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Chemistry
241
Homework Problems
Answer Key
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Crystalization:
Purification of Solids
Answers to Problems:
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1.
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- A plot
similar to line A in Figure 5.1 on page 559 will be
obtained. The line will be slightly curved.
- All of the substance A
would dissolve at 80°C. A solubility of 17.0 g in
100 mL of water is equivalent to a solubility of 0.17 g
in 1 mL of water. This is a greater solubility than is
required.
- Crystals of A should
appear around 56 OC.
- The recovery of A would
amount to 0.085 g. A solubility of 1.5g in 100 mL of
water at 0 OC is equivalent to a solubility of 0.015 g in
1 mL of water. Therefore, 0.015 g of A would remain
dissolved in the water, with the remainder being formed
as crystals.
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2.
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If a saturated hot solution
was filtered by vacuum filtration, the cooling which
occurred as the solvent was drawn through the filter paper
would cause the solute to precipitate in the form of
crystals. The result would be that the filter paper would
become clogged with crystals, and impurities would not be
removed successfully from the solution being
filtered.
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Physical
Constants: Melting Points, Boiling Points, Density
Answers to Problems:
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1.
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A mixed-melting-point
could be used. Equal quantities of A and B are intimately
mixed using a mortar and pestle. The mixture is placed in a
melting point capillary tube, and the melting point
determined. If the melting point is identical to that of
pure A (and pure B) without depression or expansion, then A
and B are identical.
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2.
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Curve 2 would be an ideal
heating rate
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8.
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Perform a
mixture-melting-point with the unknown and each of the other
substances independently. The mixture that gives no
depression or expansion of the melting point reveals the
identity of the compound.
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Extractions,
Separations, and Drying Agents
Answers to Problems:
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1.
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Following the method
given in Section 7.2, we obtain:
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2.
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3.
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One could extract the
mixture with hydrochloric acid at any point in the
separation procedure. However the sodium becarbonate
extraction must be done before the sodium hydroxide
extraction.
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4.
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8.
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The washing procedure
involves adding 1 mL of aqueous sodium bicarbonate to an
organic mixture and shaking the mixture vigorously. When
this procedure is followed, any acidic substances are
removed into the sodium bicarbonate as their sodium
salts.
Extracting an aqueous layer
three times with methylene chloride involves adding a volume
of methylene chloride to an aqueous layer, shaking it
vigorously, allowing the layers to separate and removing the
lower (methylene chloride) layer while leaving the aqueous
layer behind. Save the methylene chloride layer in another
container. Then add another portion of methylene chloride to
the aqueous layer and repeat the above procedure. Each time
save the methylene chloride layer and leave the aqueous
layer in the original container. The procedure involves a
total of three extractions with methylene
chloride.
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9.
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You should not add drying
agent at this point. First, transfer the organic layer with
a dry Pasteur pipet to a dry container and then add drying
agent to the organic layer.
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Simple
Distillation
Answer to Problems:
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1.
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(a) Approximately 92%
A, 8% B.
(b) Approximately 18% A, 82% B.
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3.
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The line connects the
compositions of the boiling liquid (lower curve) and its
equilibrium vapor (upper curve) at the same temperature.
Using line xy in Figure 8.3, if the boiling liquid has
composition W, its vapor has composition Z.
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4.
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It would boil at about
110° C.
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7.
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A good separation can be
achieved in a simple distillation only if thereis a large (
>100°C) difference in the boiling points of the
liquids to be separated.
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Fractional
Distillation, azeotropes
Answer to Problems:
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1.
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(a) 3.9 g = 0.05 mole
benzene; 4.6 g = 0.05 mole toluene
(b) Partial vapor pressure of benzene = (270 mm)(0.5) = 135
mm
(c) At 90°,
Ptotal = (1010 mm)(O.5) + (405 mm)(O.5) = 707 mm
at 100°, Ptotal = (1340 mm)(O.5) + (560
mm)(O.5) = 950 mm
The boiling point is greater
than 90°, but less than 100°. Assume a linear
relationship between the vapor pressure of each substance
and the temperature from 90 to 100°.
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The following approximate
vapor pressures are obtained at certain temperatures:
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temp
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benzene
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toluene
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90°
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1010
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405
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92°
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1076
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436
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94°
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1142
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467
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96°
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1208
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498
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98°
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1274
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529
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100°
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1340
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560
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At 92°;
Ptotal = (1076)(0.5) + (436)(0.5) =
756mm
Thus, the boiling point is
approximately 92°.
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(d) Partial vapor pressure
of benzene at 92° = (1076)(0.5) = 538 mm
Partial vapor pressure of
toluene at 92° = (436)(0.5) = 218 mm 538
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(e) 0.71 mole = 55.4 g
benzene; 0.29 mole = 26.7 g toluene
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Column
Chromatography
Answer to Problems:
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1.
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If the components of
the mixture all passed through the column, the solvent must
have been too polar. In that case, repeat the chromatography
using a less polar solvent, such as petroleum ether. If all
of the mixture stayed on the column too long, the solvent
was probably not polar enough. Switch to a more polar
solvent, such as methanol or ethanol. Other parameters which
might be adjusted include the length of the column and the
column packing.
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3.
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Order of elution:
Biphenyl (elutes first) > benzyl alcohol > benzoic
acid (elutes last).
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4.
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The person forgot to allow
the column to drain until the surface of the liquid
containing the orange compound had passed the upper surface
of the adsorbent. Then a small amount of solvent should have
been added, and that solvent should have been allowed to
drain below the upper surface of the adsorbent. Only then
would it be permissible to fill the solvent reservoir. The
solvent reservoir should not be filled until the sample has
been completely applied to the adsorbent.
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5.
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Clearly, petroleum ether is
not sufficiently polar to elute this sample. A more polar
solvent, such as methanol or ethanol, should be
selected.
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Thin-layer
Chromatography
Answer to Problems:
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1.
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The presence of only
one, highly-mobile spot does not necessarily indicatethat
the unknown material contains one pure compound. It is
possible that the material is a mixture, but that all of the
components travel all the way up the TLC plate because the
solvent is too polar. The experiment should be repeated with
a less polar solvent, such as petroleum ether or
cyclohexane. If only one spot appears in this experiment, it
is safer to conclude that the unknown is a single, pure
compound.
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2.
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Owing to differences in
adsorbent thickness, it is not safe to conclude that the two
samples are identical. The experiment should be repeated
with both samples applied to the same TLC plate. If they
both have the same Rf value on the same plate, one may
conclude that they are identical.
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5.
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This is the reverse
situation of that described in Problem 1. In this case, the
solvent is not sufficiently polar to move the components
down the plate. The experiment should be repeated in a more
polar solvent. If only one spot is observed in this second
experiment, one may conclude that the sample is
pure
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Gas
Chromatography
Answer to Problems:
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1.
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(a) I-Chloroproane
would have the shorter retention time (elutes first). The
elution order is according to boiling point.
(b) The retention times
would not be identical because it is impossibleto duplicate
exactly all the factors affecting retention times. However,
the order of elution would be the same.
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2.
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Area Peak A = 66 x 8 = 528
mm2
Area Peak B = 43 x 8 = 344 mm2
Total area= 872 mm2
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4.
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(a) Retention time would
increase.
(b) Retention time would
decrease.
(c) Retention time would
increase.
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