Chemistry 113 Quiz 2
January 25, 2001 Name ______________________________
Matching: Place the letter corresponding to the word(s) being defined in front of each of the following. (3)
_G__ 1. The ease of distortion of the electron cloud of an atom or molecule by an outside influence.
_F__ 2. Weak attractive forces caused by instantaneous dipole-induced dipole interactions.
_A__ 3. Description for the temperature at which the vapor pressure of a liquid equals atmospheric pressure.
_H__ 4. The conversion of a solid directly into a gas without passing through the liquid state.
_B__ 5. The temperature above which a substance cannot exist as a liquid regardless of the pressure.
_J__ 6. A measure of the energy needed to expand the surface area of a liquid.
J. Surface tension N. Wetting
Multiple Choice: Place the letter(s) corresponding to the best answer(s) in front of each of the following. (6)
A.B.C 7. The type of intermolecular force(s) that ethyl alcohol, H3C-C(H)2OH is expected to exhibit is (are)
A. Dipole-dipole B. H- bonding C. London
_A__ 8. The molecule(s) that is (are) expected to exhibit only London forces is (are)
A. CH4 B. H3COH C. H3CCl D. H2CF2
_B__ 9. The substance that is expected to have the highest normal boiling point. H-Bonding more
important than MW
A. CH4 B. H3COH C. H3CCl D. H2CF2
_D__ 10. At a given temperature the substance that is expected to have the highest vapor pressure is
A. CCl4 B. HCCl3 C. H2CCl2 D. H3CCl
_C__ 11. The property whose magnitude must always increase with increasing temperature is
A. density B. surface tension C. vapor pressure D. viscosity
_C,D 12. The following factor(s) that is (are) expected to affect the vapor pressure of a liquid is (are)
To cool to f.p. water = mass x sp ht x
DT = 100 g x 4.184 J/g deg x 18 deg = 7531.2 JTo freeze water = mass x
DHfus = 100 g x 334 J/g = 33,400 JHeat removed from water: 40,931.2 J = Heat absorbed by the freon = mass x
DHvap = mass x 289 J/gMass = 40931.2 J/289 J/g = 141.6 g = 142 g
(note there is no need to use MW because units cancel without need for mol)
g Freon = |_142 g_
ln (p1/p2) = (
DHvap / R) (1/T2 - 1/T1) ; normal b.p water = 100oC at 1.00 atm = 760 torrln (p1/ 760 torr) = (41.2 kJ/mol x 1000 J/kJ / 8.314 J/mol deg) ( 1/373 - 1/388) =
4.9555 x 103 ( 1.0365 x 10-3) = 0.5136
p1 / 760 torr = inv ln 0.5136 = 1.6713
p1 = 1.6713 x 760 torr = 1.2702 x 103 torr
v.p. = |_1.67 atm_
or 1.27 x 103 torr
Going across phase diagram at 1 atm to intersect liquid He-1 and vapor curve and then down to temperature, Read ~ 5.5K
There is none because the solid does not exist at 1 atm pressure.
Solid + liquid He-I and liquid He-II
Gas + liquid He-I and liquid He-II
(Note you need three phases or it would not be a triple point.)
He-I is denser. As pressure is increased (to form the denser form), He -II is changed to He-I.
Bonus. Will water rise to a greater height by capillary action in a glass tube or in a polyethylene tube of the same diameter? Explain your reasoning. (1)
The glass has silicate groups with oxygens that the water can H-bond with.
The polyethylene is nonpolar so that the intermolecular attraction would be the weaker dipole-induced dipole interaction.
The water climbs the capillary tube by interacting with the walls of the tube. Thus it would rise higher in the glass tube.