367 Version Quiz 1 laude (51635)
This print-out should have 8 questions.
Multiple-choice questions may continue on
the next column or page nd all choices
before answering.
Constants
R = 8.314 J K1 mol1
Equations
P2
H 1
1
ln
=
P1
R
T1 T2
= M RT
Tb = iKb m
Tf = iKf m
Pi = Pi i
P=
Pi
i
Pressure, bar
3
4
2
101
100
200
4. solid, gas
Explanation:
002 5.0 points
Assuming standard pressure, what physical
constants are needed to calculate the change
in enthalpy for raising a sample of water from
250 K to 300 K.
I) the heat capacity of ice
II) the enthalpy of fusion of water
III) the heat capacity of water
IV) the enthalpy of vaporization of water
V) the heat capacity of steam
2. I, III
001 5.0 points
Determine whether carbon dioxide is a
solid, a liquid, gas, or supercritical uid at
each of the following combinations of temperature and pressure (use phase diagram provided):
I) 10 bar, 275 K
II) 1000 bar, 200 K
Carbon Dioxide
4
10
1
102
3. supercritical uid, solid
1. II, IV
q = mcT
q = mH
10
1
3
250
300
350
Temperature, K
1. gas, liquid
2. gas, solid correct
400
3. III, IV, V
4. V
5. III
6. I, II, III, IV, V
7. III, V
8. I, II and III correct
Explanation:
The heat capacity of ice is necessary to
calculate heat needed to raise the liquid waters temperature from 250 K to 273 K. The
enthalpy of fusion of water is necessary to calculate the heat needed to convert 10 grams of
water at 273 K to 10 grams of liquid water
at 273 K. Finally, the heat capacity of water
is necessary to calculate heat needed to raise
the liquid waters temperature from 273 K to
300 K.
003 5.0 points
Making ice cream at home typically involves
a salt and water solution that serves as a heat
sink to cool and then freeze the cream. This
process relies on freezing point depression.
Version 367 Quiz 1 laude (51635)
Rank the following by freezing point, from
lowest to highest: 3 m NaCl, 4 m LiBr, 3 m
Na2 S, 7 m sugar.
1. 7 m sugar < 4 m LiBr < 3 m Na2 S < 3 m
NaCl
2. 3 m NaCl < 7 m sugar < 4 m LiBr < 3 m
Na2 S
3. 3 m Na2 S < 4 m LiBr < 7 m sugar < 3 m
NaCl correct
4. 3 m NaCl < 3 m Na2 S < 4 m LiBr < 7 m
sugar
Explanation:
Multiplying the stated molality by the vant
Ho factor yields the eective molality of each
solution. This can be used to predict the relative magnitude of the new depressed freezing
point, with the highest eective molality corresponding to the lowest freezing point.
004 5.0 points
Consider a membrane, permeable to water
but not sugar, which can withstand an osmotic pressure of 5 atm. If the osmotic pressure is currently just below the membranes
breaking point, and you measure a sugar concentration of 0.15 M on the left side, what is
the concentration on the right side? (Assume
RT = 25 L atm mol1 for the purposes of
this question.)
1. 0 M sugar (pure water)
2. 0.5 M sugar
3. 0.35 M sugar correct
4. M 0.2 sugar
5. 0.25 M sugar
Explanation:
The dierence in the molarities of the sugar
solution must be large enough to produce 5
atm of pressure, which would be produced by
a 0.2 M concentration gradient, so the other
2
side needs to be at least 0.35 M.
005 5.0 points
Rank the following by their miscibility in
C10 H22 (decane), from least miscible to
most: H2 O (water), C9 H20 (nonane), CH3 OH
(methanol), CH3 CH2 OH (ethanol).
1. H2 O < CH3 OH < CH3 CH2 OH < C9 H20
correct
2. H2 O < CH3 CH2 OH < CH3 OH < C9 H20
3. C9 H20 < H2 O < CH3 OH < CH3 CH2 OH
4. CH3 OH < CH3 CH2 OH < C9 H20 < H2 O
5. CH3 OH < C9 H20 < CH3 CH2 OH < H2 O
Explanation:
Decane is a non-polar solvent, and thus
ranking in terms of increasing miscibility requires ranking in terms of increasing nonpolar character, i.e. from most to least polar.
006 5.0 points
Butane is a common fuel used in cigarette
lighters, camping stoves and cr`me brulee
e
torches. Normally supplied in metal containers under pressure, the fuel exists as a mixture
of liquid and gas, so high temperatures may
cause the container to explode. At 25 o C, the
vapor pressure of butane is 2.3 atm. Which
equation below can be used to determine the
vapor pressure in the container at 150 o C?
(Butanes Hvap = 24.3 kJ/mol.)
1.
ln P2 = ln 2.3 +
24, 300
8.314
1
1
298 423
2.
ln P2 = ln 2.3 +
24, 300
8.314
1
1
298 423
24, 300
8.314
1
1
423 298
correct
3.
ln P2 = ln 2.3 +
Version 367 Quiz 1 laude (51635)
4.
ln P2 = ln 2.3 +
24.3
8.314
1
1
298 423
20 torr at 30 C and acetaldehye has a pure
vapor pressure of 1000 torr at 30 C, a mixture of 6 moles of acetic acid and 4 moles
of acetaldehyde would have what total vapor
pressure?
1. 408 torr
2. 608 torr
5.
ln P2 = ln 2.3 +
24, 300
8.314
1
1
+
298 423
3. 612 torr
4. 510 torr
Explanation:
The solution requires an algebraic rearrangement of the Clausius-Clapeyron equation and then substituting in the given values.
007 5.0 points
Consider a salt that dissolves endothermically
in water. Which of the following is a possible
set of values describing the salts lattice energy, heat of hydration and heat of solution,
respectively?
1. 800 kJ/mol, -795 kJ/mol, -5 kJ/mol
2. -800 kJ/mol, -805 kJ/mol, -1605 kJ/mol
3. -800 kJ/mol, 805 kJ/mol, 5 kJ/mol
4. 800 kJ/mol, -795 kJ/mol, 5 kJ/mol correct
5. 800 kJ/mol, 805 kJ/mol, 1605 kJ/mol
Explanation:
Hsolution = Hlattice + Hhydration
The question stipulates that Hsolution
must be positive, allowing us to eliminate 2 responses. Considering that for salts dissolving
in water Hlattice is always a positive value
and Hhydration is always a negative value, we
can eliminate to two more responses.
008 5.0 points
If acetic acid has a pure vapor pressure of
3
5. 412 torr correct
Explanation:
Ptotal = a Pa + b Pb + ...
Ptotal = 0.6 20 torr + 0.4 1000 torr
Ptotal = 412 torr