# 12: Intermolecular Forces

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## Q1a

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

## Q1b

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

1. $$LiBr$$
2. $$GeH_4$$
3. $$SO_2$$
4. $$CH_4$$

### S1b

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

## Q1c

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

### S1c

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.

## Q1d

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

### S1d

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.

## Q3a

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. $$He$$

### S3a

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. $$Ar$$

## Q3c

List the following molecules in order of increasing viscosity.

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

### S3c

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

(d)<(a)<(c)<(b)

## Q5a

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?

### S5a

$$C_3H_8O$$

## Q5b

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?

### S5b

$$H_2O$$: 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).

## Q5c

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

### S5c

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.

## Q5d

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

### S5d

$$NaCl$$ 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.

## Q5e

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

• $$C_2H_6$$
• $$O_2$$
• $$C_2H_5 OH$$
• $$CH_4$$

### S5e

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.

## Q11a

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

### S11a

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.

## Q11b

List the following molecules in order of increasing viscosity.

1. $$C_4H_{10}O$$
2. $$CH_4O$$
3. $$CCl_4$$

(c) < (a) < (b)

## Q11c

Why doesn’t oil and water mix?

### Q11c

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.

## Q12d

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

### S11d

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.

## Q11e

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.

### S11e

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.

## Q11f

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.

### S11f

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.

## Q12a

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

### S12a

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.

## Q13a

Does honey have high viscosity and Why?

### S13a

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

## Q13b

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

### S13b

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.

## Q14a

Why does detergent lower the surface tension of water?

### S14a

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

## Q14b

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

### S14b

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.

## Q19a

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

### S19a

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.

## Q19b

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.

### S19b

First conver temperature to absolution temperature

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

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

$∆H = 18.0\; kJ$

## Q20a

Why does ethyl alcohol feel cool against the skin?

### S20a

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.

## Q20b

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

### S20b

• 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.

http://chemwiki.ucdavis.edu/Physical_Chemistry/Thermodynamics/State_Functions/Enthalpy/Enthalpy_Of_Vaporization

## Q20c

At 100 C° what part of $$H_2O$$ is in which phase?

### S20c

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

## Q20d

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?

### S20d

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.

## Q21a

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?

### S21a

Vaporization: liquid to gas

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

$V=\dfrac{nRT}{P}$

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

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

$V= 238.2\ L$

## Q21b

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℃?

### S21b

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

$$100^oC + 273.15 = 373.15\ K$$

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

## Q21c

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 $$C_6H_{12}$$?

### S21C

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

## Q20E

Explain the process of vaporization?

### S20E

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

## Q21E

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?

### S21E

$$V=\frac{nRT}{P}$$

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

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

## Q25B

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?

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

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

$$CH_3OH\ B.P.= 64.7^oC$$

### S25B

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:

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

## Q12.29

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

### S12.29

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\]

## Q89c

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

### S89c

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.

## Q89d

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

### S89d

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.

## Q92a

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 =$$2.09\frac{J}{g°C}$$.

### S92a

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

=$$271.7J+4342J+5434J+29341J+135.85J=39524.55J$$

## Q92b

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

### S92b

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

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

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

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

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

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

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

$$q\ total = 20.2 + 174 + 134+ 4580 = 4910\ J$$

## Q92c

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 $$Cl_2 = -34.0 °C)$$

### S92c

$$q= mC\Delta T$$ $$q = n\Delta H$$

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

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

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

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

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

## Q92d

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: $$2.33 \frac{kJ}{mol}$$.

### S92d

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

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

$$Heat\ of\ formation= 61.32\frac{kJ}{mol}$$

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

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

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

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

## Q94a

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

### S94a

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

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

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

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

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

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

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

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

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

$$T_f=74.73^oC$$

## Q94b

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

### S94b

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

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

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

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

## Q94c

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

### S94c

Past the critical point, no phase boundaries exist.

## Q110a

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.

### S110a

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. $$[1 + (8\times (\frac{1}{8}))]$$

http://chemwiki.ucdavis.edu/Physical_Chemistry/Physical_Properties_of_Matter/Solids/Virtual%3a_Crystal_Structure/Unit_Cells/Body_Centered_Cubic

## Q110b

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.

### S110b

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

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

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

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

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

## Q110c

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.

### S110c

$$\rho = \dfrac{m}{V}$$

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

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

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

$$r = 131\ pm$$

## Q110d

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.

### S110d

$$Radius=128pm$$

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

$$(512pm)^2=L^2+1^2$$

$$L=362 pm$$

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

$$Ratio=1-0.7410=0.2589$$

$$Percent\ FCC=25.89\%$$