Soquids Answers
M/C
58. C
68. C
27. E
54. A
21. A
49. C
50. B
51. D
26. C
5. A
6. B
7. B
13. A
15. D
16. C
25. C
39. C
F/R
1977 D
The state of aggregation of solids can be described as belonging to the following four
types:
(1) ionic
(3) covalent network
(2) metallic
(4) molecular
For each of these types of solids, indicate the kinds of particles that occupy the
lattice points and identify forces among these particles. How could each type of solid
be identified in the laboratory?
Answer:
Particles
Binding Forces
Experimental Identification
ionic
+ & - ions
electrostatic attraction
conductivity of fused salt
metallic
+ ions
electrostatic attraction
conductivity of the solid
between ions and
electrons
covalent network
atoms
covalent bonds
high melting pt., extreme hardness, etc
molecular
molecules
van der Waals
low melting pt., non-conductivity of fuse
salt, etc.
1978 D
The freezing point and electrical conductivities of three aqueous solutions are given
below.
Solution
Freezing
Electrical
(0.010 molal)
Point
Conductivity
sucrose
-0.0186C
almost zero
formic acid
-0.0213C
low
sodium formate
-0.0361C
high
Explain the relationship between the freezing point and electrical conductivity for
each of the solutions above. Account for the differences in the freezing points among
the three solutions.
Answer:
sucrose
a non electrolyte
0.010 mol/kg lowers freezing point 0.0186C
formic acid
a weak electrolyte; low conductance as a result of low ion concentration
mtotal > 0.010 molal due to partial ionization and Tf somewhat greater than 0.0186C
sodium formate
a salt and strong electrolyte
approximately 100% dissociation into ions, mtotal approaching 0.02 molal
1984 C
Give a scientific explanation for the following observations. Use equations or
diagrams if they are relevant.
(a) It takes longer to cook an egg until it is hard-boiled in Denver (altitude 1 mile above
sea level) than it does in New York City (near sea level).
(b) Burn coal containing a significant amount of sulfur leads to ôacid rain.ö
(c) Perspiring is a mechanism for cooling the body.
(d) The addition of antifreeze to water in a radiator decreases the likelihood that the
liquid in the radiator will either freeze or boil.
Answer:
(a) Water boils at a lower temperature in Denver than in NYC because the atmospheric
pressure is less at high altitudes. At a lower temperature, the cooking process is
slower, so the time to prepare a hard-boiled egg is longer.
(b) S + O2  SO2 (as coal is burned)
SO2 + H2O  H2SO3 (in the atmosphere)
H2SO3 is sulfurous acid.
(c) Vaporization or evaporation of sweat from the skin is an endothermic process and
takes heat from the body and so cool the skin.
(d) Colligative properties, which depend on the number of particles present, are
involved.
Solute (the antifreeze) causes the lowering of the vapor pressure of the solvent.
When the vapor pressure of the solvent is lowered, the freezing point is lowered and
the boiling point is raised.
1987 D
In 1884 the Swedish chemist Svante Arrhenius proposed that salts dissociate into
two or more separate, independent, ionic fragments when they dissolve in water.
(a) Give one piece of experimental evidence that more than 1 mole of particles is formed
when 1 mole of a salt dissolves in water.
(b) Give one piece of experimental evidence that the particles formed when a salt
dissolves in water are charged.
(c) Explain why the heat of neutralization is always the same when 1 mole of any
monoprotic strong acid reacts with enough strong base to form a neutral solution.
(d) Explain why hydrogen chloride, HCl, dissociated when it dissolves in water but not
when it dissolves in benzene.
Answer:
(a) The freezing point depression (or any colligative effect) that occurs when a mole of a
salt is dissolved is greater than when a mole of a non-dissociated substance is
dissolved. (The greater the number of solute particles the greater the colligative
effect.)
(b) The solution of a salt conducts electricity.
(c) Every neutralization between a strong acid and a strong base involves the same
reaction:
H+(aq) + OH-(aq)  H2O
since both the strong acid and the strong base are completely dissociated. Spectator
ions have no appreciable effect.
(d) Because of the polar nature of water, it is capable of solvating the ions that result
from the dissociation, whereas the nonpolar benzene interacts very weakly with these
ions. OR
Because of the greater dielectric constant of water, it is better able to separate the
ions.
1988 D
The normal boiling and freezing points of argon are 87.3 K and 84.0 K, respectively.
The triple point is at 82.7 K and 0.68 atmosphere.
(a) Use the data above to draw a phase diagram for argon. Label the axes and label the
regions in which the solid, liquid and gas phases are stable. On the phase diagram,
show the position of the normal boiling point.
(b) Describe any changes that can be observed in a sample of solid argon when the
temperature is increased from 40 K to 160 K at a constant pressure of 0.50
atmosphere.
(c) Describe any changes that can be observed in a sample of liquid argon when the
pressure is reduced from 10 atmospheres to 1 atmosphere at a constant temperature
of 100 K, which is well below the critical temperature.
(d) Does the liquid phase of argon have a density greater than, equal to, or less than the
density of the solid phase? Explain your answer, using information given in the
introduction to this question.
Answer:
(a)
(b) The argon sublimes.
(c) The argon vaporizes.
(d) The liquid phase is less dense than the solid
phase. Since the freezing point of argon is higher
than the triple point temperature, the solid-liquid
equilibrium line slopes to the right with
increasing pressure. Thus, if a sample of liquid
argon is compressed (pressure increased) at
constant temperature, the liquid becomes a solid.
Because increasing pressure favors the denser
phase, solid argon must be the denser phase.
Ö
Ö
1989 D
Consider three unlabeled bottles, each contain small pieces of one of the following
metals.
- Magnesium
- Sodium
- Silver
The following reagents are used for identifying the metals.
- Pure water
- A solution of 1.0 molar HCl
- A solution of concentrated HNO3
(a) Which metal can be easily identified because it is much softer than the other two?
Describe a chemical test that distinguishes this metal from the other two, using only
one of the reagents above. Write a balanced chemical equation for the reaction that
occurs.
(b) One of the other two metals reacts readily with the HCl solution. Identify the metal
and write the balanced chemical equation for the reaction that occurs when this metal
is added to the HCl solution. Use the table of standard reduction potentials (attached)
to account for the fact that this metal reacts with HCl while the other does not.
(c) The one remaining metal reacts with the concentrated HNO3 solution. Write a
balanced chemical equation for the reaction that occurs.
(d) The solution obtained in (c) is diluted and a few drops of 1 M HCl is added. Describe
what would be observed. Write a balanced chemical equation for the reaction that
occurs.
Answer:
(a) Sodium is softest of the three.
Na added to water  gas and base
Ö
no rmal
B .P .
Na + H2O  H2 + NaOH
(b) Mg reacts with HCl. Mg + 2 H+  Mg2+ + H2
Reduction potentials, E: Mg = -2.37v; Ag = + 0.80v. Mg, not Ag, reacts with HCl
(c) Ag + 4 H+ + NO3-  3 Ag+ + NO + 2 H2O OR Ag + 2 H+ + NO3-  Ag+ + NO2 +
H2O
(d) A white precipitate forms: Ag+ + Cl-  AgCl(s)
1994 D
For each of the following, use appropriate chemical principles to explain the
observation.
(a) Sodium chloride may be spread on an icy sidewalk in order to melt the ice;
equimolar amounts of calcium chloride are even more effective.
(b) At room temperature, NH3 is a gas and H2O is a liquid, even though NH3 has a molar
mass of 17 grams and H2O has a molar mass of 18 grams.
(c) C (graphite) is used as a lubricant, whereas C (diamond) is used as an abrasive.
(d) Pouring vinegar onto the white residue inside a kettle used for boiling water results
in a fizzing/bubbling phenomenon.
Answer:
(a) NaCl(s)  Na+(aq) + Cl-(aq)
CaCl2(s)  Ca+(aq) + 2 Cl-(aq)
The freezing point of an aqueous solution is lower than the freezing point of water. A
higher molality of a solution lowers the freezing point more and an equimolar
amount of the two solids gives a larger molal solution from the calcium chloride as
illustrated by the above equations.
(b) Water is more polar than ammonia creating stronger attractions (IMF) between
molecules and making it a liquid.
(c) Diamond, the hardest naturally
occurring substance, has each
carbon atom surrounded by a
tetrahedral arrangement of other
carbon atoms (see drawing). The
network solid structure is
stabilized by covalent bonds,
formed by the overlap of sp3
hybridized
carbon
atomic
orbitals. A diamond has uniform
very strong bonds in all
directions in the crystal.
Graphite has a different kind of bonding based on layers of
carbon atoms arranged in fused six-member rings (see drawing). Each carbon atom
in a layer is surrounded by three other carbons in a trigonal planar arrangement with
120 bond angles. The slipperiness is caused by noting that graphite has very strong
bonds within layers but little bonding between the layers which allows the layers to
slide past one another quite readily.
(d) Calcium and magnesium carbonates are left behind from the evaporation of hard
water. These carbonates decompose and release carbon dioxide gas when reacted
with the acetic acid in the vinegar.
CaCO3(s) + MgCO3(s) + CH3COOH(aq)  Ca2+(aq) + Mg2+(aq) + H2O(l) + CO2(g) +
CH3COO-(aq)
1995 D
The phase diagram for a pure substance is shown above. Use this diagram and your
knowledge about changes of phase to answer the following questions.
(a) What does point V represent? What characteristics are specific to the system only at
point V?.
(b) What does each point on the curve between V and W represent?
(c) Describe the changes that the system undergoes as the temperature slowly increases
from X to Y to Z at 1.0 atmosphere.
(d) In a solid-liquid mixture of this substance, will the solid float or sink? Explain.
Answer:
(a) Triple point. All three states of the substance coexist (equilibrium); the solid and the
liquid have identical vapor pressures.
(b) Curve VW represents the equilibrium between the liquid and its vapor. Along this
line the liquid will be boiling. The points represent the vapor pressure of the liquid as
a function of temperature.
(c) At point X the substance is a solid, as its temperature increases (at constant pressure), at
point Y the solid is in equilibrium with its vapor and will sublime. From point Y to Z
it exist only as a vapor.
(d) Sink. A positive slope of the solid-liquid line indicates that the solid is denser than its
liquid and, therefore, will sink.
1998 D
Answer each of the following using appropriate chemical principles.
(a) Why does it take longer to cook an egg in boiling water at high altitude than it does
at sea level?
Answer
(a) At a high altitude the pressure is lower and the boiling point of water is lower. Since
the temperature of the boiling water is less, it takes longer to cook the egg.
2001 D Required
Answer the questions below that relate to the five aqueous solutions at 25C shown
above.
(a) Which solution has the highest boiling point? Explain.
(b) Which solution has the highest pH? Explain.
(c) Identify a pair of the solutions that would produce a precipitate when mixed together.
Write the formula of the precipitate.
(d) Which solution could be used to oxidize the Cl–(aq) ion? Identify the product of the
oxidation.
(e) Which solution would be the least effective conductor of electricity? Explain.
Answer:
(a) solution 1, Pb(NO3)2. This compound will dissociate into three ions with the highest
total particle molality. The greater the molality, the higher the boiling point.
Solutions 2, 3, and 5 will produce two ions while solution 4 is molecular.
(b) solution 5, KC2H3O2. The salt of a weak acid (in this case, acetic acid) produces a
basic solution, and, a higher pH.
(c) solution 1, Pb(NO3)2, and solution 2, NaCl. PbCl2
(d) solution 3, KMnO4 , ClO3–
(e) solution 4, C2H5OH. Ethyl alcohol is covalently bonded and does not form ions in
water. Therefore, the solution is not a better conductor of electricity than water,
which is also covalently bonded.
2003 D Required
For each of the following, use appropriate chemical principles to explain the
observations. Include chemical equations as appropriate.
(a) In areas affected by acid rain, statues and structures made of limestone (calcium
carbonate) often show signs of considerable deterioration.
(b) When table salt (NaCl) and sugar (C12H22O11) are dissolved in water, it is observed
that
(i) both solution have higher boiling points than pure water, and
(ii) the boiling point of 0.10 M NaCl(aq) is higher than that of 0.10 M C12H22O11(aq).
(c) Methane gas does not behave as an ideal gas at low temperatures and high pressures.
(d) Water droplets form on the outside of a beaker containing an ice bath.
Answer:
(a) limestone reacts with acid to produce a soluble substance, a gas, and water which
wash away
CaCO3(s) + H+(aq)  Ca2+(aq) + CO2(g) + H2O(l)
(b) (i) a solution made from a non-volatile solute has a higher boiling point than the pure
solvent because the solution has a lower vapor pressure than the water (Raoult’s
Law) . the temperature of the solution has be higher to produce enough vapor
pressure to equal the atmospheric pressure (i.e., boiling)
(ii) the amount of boiling point elevation depends on the number of non-volatile
particles in solution. since the salt dissociates into 2 particles for every NaCl that
dissolves, it will increase the boiling point more that an equal concentration of sugar
(a molecular cpd) that does not dissociate or ionize.
(c) at low temperatures and high pressures, the methane molecules are slow and closer
together. under these conditions, van der Waal forces become measurable and
significant and creates a deviation from ideal behavior. at high pressure the volume
of a real molecule is also significant.
(d) a water vapor molecules collide with the cool beaker, the molecules lose kinetic
energy, slow down, attract others, and condense into a liquid
1973 D
Discuss briefly the relationship between the dipole moment of a molecule and the
polar character of the bonds within it. With this as the basis, account for the
difference between the dipole moments of CH2F2 and CF4.
Answer:
In order to have a dipole moment (i.e., to be a polar molecule) a molecule must have
polar bonds and must have a molecular geometry which is not symmetrical (i.e., one
in which the vector sum of the bond dipoles  0).
In CH2F2 the C-F and C-H bonds are polar and the molecule is not symmetrical;
therefore, the molecules is polar and would show a dipole moment.
In CF4 the C-F bonds are polar, but the molecule is symmetrical; therefore, the
molecule is non-polar and would not show a dipole moment.
1974 D
The possible structures for the compound dinitrogen oxide are NNO and NON. By
experimentation it has been found that the molecule of dinitrogen oxide has a nonzero dipole moment and that ions of mass 44, 30, 28, 16, and 14 are obtained in the
mass spectrometer. Which of the structures is supported by these data? Show how
the data are consistent with this structure.
Answer:
The correct structure is NNO. N-N=O; N-O bond di-pole; non-linear structure, nonsymmetrical; molecular dipole moment.
Spectral Data (mass of molecular fragments):
44 = NNO 28 = NN 14 = N 30 = NO 16 = O
A fragment of 28 couldn’t be made if the structure was NON.
1974 D
The boiling points of the following compounds increase in the order in which they
are listed below:
CH4 < H2S < NH3
Discuss the theoretical considerations involved and use them to account for this
order.
Answer:
CH4 - weak London dispersion (van der Waals) forces
H2S - London forces + dipole-dipole interactions
NH3 - London + dipole + hydrogen bonding
1978 D
State precisely what is meant by each of the following four terms. Then distinguish
clearly between each of the two terms in part (a) and between each of the two terms
in part (b), using chemical equations or examples where helpful.
(a) Bond polarity and molecular polarity (dipole moment)
Answer:
(a) Bond polarity - resulting from unequal sharing of electrons between bonding atoms;
or from bonding of atoms with different electronegativities.
Molecular polarity - result of the separation of the centers of positive and negative
charges in an entire molecule (The dipole moment is a measure).; or the result of the
non-zero vector sum of bond dipoles and lone-pair electrons.
Distinction (normally included within the definitions).
1979 D
Butane, chloroethane, acetone, and 1-propanol all have approximately the same
molecular weights. Data on their boiling points and solubilities in water are listed in
the table below.
Boiling Solubility
Compound
Formula
Pt.(ºC) in water
Butane
CH3CH2CH2CH3
0
insoluble
Chloroethane
CH3CH2Cl
12
insoluble
O
Acetone
56 completely
||
miscible
CH C C CH
3
3
completely
miscible
1-Propanol CH3CH2CH2OH
97
On the basis of dipole moments (molecular polarities) and/or hydrogen bonding,
explain in a qualitative way the differences in the
(a) boiling points of butane and chloroethane.
(b) water solubilities of chloroethane and acetone.
(c) water solubilities of butane and 1-propanol.
(d) boiling points of acetone and 1-propanol.
Answer:
(a) Butane is nonpolar; chloroethane is polar. Intermolecular forces of attraction in
liquid chloroethane are larger due to dipole-dipole attraction; thus a higher boiling
point for chloroethane.
(b) Both chloroethane and acetone are polar. However, acetone forms hydrogen bonds to
water much more effectively than chloroethane does, resulting in greater solubility of
acetone in water.
(c) Butane is non-polar and cannot form hydrogen bonds; 1-propanol is polar and can
form hydrogen bonds. 1-propanol can interact with water by both dipole-dipole
forces and hydrogen bonds. Butane can interact with water by neither means. Thus,
1-propanol is much more soluble.
(d) Acetone molecules are attracted to each other by van der Waals attraction and
dipole-dipole attraction. 1-propanol molecules show these two types of attraction.
However, 1-propanol can also undergo hydrogen bonding. This distinguishing
feature results in the higher boiling point of 1-propanol.
1988 D
Using principles of chemical bonding and/or intermolecular forces, explain each of
the following.
(a) Xenon has a higher boiling point than neon has.
(b) Solid copper is an excellent conductor of electricity, but solid copper chloride is not.
(c) SiO2 melts at a very high temperature, while CO2 is a gas at room temperature, even
though Si and C are in the same chemical family.
(d) Molecules of NF3 are polar, but those of BF3 are not.
Answer:
(a) Xe and Ne are monatomic elements held together by London dispersion (van der
Waals) forces. The magnitude of such forces is determined by the number of
electrons in the atom. A Xe atom has more electrons than a neon atom has. (Size of
the atom was accepted but mass was not.)
(b) The electrical conductivity of copper metal is based on mobile valence electrons
(partially filled bands). Copper chloride is a rigid ionic solid with the valence
electrons of copper localized in individual copper(II) ions.
(c) SiO2 is a covalent network solid. There are strong bonds, many of which must be
broken simultaneously to volatilize SiO2. CO2 is composed of discrete, nonpolar CO2
molecules so that the only forces holding the molecules together are the weak
London dispersion (van der Waals) forces.
(d) In NF3 a lone pair of electrons on the central atom results in a pyramidal shape. The
dipoles don’t cancel, thus the molecule is polar.
While in BF3 there is no lone pair on the central atom so the molecule has a trigonal
planar shape in which the dipoles cancel, thus the molecule is nonpolar.
1989 D
The melting points of the alkali metals decrease from Li to Cs. In contrast, the
melting points of the halogens increase from F2 to I2.
(a) Using bonding principles, account for the decrease in the melting points of the alkali
metals.
(b) Using bonding principles, account for the decrease in the melting points of the
halogens.
(c) What is the expected trend in the melting points of the compounds LiF, NaCl, KBr,
and CsI? Explain this trend using bonding principles.
Answer:
(a) Alkali metals have metallic bonds: cations in a sea of electrons. As cations increase
in size (Li to Cs), charge density decreases and attractive forces (and melting points)
decrease.
(b) Halogen molecules are held in place by dispersion (van der Waals) forces: bonds due
to temporary dipoles caused by polarization of electron clouds. As molecules
increase in size (F2 to I2), the larger electron clouds are more readily polarized, and
the attractive forces (and melting points) increase.
(c) Melting point order: LiF > NaCl > KBr > CsI
Compounds are ionic. Larger radii of ions as listed. Larger radii -> smaller attraction
and lower melting points.
1992 D
Explain each of the following in terms of atomic and molecular structures and/or
intermolecular forces.
(a) Solid K conducts an electric current, whereas solid KNO3 does not.
(b) SbCl3 has measurable dipole moment, whereas SbCl5 does not.
(c) The normal boiling point of CCl4 is 77ºC, whereas that of CBr4 is 190ºC.
(d) NaI(s) is very soluble in water, whereas I2(s) has a solubility of only 0.03 gram per
100 grams of water.
Answer:
(a) K conducts because of its metallic bonding - or – “sea” of mobile electrons (or free
electrons). KNO3 does not conduct because it is ionically bonded and has immobile
ions (or immobile electrons).
(b) SbCl3 has a measurable dipole moment because it has a lone pair of electrons which
causes a dipole - or - its dipoles do not cancel - or - it has a trigonal pyramidal
structure - or - a clear diagram illustrating any of the above.
(c) CBr4 boils at a higher temperature than CCl4 because it has stronger intermolecular
forces (or van der Waal or dispersion). These stronger forces occur because CBr4 is
larger and/or has more electrons than CCl4.
(d) NaI has greater aqueous solubility than I2 because NaI is ionic (or polar), whereas I2
is non-polar (or covalent). Water, being polar, interacts with the ions of NaI but not
with I2. (Like dissolves like accepted if polarity of water is clearly indicated.)
2001 D
Account for each of the following observations about pairs of substances. In your
answers, use appropriate principles of chemical bonding and/or intermolecular forces. In
each part, your answer must include references to both substances.
(a) Even though NH3 and CH4 have similar molecular masses, NH3 has a much higher
normal boiling point (-33C) than CH4 (-164C).
(b) At 25C and 1.0 atm, ethane (C2H6) is a gas and hexane (C6H14) is a liquid.
(c) Si melts at a much higher temperature (1,410C) than Cl2 (-101C).
(d) MgO melts at a much higher temperature (2,852C) than NaF (993C).
Answer:
(a) NH3 exhibits hydrogen bonding (H attached to nitrogen, attracted to N in adjacent
molecule) between molecules which creates a larger IMF than CH4 which doesn’t
exhibit H-bonding, only weak London dispersion forces). More energy is required to
overcome this higher IMF in NH3 and, therefore, has a higher boiling point.
(b) Both ethane’s and hexane’s IMF consist mainly of weak London dispersion forces.
The greater number of electrons in hexane (50 vs. ethane’s 18) creates a greater IMF,
enough to make it a liquid at 25oC but for ethane, the fewer electrons make a smaller
IMF and that is not strong enough to cause ethane to condense.
(c) Si forms strong network covalent bonds (4 per atom) to create a high melting solid.
The non-polar molecules of Cl2 (covalent bond, Cl–Cl) do not form strong IMF, only
weak London dispersion forces and that makes it easy to melt at a low temperature.
(d) Magnesium oxide is a Mg2+O2– ionic compound while sodium fluoride is a Na+F–
ionic compound. The larger ionic charge creates a stronger Coulombic attraction
between the anion and cation in MgO and a higher temperature is required to
overcome it and melt it.