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Chemistry 113 |
Midterm Examination |
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February 8, 2001 |
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Name ____________________________ |
following (12)
_C__ An apparatus for measuring atmospheric pressure.
_R__ The S.I. unit for pressure = 1 N/m2
_M_ An extrastong dipole-dipole attraction of a hydrogen covalently bonded to a very electronegative atom.
_I__ The spontaneous intermingling of one substance with another.
_A__ Description for a noncrystalline solid, i.e. a glass.
_X__ The smallest portions of a crystal that can be repeated over and over in all directions to give the crystal
lattice.
_S__ The conversion of a solid directly into a gas without passing through the liquid state
_D__ The temperature at which the vapor pressure of a liquid equals atmospheric pressure.
_Y__ The resistance of a fluid to a change in its form (to flow)
_V__ The temperature and pressure at which the liquid, solid, and vapor state of a substance can coexist in
equilibrium
_L__The enthalpy change associated with gaseous ions or molecules as they dissolve in water.
_P__ The passage of solvent molecules through a semipermeable membrane
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A. Amorphous |
G. Crystal |
M. Hydrogen bond |
T. Surface tension |
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B. Atmosphere |
H. Dialysis |
N. London force |
U. Torr |
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C. Barometer |
I. Diffusion |
O. Manometer |
V. Triple |
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D. Boiling point |
J. Effusion |
P. Osmosis |
W. Vaporization |
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E. Colloid |
K. Fusion |
R. Pascal |
X. Unit cell |
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F. Critical |
L. Hydration |
S. Sublimation |
Y. Viscosity |
each of the following. (28)
_D___ Of the following gases, the one with the greatest density at STP is
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A. Ar |
B. He |
C. Ne |
D. Xe |
_D___ If a gas’s volume is doubled but the temperature remains constant
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A. the molecules move faster |
B. the molecules move slower |
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C. the pressure is doubled |
D. the pressure is halved |
_D___ The volume correction term in the van der Waals equation is present because
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A. manometers are inaccurate |
B. molecules are not usually monatomic |
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C. molecules attract each other |
D. molecules occupy space |
_D___ The boiling points of the rare gases increases going from He to Xe. The type of intermolecular force
that is most responsible for this trend is
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A. dipole-dipole attraction |
B. hydrogen bonding |
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C. ion-ion attraction |
D. London dispersion forces |
_B___ The intermolecular forces most responsible for the relatively high heat of vaporization for ethanol,
C2H5OH are
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A. dipole forces |
B. hydrogen bonding |
C. London forces |
D. none of these |
_A___ The molecules below all have the same number of electrons and nearly equal molecular weights.
Which one is likely to have the highest vapor pressure?
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A. CH3-CH2-CH3 |
B. CH3CH2-F |
C. CH3-CH2-OH |
D. CH3-O-CH3 |
_B___ Bromine, Br2, has a triple point at -7.3oC and 44 mmHg and a critical point at 315oC and 102 atm.
The density of the solid is 3.4 g/cm3, and the density of the liquid is 3.1 g/cm3. When cooled
sufficiently at 40 mmHg, bromine vapor
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A. condenses to liquid |
B. condenses to solid |
C. does not condense |
_A___ Perovskite is a mineral containing calcium, titanium, and oxygen. The unit cell for the mineral has a
Ti ion in the center of the cell, Ca ions at the 8 corners of the cell, and Oxide ions in the 6 faces. The
chemical formula for perovskite is
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A. CaTiO3 |
B. CaTiO6 |
C. Ca8TiO6 |
D. none of these |
_D____ The type of crystal that is soft, low melting, and does not conduct electricity is
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A. covalent network |
B. ionic |
C. metallic |
D. molecular |
_B,(C)_ The most water soluble of the following compounds would be
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A. C6H6 (l) |
B. C2H5OH (l) |
C. NaCl (s) |
D. SiO2 (s) |
_C___ The lattice energy of SrCl2 is (–)2114 kJ/mol. The heat of hydration of SrCl2 (infinite dilution) at
298 K is –2161 kJ/mol. When a very dilute solution of SrCl2 is prepared, the temperature change
expected would be
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A. none |
B. a decrease |
C. an increase |
_C___ An ideal liquid solution is prepared having equal mole fractions of two volatile components, A
and B. In the vapor above the solution
_B___ The aqueous solution with the lowest freezing point of the following group is
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A. 0.01 m CH3OH |
B. 0.01 m CaCl2 |
C. 0.01 m CaSO4 |
D. 0.01 m NaCl |
_C___ The best method for determining the molecular weight of a relatively large bio-molecule generally
involves measurement of
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A. freezing point depression |
B. boiling point elevation |
C. osmotic pressure |
D. vapor density |
Problems:
PV / T = P’V’ / T’ with P constant; 5.83 L / 273 K = v’ / 298 K
V’ = 5.83 L x 298 / 273 = 6.364 L
V = |__6.36 L____
The volume of CO2 formed is 154 ml at 23oC and 731 mmHg. What is the molarity of the HCl solution? PV = nRT; n = PV/RT;
18.6 mL/1000 mL/L = 0.0186L; 23oC + 273 o = 296 K (8)
n = [(731 torr/760 torr/atm) x 0.154 L] / [0.08206 Latm/mol deg x 296 K] = 0.0060982 mol CO2
0.
0060982 mol CO2 x 2 mol HCl/1 mol CO2 = 0.01296 mol HCl consumedM = mol/L = 0.012196 mol / 0.0186 L = 0.065572 mol/L
M = |_0.656 M__
Since it takes longer for the unknown gas to effuse, it must be heavier than the hydrogen.
Time unknown / Time H2 = (MWunk)^0.5 / (MWH2)^0.5 = 4.67 / 1 = (MW)^0.5 / (2.016)^0.5
(4.67)2 / 12 = MW / 2.016
MW = 21.8089 x 2.016 = 43.967
MM = |_ 44.0 g/mol
Bcc = 1/8 x 8 corners + 1 x 1 center = 2 atoms/unit cell
Edge = 350.9 pm x 10-12 m/pm x cm/10-2m = 3.509 x 10-8 cm
D = mass unit cell/ volume unit cell
= [6.941 g/mol x (1 mol/6.022 x 1023 atoms) x 2 atoms/unit cell]/(3.509 x 10-8 cm)3
D = 2.305 x 1023 / 4.3207 x 10-23 = 0.53353 g/cm3
D = |_0.5335 g/cm3
21.6 + 273.2 = 294.8; 60.6 + 273.2 = 333.8 (8)
ln (p1 / p2) = (DHvap / R) (1/T2 – 1/T1)
= ln 400.0 torr/100.0 torr = (DHvap /8.314 J/mol deg) (1/294.8 – 1/333.8)
D
Hvap = (ln 4.000 x 8.314 J/moldeg) / 3.963 x 10-4 deg-1 = (1.2863 x 8.314 J/moldeg) / 3.963 x 10-4 deg= 2.908 x 10-4 J/mol = 29.1 kJ/mol
D
HVap = |_29.1 kJ/mol_
106 mL x 1 L / 1000 mL = 0.106 L; 1.47 torr x 1 atm/760 torr = 0.001934 atm
P
= MRT = (n/V)RT = (mass/MW)/V x RT; MW = mass x RT / V x PMW = [0.582 g x 0.08206 Latm/mol deg x 298] / 0.106 L x 0.001934 atm = 6.942 x 104 g/mol
(if you solve for M instead it = 7.909 x 10-5 mol; n = M x L = 8.384 x 10-6 mol;
MW = 0.582 g/ 8.384 x 10-6 mol)
MM = |_6.94 x 104 g/mol
MW ethylene glycol = 2(12.01) + 6(1.008) + 2(16.00) = 62.07 g/mol; for H2O = 18.02 g/mol
The two most straight-forward approaches are :
Assume 1.00 L of solution; 1.05 g/mL x 1000 mL/ L = 1050g/L solution;
1050 g x 40.0/ 100.0 = 420 g E.G.; 1050 – 420 = 630 g H2O
Mol E.G. = 420 g/62.07 g/mol = 6.767 mol; mol H2O = 630 g/ 18.02 g.mol = 34.96 mol
(1 mL of solution works similarly but with appropriate change of decimal place)
Assume 100. g soln; 40.0 g E.Gà 40.0 g/ 62.07 g/mol = 0.6444 mol ; 60.0 g H2O à 60.0/18.02 = 3.330 mol
D = M/V, V = M/D = 100 g solution / 1.05 g/mL = 95.24 mL = 0.09524 L
(1 g of solution works similarly but with appropriate change of decimal place)
molality = mol solute / kg solvent = 6.767 mol / 0.630 kg = 0.6444 mol / 0.0600 kg = 10.74 m
m = |_10.74 molal___
Molarity = mol solute / L solution = 6.767 mol/L = 0.6444 mol/0.9534 L = 6.766 M
M = |_6.77 Molar____
X = mol solute/ mol solution = 6.767 mol /(6.767 + 34.96) mol = 0.644 mol (0.644 mol + 3.330 mol)
= 0.1622
X = |_0.162______
1 + 4 = 5 mol total
PT = XP Ppo + XH PHo = 1/5(441 torr) + 4/5(121 torr) = 88.2 torr + 96.8 torr = 185.0 torr
Ppentatne = |_88.2 torr___
Phexane = |_96.8 torr____
n proportional to P (ideal gas law); X = pp/PT
XP = 88.2 / 185 = 0.4767
XH = 96.8 / 185 = 0.5332
Xpentane = |_0.477 ____
Xhexanae = |_0.523 ____
Note how the vapor phase is enriched in the more volatile substance.
Bonus. You decide to cool a can of soda pop quickly in the freezer compartment of a refrigerator. When you take out the can, the soda pop is still liquid; but when you open the can, the soda pop immediately freezes. Explain why this happens. (continue on back of page if necessary) (2)
The answer for this bonus must deal with a solution and in particular the freezing point depression of that solution. Two factors would seem important.
The first is that when the can is opened and the gas released from solution, the concentration of the remaining solute is dramatically decreased so that the freezing point of the remaining solution would be raised so that the remaining solution would freeze at a higher temperature than the original.
The second is that when the solute gas escapes from the solution when the can is opened, it must overcome the intermolecular forces it experiences in solution. The escape is thus an endothermic process that will leave the remaining solution colder so that the freezing point is more likely reached.