Name: AP Chemistry November 23, 2010 Unit 6 Test: Intermolecular

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Name: ____________________________________________
AP Chemistry
November 23, 2010
Unit 6 Test: Intermolecular Forces, Liquids, & Solids
Part I Directions (1 –13): Each set of lettered choices below refers to the numbered statements immediately
following it. Select the one lettered choice that best fits each statement.
1. Fe (s)
____E
Use the following answers for questions 1 through 6.
2. CH4 (s)
____A
(A) molecules held together by London dispersion forces, only
(B) positive and negative ions held together by electrostatic attractions 3. NaCO3 (s) ____B
4. SiO2 (s)
____C
(C) macromolecules held together by strong bonds
5.
NH
(s)
____D
3
(D) molecules held together by dipole-dipole interactions
(E) atoms held together by delocalized electrons
6. SO2 (s)
____D
7. The normal boiling point of water is often depressed at high altitudes. Which of the following explain this
phenomenon?
(A) At high altitudes, the lower atmospheric pressure equals the equilibrium water vapor pressure at a lower
temperature.
(B) There are fewer oxygen molecules present at high elevations.
(C) Water molecules have greater average kinetic energies at higher altitudes because the pressure is lower.
(D) All polar substances (such as water) have lower boiling points at higher elevations.
(E) Water at this altitude has more nitrogen dissolved in it, lowering the boiling point.
8. Which cannot be determined from a phase diagram?
(A) critical pressure
(B) critical temperature
(D) conditions needed for sublimation
(E) heat of vaporization
(C) triple point
9. In the phase diagram of water, graphing pressure versus temperature, why does the equilibrium between the
solid and liquid phases curve to the left?
(A) The solid phase is more dense than the liquid phase.
(B) An increase in pressure causes the liquid to vaporize.
(C) The liquid phase is more dense than the solid phase.
(D) The liquid phase has a higher vapor pressure than the solid phase.
(E) At higher temperatures, the liquid is more likely to vaporize.
Questions 10 through 13 refer to the following terms.
(A) allotropes
(B) isotopes
10. O2 and O3
_____A
11. U-235 and U-238 _____B
(C) isomers
12. diamond and graphite
13. CH3CH2OH and CH3OCH3
14. Which of the following molecules has a higher boiling point than the others?
(A) CO2
(B) SO2
(C) C6H6
(D) CH4
(E) BF3
15. The critical pressure of a substance is the
(A) pressure above which it cannot be liquefied at any temperature
(B) pressure where the solid, liquid, and gas phases can coexist in equilibrium
(C) pressure that is present at absolute zero
(D) pressure at which the substance would boil at a constant temperature
(E) vapor pressure of the substance at 0ºC
16. Which of the following organic compounds is LEAST soluble in water at 298 K?
(A) C6H6, benzene
(B) CH3Cl, chloromethane
(C) C2H5OH, ethanol
(D) C2H5COOH, propanoic acid
(D) C5H10O5, fructose
____A
____C
The diagram below represents the heating curve for a pure crystalline substance.
17. The solid is the only phase present up to point
(A) C
(B) B
(C) E
(D) A
(E) D
18. Which of the following is true about the substance from
points C to D?
(A) Kinetic energy is increasing; potential energy is decreasing
(B) Kinetic energy is decreasing; potential energy is increasing
(C) Kinetic energy is constant; potential energy is decreasing
(D) Kinetic energy is constant; potential energy is increasing
(E) Kinetic energy is increasing; potential energy is constant
Part II Directions: Answer the following questions on a separate sheet of graph paper and attach.
19. The normal boiling and freezing points of krypton are -153.4 ºC and -157.0 ºC, respectively. (The normal boiling
and freezing points of any substance are the temperatures of boiling at freezing at Standard Pressure, 1.00 atm.) Its triple
point occurs at -157.4 ºC and 0.720 atm.
(a) Sketch a properly labeled phase diagram for krypton. [2]
(b) What changes would be observed if a sample of solid krypton at 0.600 atm is heated from 50K to 150K?
[1]
(c) What changes would be observed if a sample of gaseous krypton at 120 K, which is below the critical
temperature, is compressed from 0.900 atm to 1.500 atm? [1]
(d) Which is the densest phase of krypton? [1]
20. Answer the following questions using principles of molecular structure and intermolecular forces.
(a) Structures of the pyridine molecule and the benzene molecule are shown below. Pyridine is soluble in
water, whereas benzene is not soluble in water. Account for the difference in solubility. You must
discuss both of the substances in your answer. [2]
(b) Structures of the dimethyl ether molecule and the ethanol molecule are shown below. The normal
boiling point of dimethyl ether is 250 K, whereas the normal boiling point of ethanol is 351 K. Account
for the difference in boiling points. You must discuss both of the substances in your answer. [2]
(c) SO2 melts at 201 K, whereas SiO2 melts at 1,883 K. Account for the difference in melting points. You
must discuss both of the substances in your answer. [2]
(d) At a pressure of 1 atm, the boiling point of NH3(l) is 240K, whereas the boiling point of NF3(l) is 144K.
(i) Identify the intermolecular forces(s) in each substance. [1]
(ii) Account for the difference in the boiling points of the substances. [1]
(e) The melting point of KCl(s) is 776˚C, whereas the melting point of NaCl(s) is 801˚C.
(i) Identify the type of bonding in each substance. [1]
(ii) Account for the difference in the melting points of the substances. [1]
(f) At 25˚ C and 1 atm, F2 is a gas whereas I2 is a solid. [1]
Part II Answers:
20. (a) water is polar and can form hydrogen bonds since it has a hydrogen attached to an oxygen; the
lone pair of electrons on the nitrogen creates a slightly polar nitrogen and it can hydrogen bond to
the hydrogen in the water. A C-H bond, as those in benzene, is non-polar and cannot hydrogen bond
with water. Since there is little attraction between water and benzene (a non-polar molecule) and
“like dissolves like”, benzene will not dissolve in water but the polar pyridine will.
(b) Identical in formula, C2H6O, and molar mass, the dimethyl ether cannot form hydrogen bonds
with other molecules and only has weak London dispersion forces that keep it together as a liquid.
The ethanol has similar London forces but it also has a polar end (-OH) and can hydrogen bond to
other molecules. The stronger intermolecular forces hold it together better as a liquid and it takes
more energy to break these IMF, thus a higher boiling temperature.
(c) SO2 molecules are held together by dipole-dipole forces and London dispersion forces. These
two types of IMF are very weak compared to the covalent bonds present in SiO2, which is a network
solid.
(d)
(i)
force
NH3
NF3
London dispersion +
+
polar attraction
+
+
hydrogen bonding +
-
ionic attraction
-
-
(ii) the ability of ammonia to create intermolecular hydrogen bonds, leads to higher amount of
energy to separate the molecules by boiling them.
(e) (i) both compounds have ionic bonding
(ii) the sodium ion in NaCl is a smaller size than the corresponding potassium ion in KCl. This
smaller size creates a greater force of attraction in the NaCl, according to Coulomb’s Law, making it
harder to melt.
(f) The I2 molecule has 106 electrons to the 18 of the F2 and, therefore, exhibits stronger London
dispersion forces.
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