# Practice Problems for Exam 1

Practice problems for exam 1 are listed below. This exam will cover sections 10.1 - 11.8

## Topic: Gas Laws

1. #### Properties of Gases

Given a balloon of H2 and O2 which of the following statements are not true?

• The balloon can be easily compressed to a smaller volume.
• Since two gases are present strong interactions between them will result in H2 and N2 condensing into liquids.
• The gas molecules are in a rigidly defined structure.
• The gas molecules are randomly mixed.
• True
• False
• False
• True
##### Explanation
• Gases are easiliy compressible
• The interaction between gas molecules are weak
• Gas molecules are in random motion
• Gases form homogeneously mixtures with each other
2. #### Pressure

You move into an apartment with a new hardwood floor. The landlord informs you that heavy furniture may damage the floor, the maximum pressure the floor can withstand without damage is 40 atm2. The next day one of your friends comes to visit. She has a mass of 50 kg but is wearing stilleto heals with a cross section of 1 cm2. Are you going to get your security deposit back?
Your friend put holes in the floor, you will not get your money back
##### Explanation
Your friend has a mass of 50 kg while walking her entire weight is occasionally on one heel with an area of 1 cm2 (0.0001 m2). Pressure is defined as F/A and in this example the pressure applied to the floor is (50 kg * 9.8 m/s) / 0.0001 m2 = 490 N / 0.0001 m2 = 4900000 Pa = 48 atm.
3. #### Gas laws and changes in pressure and volume

Given a steel piston at 298 K 10 L in size filled with 4 atm of N2 what happens to the pressure when
• The piston is compressed by a factor of 4
• The temperature is halved
• The number of moles of N2 is tripled
• The pressure is halved
• The pressure is tripled
##### Explanation
• According to Boyle's Law pressure and volume are inversely related
• According to Charles's Law temperature and pressure are directly proportional
• According to Avogadro's Law pressure is proportional to the number of moles

Consider the three mecruy manometers pictured below. For each manometer, what is the pressure of the gas for a value of h1 = h2 = h3 = 20 cm, assuming the manometer is under standard atmospheric pressure.

The gas pressures are:
• Left - 200 mmHg
• Center - 560 mm Hg
• Right - 960 mm Hg
##### Explanation
• Left - This is a close-ended manometer. The column height in mm is the gas pressure.
• Center - This is an open-ended manometer. The gas pressure is less than atmospheric pressure since the mercury level on the flask side is higher than that on the side open to atmosphere. The gas pressure is Patm - Ph.
• Right - This is an open-ended manometer. The gas pressure is more than atmospheric pressure since the mercury level on he flask side is lower than that on the side open to atmosphere. The gas pressure is Patm + Ph.

## Topic: Ideal Gas Laws

1. #### Ideal Gas Constant

True or False: If a gas pressure is expressed in atm, the volume is in L, temperature in K, and the quantity of the gas is given in moles the ideal gas law constant has units of L atm / K.
False
##### Explanation
The ideal gas law is

and the units of R are (L atm) / (K mol)
2. #### Ideal Gas Law

A piece of dry ice [CO2(s)] is allowed to sublimate (convert from a solid to a gas) inside a large balloon. If 92 g of CO2 is allowed to sublimate what is the volume of the ballon if the temperature is 25 o.
48.9 L
##### Explanation
From the ideal gas law, PV = nRT, the pressure is assumed to be 1 atm. The temperature is 298 K. The number of moles of CO2 is

92g CO2 * (1 mol / 46 g) = 2 mol

Using the ideal gas equation the volume of the ballon is

V = nRT/P

V = 2 mol * 0.08206 (Latm/Kmol) * 298 K / 1 atm

V = 48.9 L

3. #### Breakdown of the Ideal Gas Law

In which of the following conditions would you expect the ideal gas law to breakdown?
• The temperature of a gas is close to 0 K
• A balloon of N2 at STP.
• The volume is small enough that the gas molecules occupy approximately 20% of avialable volume
• Breakdown
• Ideal Gas Law applies
• Breakdown
##### Explanation
• As the temperature decreases the individual gas molecues slow down and intermolecular forces increase which increase the deviations from ideal gas behavoir.
• The ideal gas law is applicable in this situtation.
• The the individual gas molecules occupy a greater percentage of the volume of a container intermolecular forces play an increasing role limiting the usefullness of the ideal gas equation.
4. #### Gas Density

A hot air balloon rises because the density of the air inside the balloon is less than the density of the air around the balloon. Calculate the density of air at 25 o and 1 atm (assume for simplicity that air is 100% N2). What temperature would the balloon need to be to lower the gas density to 0.03 mol/L?
gas density = 0.0409 mol/L, T = 406 K
##### Explanation
The density of air can be calculated by rearranging the ideal gas law, PV = nRT to

n/V = P/RT

At a temperature of 298 K and a pressure of 1 atm the density of air is

n/V = 1 atm / (0.08206 Latm/Kmol * 298 K) = 0.0409 mol/L

To lower the gas density to 0.03 mol/L would require raising the temperature to

T = P / (R * n/V) = 1 atm / (0.08206 Latm/Kmol * 0.03 mol/L) = 406 K

5. #### Gaseous Chemical Reactions

Methanol can be created according to the following reaction

CO(g) + 2 H2 → CH3OH(g)

What volume of Hydrogen gas at 748 torr and 86oC is required to synthesize 25.8 g of methanol? What volume of CO gas is required?
48.2 L of H2 and 24.1 L of CO.
##### Explanation
First convert the mass of CH3OH into the number of moles of CH3OH.
25.8 g CH3OH * 1 mol / 32 g = 0.806 mol H2
To produce 0.806 moles of CH3OH2 requires
2 * (0.806) = 1.612 moles of H2.
Using a pressure of 0.9842 atm and the ideal gas law the volume of H2 required is
V = nRT/P
V = (1.612 mol)(0.08206 Latm/Kmol)(359 K) / 0.984 atm = 48.2 L
The volume of CO required is
V = nRT/P
V = (0.806 mol)(0.08206 Latm/Kmol)(359 K) / 0.984 atm = 24.1 L

## Topic: Partial Pressures

1. #### Partial Pressures

A 22.40 L container holds one mole of N2 and one mole of NH3 at a temperature of 0 oC. Both N2 and NH3 have a molar volume of 22.40 L at STP (standard temperature and pressure). Determine whether each of the following statements is true, false, or approximately true.
1. The partial pressure of nitrogram is 1.000 atm
2. The partial pressure of ammonia is 1.000 atm
3. The total pressure is 1.000 atm
1. Approximately true
2. Approximately true
3. False
##### Explanation
1. The ideal gas equation makes assumptions regarding the behavoir of a gas, therefore, the equation P = (nRT)/V is an approximation
2. The ideal gas equation makes assumptions regarding the behavoir of a gas, therefore, the equation P = (nRT)/V is an approximation
3. The total approximate pressure is 2.000 atm. Ptot = (ntotRT)/V
2. #### Three Gas Mixture

A mixture containing 0.304 mol Rn(g), 0.692 mol of Cl2(g), and 0.653 mol CO2 is confined in a 23.00 L vessel at 8.0 oC. Calculate the partial pressure of Rn(g) and the total pressure assuming ideal gas behavior.
Partial pressure of Rn(g) is 0.305 atm, total pressure is 1.65 atm.
##### Explanation
The partial pressure of Rn(g) is

PRn = nRnRT/V,

were nRn = 0.304 mol, R = 0.08206 L-atm/mol-K, T = 281.15 K and V = 23.00 L The total pressure is

Ptot = ntotRT/V,

where ntot = nRn + nCl2
3. #### Partial Pressure and mole fraction

5.00 L of He gas at 1.2 atm is mixed with 4.00 L of Ne gas at 1.25 atm at a temperature of 35 oC to make a total volume of 9.00 L of a mixture. Assuming no temperature change and that the He and Ne can be approximated as ideal gases, what is the mole fraction of the He and Ne in the mix?
XHe = 0.55, XNe = 0.45
##### Explanation
The mole fraction of He is

XHe = nHe/ntot

where the number of moles of moles of He is

nHe = (PHeVHe)/RT = 0.2372 mol

The number of moles of Ne is

nNe = (PNeVNe)/RT = 0.1977 mol

The total number of moles is

ntot = nHe + nNe = 0.4349 mol

and the mole fraction of He and Ne are

XHe = 0.2372/0.4349 = 0.55, XNe = 1 - XHe =0.45

## Topic: Kinetic Molecular Theory

1. #### Principles of Kinetic Molecular Theory

Which of the following statements are ture in relation to Kinetic Molecular Theory of an ideal gas?
• The average kinetic energy of a gas is independent of absolute temperature
• The attractive forces between gas molecules can be ignored
• The pressure exerted by a gas is a consequence of molecular collisions between individual gas molecules within the container
• The finite volume of individual gas molecules contributes to the ideal behavoir of the gas
• The average molecular speed of a gas is dependent on the absolute temperature and molar mass
• Gas molecules are in continuous, random motion
• False
• True
• False
• False
• True
• True
##### Explanation
• The average kinetic energy of a gas is directly proportional to temperature
• Attractive and repulsive forces between gas molecules are negligible in Kinetic Molecular Theory for an ideal gas
• The pressure is due to collisions of gas molecules with the wall of the container
• The combined volume of all the molecules of the gas is negligible relative to the total volume in which the gas is contained
• The average molecular speed is the SQRT[(3RT)/M] where R is the gas constant, T is the temperature, and M is the molar mass
• Gases consist of large numbers of molecules that are in continuous, random motion
2. #### Diffusion of Two Gases in a Reaction Tube

Ammonia and hydroiodic gas can mix and react to form ammonium iodide

NH3(g) + HI(g) -> NH4I(s)

The product is a white solid and the formation of white smoke indicates where the gases mix and react. Equal molar amounts of HI(g) and NH3 are placed at opposite ends of a 30.0 cm tube kept at 20oC. Calculate where along the tube the NH4I(s) will form.
22 cm from the end where NH3 was placed
##### Explanation
The average velocity of each gaseous species can be determined from the temperature and molar mass

u(NH3) = SQRT[(3RT)/M]

Using R=8.314 J/mol-K, T = (20oC + 273.15) = 293.15 K and M = 17.03 g/mol (0.01703 kg/mol) for the molar mass of NH3 and 127.91 g/mol (0.12791 kg/mol) for the molar mass of HI the average velocity of both gases are:

u(NH3) = 655 m/s, u(HI) = 239 m/s

Since the travel time of both gases will be the same

u(NH3) = 655 m/s = d(NH3)/t

u(HI) = 239 m/s = d(HI)/t

d(NH3)/655 m/s = d(HI)/239

The reaction is confined to take place inside a tube with a length of 30 cm

0.30 m = d(NH3) + d(HI)

There are two equations and two unknowns that can then be solved for the interaction distance.
3. #### Graham's Law and unknown noble gas

An unknown noble gas effuses at a rate that is only 0.219 times that of He at the same temperature. What is the identity of the unknown gas?
Kr
##### Explanation
The rates of effusion of two substances are governed by Graham's Law:

runknown/rHe = SQRTMHe/Munknown

where r is the rate and M is the molar mass. Using MHe = 4.002 g/mol the molar mass of the unknown is 83.9 g/mol.
4. #### Effusion through a pinhole

A container of argon gas a 870 torr is evacuated through a tiny pinhole. After 6.80 s, the pressure drops to 100 torr. The container is subsequently completely evacuated and refilled with an unknown gas to a pressure of 870 torr. Once again, the container is evacuated through the same pinhole, but this time it takes 9.63 s for the pressure to drop to 100 torr. Determine the molar mass of the unknown gas.
80.1 g/mol
##### Explanation
The rates of effusion of two substances are governed by Graham's Law:

runknown/rAr = SQRTMAr/Munknown

where r is the rate and M is the molar mass. The rate of the unknown is 770 torr/9.63 s and the rate of Ar is 770 torr / 6.80 s Therefore

runknown/rAr = 0.706

using MAr = 39.948 g/mol the molar mass of the unknown is 80.1 g/mol.
5. #### Average Speed as a Function of Temperature

Consider the distribution of speeds for oxygen molecules at T = 5000 K shown below. Sketch on the same figure the distribution of speed if the temperature is increased to T=10000 K.

##### Explanation
An increase in temperature means an increase in the average kinetic energy of the molecules and thus an increase in the average speed.
6. #### Molecular Collisions

Which of the following statements are true in relation to a real gas?
• Collisions between gas molecules result in the exchange of energy
• Elastic collisions between gas molecules transfer a fixed amount of energy
• All molecules in a gas have the same kinetic energy
• Molecular collisions occur frequently in gases at atmospheric pressure
• True
• False
• False
• True
##### Explanation
• Energy can be transferred between molecules during collisions
• The energy transferred during an elastic collision will depend on the geometry of the reaction (impact parameter)
• Although the molecules in a sample of gas have an average kintetic energy each molecule is moving at a different speed
• Moving molecules in a gas collide frequently with other molecules even at atmospheric pressure
7. #### Deviations from ideal behavior

A plot of PV/RT for 1 mol of nitrogen gas at three different temperatures is shown below. Which of the following statements are true keeping in mind the van der Waals equation:

[P + (n2a)/V2][V - nb] = nRT

• The curve for T = 1000 K shows a small positive deviation from the ideal behavior at pressure below 300 atm which indicates that attraction between individual molecules is more significant at higher temperatures
• The van der Waals constant b is a measure of the attractive force between gas molecules
• Gas molecules have a finite volume which results in a negative deviation from ideal behavoir at high pressures
• All gases deviate to some degree from the ideal gas law
• The curve at T = 1000 K best approximates ideal gas behavoir
• False
• False
• False
• True
• Ture
##### Explanation
• Attractive interactions between constituent gas molecules should become more significant at lower temperatures
• The constant b is a correction factor for the finite volume of the gas molecules
• The finite volume correction leads to a positive deviation from ideal behavoir
• The ideal gas equation is an approximation. Real molecules do have finite volumes and they do attract one another
• As temperature is increased the behavoir of a gas more nearly approaches that of an ideal gas
8. #### van der Waals equation

Calculate the pressure that H2 will exert at 89.0 oC if 36.75 mol of the gas are enclosed in a 26.0 L container, assuming both ideal and real behavoir of the gas (a = 0.244 L2atm/mol2, b = 0.0266 L/mol).
Pideal = 42.0 atm, Preal = 43.2 atm
##### Explanation
For an ideal gas

P = nRT/V

For a real gas

P = nRT/(V - nb) - (n2a/V2)

## Topic: Intermolecular Forces

1. #### Intermolecular Interactions

Consider each of the following statements regarding intermolecular interactions. Determin whether each statement is true or false.
• SiCl4 is more readily polarized than SiH4
• Oxygen is expected to be more polarizable than tellurium
• Hydrogen bonding occurs between a hydrogen atom of a nonpolar bond and a highly electronegative atom on another molecule
• Polarizability relates the ease with which a molecule can aquire an instantaneous electric dipole moment
• Liquid CO has a slightly lower boiling point than liquid N2 due to dipole-dipole interactions between CO molecules
• True
• False
• False
• True
• False
##### Explanation
• Since the Cl atoms have a larger distribution of electrons, SiCl4 will be more readily polarizable
• Since Te has a larger distribution of electrons, SiCl4 will be more readily polarizable
• A hydrogen involved in hydrogen bonding should form part of a polar bond
• Polarizability relatesto the distortion of the charge cloud of an atom or molecule which results in an instantaneous electric dipole moment
• Since the CO molecule has a pemanent electric dipole moment the dipole-dipole interaction will lead to a greater boiling point (larger intermolecular interactions) and in N2, which is non-polar
2. #### Relative strength of intermolecular forces and boiling point

Based upon the strength of the intermolecular attraction and relative molecular mass, rank the following in order of increasing boiling point.
• HBr
• HCl
• HF
• HI
From lowest to highest boiling point: HCL (-85 oC), HBr (-67 oC), HI (-35 oC), HF (20oC)
##### Explanation
The boiling point increases going from HCl to HBr to Hi due to relative molar mass. The boiling point of HF is the highest since HF has the strongest hydrogen bonding
3. #### Vapor pressure curve

The Handbook of Chemistry and Physics provides the following equation for calculating the vapor pressure of solids and liquids:

log10 P = b - (0.05223a/T)

where P is the vapor pressure in mm Hg at the absolute temperature T and a and b are constants for a given substance. Using Excel, plot the vapor pressure curve for solid arsenic (a = 133000, b = 10.8). From your plot, determine the vapor pressure of solid arsenic at 550 oC.
229 mm Hg
##### Explanation
The excel sreadsheet can be found here. Temperature is the independent variable (x axis) and vapor pressure is the dependent variable (y axis). Since T is in absolute temperature, you need to convert to Kelvin scale. This is done in the spread sheet.t
4. #### Vapor pressure primer

Complete each of the following statements pertaining to equilibrium vapor pressure. Fill in the blank space using the words: increase, decrease, or not change.
• If the intermolecular forces increase in going from one liquid to another, the equilibrium vapor pressure will ...
• The equilibrium vapor pressure will ... when the temperature of the liquid is lowered.
• The vapor pressure will ... if the volume of the liquid increases.
• The normal boiling point will ... if the intermolecular forces get larger.
• If a super cooled liquid freezes at a constant temperature, its vapor pressure will ...
• Decrease
• Decrease
• Not Change
• Increase
• Decrease
##### Explanation
• With larger intermolecular forces, it is more difficult for liquid molecules to enter the vapor phase.
• At lower temperatures, the average kintetic energy of the molecules of the liquid is decreased. Therefore, it is harder for the liquid molecules to enter the vapor phase.
• The volume of the liquid has no impact on the vapor pressure.
• If the intermolecular forces are increased, then more energy is required to moce liquid molecules to the vapor phase. The temperature where boiling occurs will therefore decrease.
• Since the phase is changing from liquid to solid, the intermolecular forces are increased and therefore the vapor pressure will decrease.
5. #### Pressure cooker

A pressure cooker contains 1000 g of water. The safety valve on the cooker is set to relieve the inside pressure if it exceeds 3.4 atm. Use the vapor pressure curve for water shown below to estimate the maximum temperature attainable inside the pressure cooker.

138 oC
##### Explanation
Since the pressure inside the cooker is above atmospheric pressure, the molecules of the liquid will need a higher average kinetic energy to escape the gas phase. This requires a higher temperature for boiling.

## Topic: Phases of Matter

1. #### Phase changes of matter

Below are statements pertaining to the phase changes of matter. Decide if each statement is true or false.
• There can be only one liquid phase in a pure substance
• If a glass of water at 20 oC is heated, its temperature will increase
• Liquifying nitrogen releases heat
• Ammonia can be liquified at room temperature by compressing it at high pressures
• If a glass of ice water is heated, the temperature will decrease while the ice melts because melting is an endothermic process
• True
• True
• True
• True
• False
##### Explanation
• Gases and liquids of pure substances have only one phase. Solids can obtain several phases due to the different orderings of the solid lattice
• Since the temperature of the water is not at a phase transition, the heat added to the water will go to increasing the average kinetic energy, and temperature, of the liquid
• The phase change from gas to solid is an exothermic process
• As long as the temperature of the gas is below the critical temperature, high pressure can be used to liquify the gas. The critical temperature for ammonia is 405.6 K
• The temperature will remain the same. While the ice is melting the heat applied is going into the phase change and not to increase or decrease the average kinetic energy of the liquid
2. #### Naming of phase transitions

The digram below represents the energy changes accompanying phase changes between the three states of matter. Match the appropriate letter on the diagram with the name associated with the energy change.

• T - deposition
• U - melting
• W - condensation
• X - vaporization
• Y - freezing
• Z - sublimation
##### Explanation
• deposition - gas to solid phase transition (exothermic)
• melting - solid to liquid phase transition (endothermic)
• condensation - gas to liquid phase transition (exothermic)
• vaporization - liquid to gas phase transition (endothermic)
• freezing - liquid to solid phase transition (exothermic)
• sublimation - solid to gas phase transition (endothermic)
3. #### Comparison of energy changes accompanying phase transitions

Based on the phase diagram below, determine whether each statement is true or false.

• The enthalpy change in process W can be calculated from the heat of fusion
• Process T and Z represent the same magnitude of change in energy
• The heat of fusion can be calculated if the enthalpy changes in process W and X are known
• The heat of sublimation equals U + X
• T is an endothermic process
• False
• True
• False
• True
• False
##### Explanation
• The enthalpy change W corresponds to the heat of vaporization
• T and Z represent phase changes between solid and liquid. The magnitude of this energy change is the same (but the sign is opposite
• W and X represent phase changes between gas and liquid phases. Therefore, they have no realtion to the heat of fusion (phase change between solid and liquid)
• Adding the two vectors U and X will produce the vector Z, which represents the heat of sublimation (solid to gas phase transition)
• The phase transition from gas to soild is an exothermic process (heat is released)
4. #### Cooking an egg

A student in East Lansing boils a raw egg in water for three minutes and finds that the egg is cooked to perfection. The student then travels to the top of Mt. McKinley (elevation 6193 m) and gain boils the a raw egg for three minutes. Is the egg cooked to perfection?
No
##### Explanation
At higher elevations, the atmospheric pressure is lower than in East Lansing. Therefore, the water will boil at a lower temperature.
5. #### Enthalpy change from steam to ice

Calculate the enthalpy change upon converting 9.00 mol of steam at 100 oC to ice at -19 oC under a constant pressure of 1 atm. The molecular weight of water is 18.0 g/mol. The following thermodynamic data may be useful:
DHvap = 40.67 kJ/mol
DHfus = 6.01 kJ/mol
Tb = 100.0 oC
Tf = 0.0 oC
Specific heat of H2O(g) = 1.84 J/g-K
Specific heat of H2O(l) = 4.18 J/g-K
Specific heat of H2O(s) = 2.09 J/g-K
DH = -498 kJ
##### Explanation
A detailed explanation can be found in Sample Exercise 11.4 in the textbook on page 457. The sign of the enthalpy change is negative since heat is released in the phase change from gas to solid.
6. #### Phase diagram of carbon dioxide

The phase diagram of CO2 is given below.

Which of the following statements are true?
• The sublimation curve for liquid carbon dioxide is given by line segment EC
• The critical point is labelled C
• F is a triple point
• At atmospheric pressure, carbon dioxide will sublime. This pahse change is represented by B
• The density of solid carbon dioxide is greater than tate of vapor carbon dioxide at the same temperautre
• False
• True
• False
• True
• True
##### Explanation
• The sublimation curve divides the solid and gaseous phases of carbon dioxide
• The critical point deinfes the end of the vapor pressure curve
• The triple point is at E
• The process of sublimation is a phase change from solid to gas and is represented by B
• Since the sublimation curve has a positive slope the density of solid carbon dioxide is greater than that of the vapor at a given temperature.
7. #### Phase diagram of sulfur

The phase diagram of sulfur is given below.

Determine whether each of the following statements pertaining to the phase digram above is true or false.
• There are three triple points in the phase diagram
• Elemental sulfur can exist in two different liquid phases
• Rhombic sulfur can melt upon heating at atmospheric pressure
• Monoclinic and rhombic represent two different solid crystal structures of sulfur
• The density of monoclinic sulfur is greater than the density of rhomic sulfur at the same temperature
• True
• False
• False
• True
• False
##### Explanation
• It is possible for elements and compounds to have multiple triple points; regions where three phases coexist
• A pure substance can only have one liquid and one gaseous phase
• Upon heating at atmospheric pressures, rhombic sulfur will undergo a crystal structure change to monoclinic sulfur before melting
• Sulfur has two different crystal strucutres
• Since the phase transition line between rhombic and monoclinic sulfur has a positive slope, the rhombic sulfur will have a higher density at the same temperature

## Topic: Solid Structures

1. #### Crystal Stucture primer

Consider the following statements partaining to the crystal structure of solids. Determine whether each statement is true or false.
• The largest volume or a crystal solid which can be used to build up the entire solid is called a unit cell
• In a cubic crystal system all the sides are of equal dimension
• In a face centered cubic unit there are four atoms per cell
• A crystalline solid consists of a regular array of atoms, molecules, or ions
• Amorphous solids have well defined faces and shapes
• False
• True
• True
• True
• False
##### Explanation
• The unit cell is the smallest volume of a crystal structure that can be used to build the entire solid
• A cubic crystal has all the sides of equal length and all angles at 90 degrees
• An fcc structure has 1/8 of an atom at 8 corners and 1/2 of an atom on 6 faces, giving 4 atoms per unit cell
• Atoms, ions, or molecules are ordered in well defined arrangements in a crystalline solid
• An amorphous solid is a solid whose particles have no orderly structure
2. #### Close packing structures

Consider whether eeach of the follwoing statements on the crystal structure of solids is true or false
• The most efficient means of arranging a layer of equally-sized spheres is close packing
• Hexagonal close packing has the repetitive pattern ABCABCABC
• The coordination number for each sphere in a close-packed structure is 18
• A cubic close packed crystalline solid is 26% open space
• The ions of NaCl form a hexagonal close packed structure
• True
• False
• False
• True
• False
##### Explanation
• Close packing, either using hexagonal or cubic arrangment, is the most efficienct arrangement of equally-sized spheres.
• The hexagonal close packed arrangement repeats as ABABAB
• The number of nearest neighbors to an atom in a close packed crystal arrangement is 12
• The volume of close packed structure are occupied by 74% atoms and 26% empty space
• NaCl forms and interleavened fcc structure
3. #### Atmoic weight of element with bcc structure

An element crystallizes in a body-centered cubic lattice. The edge of the unit is 2.86 angstroms and the density of the crystal is 7.92 g/cm3. Calculate the molecular weight of the element.
55.8 g/mol
##### Explanation
The volume of the cell is

(2.86e-8 cm)3 = 2.34e-23 cm3

The mass of the unit cell is density x volume

7.92 g/cm3 * 2.34e-23 cm3 = 1.85e-22 g

There are 2 atoms per unit cell for BCC, so the mass per atom is

1.85e-22 g/2 atoms = 9.26e-23 g/atom

To get g/mol (Atomic weight) multiply the mass/atom by Avogadro's number
4. #### Density of nickel oxide

Nickel oxide crystallizes in the NaCl type of crystal structure. The length of the unit cell of NiO is 4.18 angstroms. Calculate the density of NiO
6.79 g/cm3
##### Explanation
To determine density, need to calculate mass per unit cell and volume per unit cell for NiO

Volume per unit cell = (4.18e-8 cm)^3 = 7.30e-23 cm3

Since there are 4 atoms of Ni and 4 atoms of O per unit cell:

Mass per mole of unit cell = 4*(15.9994 g/mol) + 4*(58.6934g/mol) = 298.8 g/mol unit cell
Mass per unit cell = (298.8 g/mol unit cell)/(6.022e+23 unit cell/mol) = 4.92e-22 g/unit cell
Density = (4.92e-22 g/unit cell)/(7.30e-23 cm3/unit cell) = 6.79 g/cm3

## Topic: Bonding in Solids

1. #### Bonding in solids

Determine whether each statement below pertaining to bonding in solids is true or false.
• Diamond is an example of a metallic solid
• A covalent network solid has excellent thermal and electric conductivity
• Molecular solids are formed through colvalent bonding between constituent molecules
• Electrostatic interactions are responsible for confining the positive and negative ions in an ionic solid
• The hard nature of quartz is due to the atoms being connected in a network of covalent bonds
• False
• False
• False
• True
• True
##### Explanation
• Diamond is an example of a covalent solid. Metallic solids are made by metal bonding of metallic elements
• Covalent network solids typically have poor eletrical and thermal conduction
• Molecular solids are formed through non-bonding interactions between atomes or molecules
• Ionic solids are formed from positive and negative ions and their strong electrostatic interactions
• Quartz is a covalent network soild and the covalent bonds provide added strength to the solid

## Topic: Previous Exams (Spring 2010)

1. #### Vapor Pressure

The vapor pressure curve of ammonia is shown below. What is the normal boiling point of ammonia? If the temperature were decreased to 205 K what is the pressure at which ammonia will boil?
~238 K, 100 torr.
##### Explanation
A liquid will boil when the vapor pressure is equal to the external pressure. The normal boiling point is the temperature at which the vapor pressure of a liquid is equal to 1 atmosphere, ~238 K. At 205 K the liquid will boil at 100 torr.
2. #### Gaseous Effusion

In class we performed a reaction the following reaction:

NH3 (g) + HCl (g) → NH4Cl (s)

A few drops of NH3 was placed five inches from a few drops of HCl. Was the ammonium chloride solid formed closer to the drops of NH3 or the drops of HCl and why?
The Nh4Cl was formed closer to the drops of HCl
##### Explanation
The solid is formed closer to the HCl since HCl has a higher molecular mass than NH3 and thus effuses slower according to Graham's Law of Diffusion
3. #### Solid Density

An element crystallizes in a body centered cubic lattice. The edge of the unit cell is 2.86 Å and the density of the crystal is 7.92 g/cm3. Calculate the weight (in amu) of the element.
55.8 amu (Fe)
##### Explanation
A body-centered cubic unit cell has 8 atoms at the corners and one atom in the center for a total of 2 atoms in the unit cell. Assuming the mass of each atom is x amu
density = mass / volume = 2 * x amu / (2.86x10-8 g/cm3) * 1 g / 6.022x1023 amu = 7.92 g/cm3 x = 55.8 amu
4. #### Partial Pressures

Gaseous iodine pentaflouride, IF5, can be prepared in the following reaction:

I2 (s) + 5 F2(g) → 2 IF5 (g)

A 5 L flask containing 20 g of I2 is filled with 10 g of F2 and the reaction proceeds until one of the reactants is completely consumed. After the reaction, the temperature is 125 oC. What is the partial pressure of IF5 in the flask and what is the mole fraction of IF5 in the flask.
PIF5 = 0.687 atm, XIF5 = 0.801
##### Explanation
before the reaction:
mol I2 = 20 g * 1 mol / 2(126.9) g = 0.0788 mol
mol F2 = 10 g * 1 mol / 2(19.0) g = 0.2631 mol For every 1 mol of I2 consumed, 5 moles of F2 are consumed. F2 is the limiting reagent and will be completely consumed in the reaction. After reaction
mol F2 = 0
mol I2 = 0.0788 - 0.2631 mol * 1 mol I2/5 mol F2 = 0.02618 mol
mol IF5 = 0.2631 mol F2 * 2 mol IF5/5 mol F2 = 0.1052 mol
PIF5 = nRT/V = (0.1052 mol)(0.08206 Latm/Kmol)(398K) / 5L = 0.687 atm
XIF5 = 0.1052 / (0.1052 + 0.02618) = 0.801
5. #### Phase Diagram

Give the phase diagram of carbon below answer the following questions.
1. What is the stable form of carbon at 0oC and 1 atm?
2. How many triple points are on the phase diagram shown ?
3. Under what conditions, if any, can diamond be transformed directly to liquid?
4. Under what conditions, if any, can diamond be transformed into a vapor?
5. Is the phase transition marked by the arrow condensation, if not, what is the transition called?

1. graphite
2. 2
3. T ~ 4000 K, P > 1010 Pa
4. none
5. no

#### Explanation

1. graphite is the stable phase at these conditions
2. triple points are where three phases can coexist and two are present in the diagram, 1- graphite, vapor and liquid and 2 - graphite,diamond and liquid
3. read off the temperature and pressure from the diamond/liquid equilibrium line
4. can't go directly from diamond into a vapor
5. the phase transition marked is melting, solid to a liquid
6. #### Real Gases

The figure below plots the ratio PV/RT for nitrogen gas at three different temperatures
1. Why to real gases deviate from ideal behavior?
2. For the nitrogen curve at 200 K, why is there a negative deviation from an ideal gas between 100 and 300 atm?
3. For the nitrogen curve at 200 K, why is there a positive deviation from an ideal gas above 600 atm
4. Which curve best approximates an ideal gas?
1. gas molecules have finite volumes and experience attractive interactions
2. attractive interactions
3. finite volumes
4. 1000 K
##### Explanation
1. real gases deviate from ideal behavoir because real gas molecules have a finite volume and experience attractive interactions
2. negative deviations from ideal gas behavior at moderate temperatures are caused by attractive interactions that decrease the total pressure of the gas
3. positive deviations from ideal gas behavoir at moderate temperatures are cause by the finite volume of the gas molecules resuling a smaller total volume for the gas container
4. the curve at 1000K is closest to the ideal gas law across a range of pressures.
7. #### Phase Changes

In hot climates, water in a canvas bag can be cooled by evaporating water of the surface of the bag. How many grams of water inside the bag can be cooled from 35 oC to 20oC by the evaporation of 30 g of water from the surface of the bag? The heat of vaporization of water in this temperature range is 43.2 kJ/mol. The specific heat of water is 4.18 J/gK.
1.15x103 g
##### Explanation
Heat removed from the evaporation of 30 g of water
(30 g)(1 mol/18g)(43.2 kJ/mol) = 72 kJ
cooling a mass of water from 35 oC to 20 oC
(x g)(15 k)(4.18 J/gK) = 72000 J x = 1.15x103
8. #### Vapor Pressure

A sample of liquid is placed in a beaker and its vapor pressure is measured. Please complete the following statements with either; not change, increase, decrease
1. If the volume of the liquid is increased the vapor pressure will __________
2. If the surface area of the liquid is decreased the vapor pressure will ___________
3. If the intermolecular attractive forces of the liquid increase the vapor pressure will ___________
4. If the temperature is increased the vapor pressure will ____________
1. not change
2. not change
3. decrease
4. increase
##### Explanation
1. The vapor pressure does not depend on the volume of liquid
2. The vapor pressure does not depend on the surface area of the liquir
3. If the intermolecular forces are increased then the molecules of the liquid experience a stronger attraction to each other making it more difficult to escape into the gas phase
4. If the temperature is increased then the molecules in the liquid have a higher energy and a larger percentage of them will be able to escape the liquid
9. #### Modern Materials

Determine whether each of the following statements are true or false.
1. Semiconductors have a larger band gap than insulators.
2. Metals have delocalized electrons.
3. A superconducting has zero resistance to the flow of electricity
1. False
2. True
3. True

## Topic: Previous Exams (Spring 2011)

1. #### Gaseous Reactions

The first step in the purification of crude Ni is the reaction with carbon monoxide gas, CO, according to the following reaction:

Ni (s) + 4 CO (g) → Ni(CO)4 (g)

Since Ni(CO)4 is incredibly toxic, potential exposure to it should be limited. If you started with 1 metric ton of solid Ni (1 metric ton = 1000 kg) and wanted to ensure that only 1 g of Ni(CO)4 is created, what volume of CO is required, assuming the reaction is carried out at 300 K and 1 atm?
0.58 L
##### Explanation
MW Ni(CO)4 = 58.69 g/mol + 4*12g/mol + 4*16g/mol = 170.69 g/mol
mol Ni(CO)4 = 1 g * 1mol/170.69 g = 0.00586 mol
mol CO = mol Ni(CO)4 * 4 mol CO/mol Ni(CO)4 = 0.0234 mol CO
V = nRT/P = (0.0234 mol)(0.08206 Latm/Kmol)(300K)/(1 atm) = 0.577 L
2. #### Real gas behavior

Determine whether each of the following statements is true or false. Write the entire word TRUE or FALSE or else no credit will be given.
• The lower the temperature, the closer a real gas will approximate ideal behavior.
• Real gases deviate from ideal behavior because real gas molecules have a finite size and significant dipole moments
• The van der Waals constant b is a measure of the attractive force between gas molecules.
• The ideal gas law is an accurate description of gases above 1000 atm.
• False
• False
• False
• False
##### Explanation
• Due to attractive interactions the lower the temperature the larger the deviations from ideal behavior
• Real gases deviate from ideal behavior due to attactive interactions between gas molecules and the fact that real gas molecules occupy volume
• b measures the volume occupied by a gas molecule
• Real gases deviate from ideal behavior at high pressures
3. #### Partial Pressures

A heliox deep-sea diving breathing mixture contains 2.0 g of oxygen to every 98.0 g of helium. What is the partial pressure of oxygen when this mixture is delivered at a total pressure of 8.5 atm?
PO2 = 0.0216 atm
##### Explanation
mol O2 = 2 g O2 * mol/32 g = 0.0625 mol
mol He = 98.0 g He * mol/4gHe = 24.5 mol He
nT = 0.0625 mol + 24.5 mol = 24.5625 mol
XO2 = 0.0625 mol / 24.5625 mol = 0.00254
PO2 = XO2PT = (0.00254)(8.5atm) = 0.0216 atm
4. #### Solids

Determine whether each of the following statements is true or false. Write the entire word TRUE or FALSE or else no credit will be given.
• Molecular solids are formed through covalent bonding between constituent molecules
• At room temperature Fe crystallizes into a body-centered cubic structure, and is called alpha-iron. Above 912 oC, a phase transition to gamma-iron occurs and changes the crystal structure from body-centered to face-centered cubic. Assuming that the volume of the unit cell remains constant, gamma-iron is more dense than alpha-iron.
• Diamond is a covalent solid
• Hexagonal close packing is represented by the repetitive pattern ABCABC.
• The smallest volume of a crystalline solid which can be used to build up the entire solid is called a unit cell
• False
• True
• True
• False
• True
##### Explanation
• Molecular solids are held together with weaker intermolecular forces such as dipole-dipole, hydrogen bonding, ...
• If the volume of the cell remains constant, gamma iron has more atoms in its unit cell
• Diamond is held together by covalent bonds
• Hexagonal close packing is ABABAB
• The smallest repeatable unit in a crystalline solid is the unit cell
5. #### Gaseous Diffusion

Two gaseous chemicals A and B are separated by a tube 10 cm long. When the two gases meet, a gas phase reaction occurs between A and B forming a solid black compound that is deposited close to the location of the solid A. Which gas, A or B, has the lower molecular weight and why
B
##### Explanation
The relative rates of diffusion can be approximated by Graham's Law. The gas molecules of B travel further thus have the higher molecular speed and lower molecular weight.
6. #### Pressure

You attach a sample of gas contained in a glass jar to an open-ended mercury manometer. The pressure within the glass jar is able to support a 220 mm column of mercury in the open-ended side of the manometer. What is the pressure of the gas inside the container assuming atmospheric pressure is 1 atm?
1.289 atm
##### Explanation
pressure from the mercury column = PHg = 220 mm Hg * 1atm/760mmHg = 0.289 atm
Ptotal = PHg + Patm = 1.289 atm
7. #### Vapor pressure

The vapor pressure as a function of temperature for a generic gas is described by the following equation:

ln(P) = -2000*(1/T) + 10.

• What is the enthalpy of vaporization of the gas?
• 16.6 kJ
##### Explanation
• -ΔHvap/R = -2000 K
Δ = (2000 K)(8.3145 J/Kmol) = 16,628 J/mol = 16.6 KJ/mol
8. #### CO2 Phase diagram

Concerning the phase diagram of CO2:
• What is the normal boiling point of CO2?
• How many triple points are shown in the diagram?
• Modern dry cleaners have moved to using supercritical CO2 as a cleaning solvent. What is the minimum temperature required for this cleaning method to be successful?
• If a sample of solid CO2 at 2 atm and -78.5 oC is warmed to 0 oC what phase transition occurs?
• At what temperature and pressure can three phases of CO2 coexist?