and the fcc structure in Exercise 139, determine the percent voids in the packing-of-spheres arrangement found in the fcc crystal structure.



\(Diagonal\ of\ the\ face=4r=4(128)=512pm\)


\(L=362 pm\)

\(ratio\ volume\ to\ unit\ cell\ volume=\frac{Vol\ sphere}{Vol\ unit\ cell}=\frac{4\times (\frac{4}{3})\pi(128pm)^3}{(362pm)^3}=0.7410\)


\(Percent\ FCC=25.89\%\)