|
Further
exercises
For pages 185–6
| A1. |
Explain
why lakes and rivers freeze from the top down and not from the bottom up.
Answering the following questions may be the best way to do this. If the
air above a shallow lake gradually cools from say 10oC
downwards, what happens to the temperature of the water at the surface of
the lake? What mixing, if any does this cause? Why? Does there come a
point at which this mixing stops? Why? As the air temperature cools to
below 0oC, where does ice form first? Does it stay there? Why?
Data in Exercise 2 on page 190 may be useful. Note also that the density
of ice at 0oC is 0.917 g/mL.
What is the significance of this for fish?
|
| A2. |
Some
students made up solutions of sodium chloride in water of known
composition (grams of NaCl per 100 g solution). They then pipetted 5.00 mL
of each solution into a weighed beaker and determined the mass of the
solution. The experiment was performed at 25oC. Their results
were:
Composition
(g NaCl
per 100 g solution) |
5.00 |
10.00 |
15.00 |
20.00 |
| Mass
of 5.00 mL of solution (g) |
5.16 |
5.32 |
5.54 |
5.73 |
|
|
|
(a)
|
Draw
a graph of density of solution versus per cent composition. They also knew
that the density of pure water at 25oC was 0.997 g/mL. Include
this point on your graph also. How does density vary with composition?
|
|
(b) |
Sea
water is 3.5% sodium chloride. What is its density at 25oC? |
|
(c) |
A
sample of bore water had a density of 1.01 g/mL at 25oC.
Assuming that the only salt it contained was sodium chloride, estimate its
per cent composition. |
|
(d) |
Using
your answer in (b), what happens to a ship when it moves out of fresh
water into sea water? Does it sink further into the water, rise further
out of the water or remain unaffected? Explain.
|
| A3. |
(a) |
Use the data below
to draw a graph of freezing point versus per cent composition (g NaCl per
100 g solution) for solutions of sodium chloride.
| Composition |
0 |
2.0 |
6.0 |
10.0 |
16.0 |
20.0 |
23.0 |
| Freezing
point (oC) |
0.0 |
–0.4 |
–1.6 |
–3.6 |
–9.0 |
–15.0 |
–21.2 |
|
|
(b)
|
What is the
freezing point of (i) sea water (3.5% NaCl) (ii) a 17.5%
NaCl solution? What is the composition of a solution that freezes at (iii)
12.5oC (iv) 22oC?
|
|
(c) |
From your graph,
what would be the freezing point of a 25% solution? How confident are you
about your prediction? Why? |
|
(d) |
In countries that
have regular snowfalls leading to ice formation on roads, the roads are
often sprinkled with salt to cause the ice to melt. Explain how this
works.
|
| A4. |
If a sample of pure
water is boiled in an open beaker, the temperature remains constant
throughout the process. However if a sample of sea water is similarly
boiled in an open beaker the temperature gradually increases during the
boiling process. Explain these different observations.
|
| A5. |
Is clean drinking
water a renewable resource? Explain. Relate your answer to the current
efforts being made by Sydney Water to persuade people to use less water.
|
For page 190
| B1. |
Draw
electron-dot diagrams and structural formulae for the molecules formed
from the following atoms: |
|
(a)
two fluorine atoms |
(c)
silicon and hydrogen atoms |
|
(b)
an oxygen and two chlorine atoms |
(d)
arsenic and fluorine atoms |
| B2. |
Which
(if any) of the molecules in Exercise B1 is(are) polar? The shapes of
these molecules are: (a) linear (b) bent (c)
tetrahedral (d) pyramidal. Explain your reasoning. |
| B3. |
Place
molecular chlorine, sodium chloride and trichloromethane, CHCl3
in order of increasing boiling points. Give reasons for your
predictions. |
| B4. |
Both
the B–Cl and P–Cl bonds are polar. However boron trichloride is
non-polar while phosphorus trichloride is polar. Offer an explanation
for this. |
For page 194
| C1. |
The molecular
weight of hydrogen fluoride is the same as the atomic weight of neon Give
a reason for the boiling point of hydrogen fluoride being so much greater
than that of neon (HF 19oC, Ne –246oC)
|
| C2. |
Methanol, CH3OH has
an OH group and three H atoms attached to a central C atom; methanal, CH2O, has an O atom and two H atoms attached to a central C atom. Which
compound would you expect to have the higher boiling point? Explain why.
|
| C3. |
What, if anything,
is wrong with the following statement: despite having similar molecular
weights (32 and 34 respectively) hydrazine, H2N–NH2,
has a higher boiling point than phosphine, PH3, because the
chemical bonds in hydrazine are stronger than those in phosphine.
|
| C4. |
Hydrogen peroxide,
HOOH, is a liquid at room temperature whereas oxygen, O2 or OO, is a gas;
hydrogen peroxide is readily soluble in water while oxygen is only
sparingly soluble. Offer explanations for these facts.
|
| C5. |
Give an
explanation for the difference in boiling points for the following pairs
of substances. |
|
(a) |
sulfur dichloride
(59oC) and magnesium chloride (1437oC) |
|
(b) |
Methanol, H3C–OH
(64.7oC) and bromomethane, H3C–Br (3.6oC)
|
| C6. |
(a) |
Using values from
Table 7.2 on page 185 and others given here, list the following compounds
in order of decreasing surface tension.
water, ethanol, H3C–CH2–OH, methanol, H3C–OH
(23 mN m–1), ethylene glycol, HO–CH2–CH2–OH,
hexane, C6H14 (18), pentane, C5H12, (16). |
|
(b) |
What, if any,
correlation is there between surface tension and magnitude of
intermolecular forces in these compounds. Offer an explanation for any
correlation that exists.
|
For page 200
| D1. |
Draw
diagrams on the molecular or ionic level of aqueous solutions of
(a) magnesium sulfate (b) urea, H2N–CO–NH2
Identify the ions, if any, in each solution.
|
| D2. |
Which
of the following substances do you expect to be soluble in water? Give a
reason for your prediction for each compound in the list.
calcium bromide, tetrafluoromethane, CF4, nickel, ammonia,
carborundum, SiC, nitrous acid, HONO, glucose, neon.
|
| D3. |
Nickel
chloride crystallises with six moles of water per mole of nickel
chloride. Give the formula and name for the hydrated compound.
If 4.82 g of the hydrate were heated until all the water had been
vaporised, what mass would remain?
|
| D4. |
Magnesium
chloride exists as a hydrate. When the compound is heated it loses its
water of crystallisation. 2.64 g of this hydrate was heated to a constant
mass of 1.24 g. Calculate the formula of the hydrate and name it.
|
| D5. |
(a) |
If a sewing needle
is very gently placed horizontally on the surface of water in a large beaker, the
needle floats. Explain why, bearing in mind that the density of the needle
is much greater than that of the water. |
|
(b) |
However if one
repeats the experiment using water that contains a small amount of
detergent (surfactant), the needle does not float, no matter how
carefully one does the experiment. Explain why.
|
| D6. |
Dimethyl
ether and ethanol both have the same molecular formula, C2H6O,
but they have different structures:
Each dash represents a shared pair of bonding electrons.
Ethanol is very soluble in water whereas dimethyl ether is insoluble.
Explain why.
|
For page 207
| E1. |
Give
the name and formula for the precipitate (if any) that forms when the
following pairs of solutions are mixed: |
|
| (a) |
silver
nitrate and potassium iodide |
(f) |
sodium
carbonate and potassium chloride |
| (b) |
aluminium
chloride and sodium hydroxide |
(g) |
potassium
carbonate and iron(II) sulfate |
| (c) |
ammonium
nitrate and copper sulfate |
(h) |
silver
nitrate and calcium chloride |
| (d) |
lead nitrate
and sodium sulfide |
(i) |
zinc nitrate
and barium hydroxide |
| (e) |
barium
chloride and sulfuric acid
|
(j) |
ammonium
sulfide and iron(III) sulfate
|
|
| E2. |
Write
net ionic equations for all reactions that occurred in Exercise E1.
|
| E3. |
(a) |
For the first,
third and fifth equations you wrote in Exercise E2 rewrite them as
complete ionic equations. |
|
(b) |
For the second,
fourth and eighth equations rewrite them as neutral species equations.
|
| E4. |
How
would you prepare samples of the following substances using precipitation
reactions? |
|
(a)
lead sulfate (b) magnesium carbonate (c) copper sulfide
|
For page 212
| F1. |
100.0 g
ammonium chloride was stirred with 200.0 g water for a considerable time
then allowed to stand overnight. The solid remaining at the bottom of the
container was filtered off, dried and weighed. It had a mass of 21.3 g.
Calculate the solubility of ammonium chloride in g per 100 g water.
|
| F2. |
The
solubility of potassium nitrate is 37 g / 100 g water at 25oC.
If you mixed
(a) 22 g potassium nitrate with 50.0 g water (b)
71 g with 200 g water, would all the solid dissolve? Explain.
|
| F3. |
(a) |
Use the data below
to draw a graph of solubility of potassium chlorate, KClO3, as
a function of temperature. Solubility is given as grams of solute per 100
g water. |
|
|
| Temperature
(oC) |
10 |
30 |
50 |
70 |
80 |
90 |
100 |
| Solubility
(g / 100 g) |
5 |
10 |
18 |
28 |
34 |
42 |
52 |
|
|
(b) |
What is the
solubility of potassium chlorate at (i) 14oC (ii)
87oC? |
|
(c) |
At what
temperature is the solubility of potassium chlorate (i) 15 g / 100
g (ii) 48 g / 100 g? |
|
(d) |
If 70 g KClO3
was mixed with 200 g water at 90oC, would a clear solution result? Why? If the
mixture was cooled to 10oC, what would you observe? Be as
quantitative as possible. Explain your observation.
|
| F4. |
Solutions
of sodium hydroxide and calcium nitrate were mixed; the total volume was
100 mL. A precipitate of calcium hydroxide formed. After many hours
analysis showed that 0.016 mol Ca2+ ions remained in solution.
2.00 g solid calcium hydroxide was shaken with 100 mL water. How many
moles of calcium ion would you expect to be in the solution after several
hours? Explain why. Why was a long time allowed before the analysis was
performed in the first experiment and why could you only make a prediction
for a long time after mixing the solid and water in the second experiment?
|
| F5. |
The table below gives relative
electrical conductivities (R.E.C.) for aqueous solutions (of suitable concentrations) of several
substances.
| Substance |
sodium
chloride |
sucrose |
A |
B |
C |
D |
E |
F |
| R.E.C. |
100 |
0.01 |
200 |
0.02 |
80 |
7 |
0.01 |
4 |
|
|
(a)
|
First classify substances A, B, C, D, E
and F as electrolytes or non-electrolytes. Then,
for the electrolytes, decide whether they are strong or weak.
|
|
(b) |
State whether
substances A, B, C, D, E and F would
be present in aqueous solution as:
(i) molecules (ii) ions (iii) mostly molecules but a few
ions.
|
| F6. |
The
weak electrolytes in the table in Section 3 of the Supplementary material
section above were acetic acid, CH3CO2H and ammonia
NH3. They were weak electrolytes because a small percentage of
the molecules reacted with water to form H3O+ and CH3CO2–
ions (from acetic acid) and NH4+ and OH–
(from ammonia). On the other hand nitric acid was a strong electrolyte,
meaning that all of its molecules reacted with water to form H3O+
and NO3– ions. For these three ionisations write
chemical equations that show the strength or weakness of the ionisations.
Which (if any) of these ionisations is (are) an equilibrium reaction?
|
| F7. |
Sodium
hydroxide reacts with nitrous acid, HNO2, to form sodium
nitrite. Write an equation for this reaction. The relative electrical
conductance of a nitrous acid solution was 12 (on a scale where
hydrochloric acid was 250 and sodium hydroxide 150). After addition of
just sufficient sodium hydroxide to convert all the nitrous acid to sodium
nitrite, the relative electrical conductance was 90. Is (a) nitrous
acid (b) sodium nitrite a strong or weak electrolyte? Explain how
you decided this. What does this tell you about the species present in a
solution of (a) nitrous acid (b) sodium nitrite?
|
For page 217
| G1. |
Calculate
the per cent composition of a saturated solution of sodium chloride which
contains 36 g solute per 100 g water.
|
| G2. |
(a) |
The solubility of
Epsom salts (magnesium sulfate) in water is 0.36 g / g water. Express this
as grams per kilogram of water and as mass percent, %(w/w). |
|
(b) |
The solubility of
copper sulfate is 220 g / kg water. Express this in g / g water and as
mass per cent. |
|
(c) |
Using an
approximate density of the solutions of 1.0 g / mL, express the
concentrations in (a) and (b) in grams per litre.
|
| G3. |
(a) |
What mass of an
8.0% aqueous sodium chloride solution do you need to take in order to have
40 g NaCl? |
|
(b) |
If an experiment
requires 5.0 g sulfuric acid, what mass of a 1.0% solution of sulfuric
acid in water should you take? |
|
(c) |
A saturated aqueous
solution of ammonium chloride at 25oC contains 39 g solute per
100 g water. The density of the solution is 1.1 g /mL. Calculate the
solubility in g / L.
|
| G4. |
(a) |
To make 200 g of a
5% solution of sodium chloride in water, what mass of NaCl is needed? |
|
(b) |
How much sucrose
should be dissolved in 25 mL water to make a 15%(w/v) solution? Assume
negligible volume change from water to solution upon mixing.
|
| G5. |
(a) |
12.3 g lead nitrate
was dissolved in water and the volume make to 250.0 mL Calculate the
concentration of the solution in g / 100 mL and in g / L. |
|
(b) |
10 mL (by pipette)
of this solution was diluted to 500 mL (volumetric flask). Calculate the
concentration of this diluted solution in g / L and in ppm. Assume the
density of the diluted solution is 1.00 g/mL.
|
| G6. |
(a) |
A solution was 24
ppm in magnesium ion. Calculate the concentration in
g / L (assuming a
density of 1.00 g/mL). |
|
(b) |
What mass of copper
sulfate pentahydrate would you dissolve in 2.00 L of water to make a
solution that was 16 ppm in Cu2+?
|
| G7. |
If 2.00 mL of a
4.8%(w/v) solution of lead nitrate was diluted to 100 mL, what would the
new concentration be? What volume of this diluted solution would you
dilute to 1.00 L to make a solution that was 3.0 ppm in Pb2+
ion? |
For pages 220–1
| H1. |
Calculate
the concentrations in mol/L of solutions that contain: |
|
| (a) |
0.048 mol
Ca(OH)2 in 2.00 L |
(c) |
37.0 g KCl in
100 mL |
| (b) |
0.214 mol
MgSO4 in 250 mL |
(d) |
3.97 g Pb(NO3)2
in 500 mL
|
|
| H2. |
(a) |
How many moles of
sodium hydroxide are needed to make 250 mL of a 0.500 mol/L solution? |
|
(b) |
How many moles of
ammonia are needed to make 100 mL of a 0.355 mol/L solution? |
|
(c) |
What masses of the
solutes are needed for the solutions in (a) and (b)?
|
| H3. |
(a) |
What mass of sodium
nitrate is required to make 500 mL of a 0.133 mol/L solution? |
|
(b) |
What mass of barium
hydroxide is required to make 2.00 L of a
4.5 x 10–3 mol/L solution?
|
| H4. |
How
many moles of solute are there in |
|
| (a) |
37.5 mL
of 0.207 mol/L Na2CO3 |
(c) |
25 mL of 2.33
x 10–3 mol/L AlCl3 |
| (b) |
18.7 mL of
0.022 mol/L Ca(OH)2 |
(d) |
0.250 L of
1.50 mol/L KI
|
|
| H5. |
In the
solutions in Exercise H4 how many moles are there of |
|
(a)
Na+ ion in (a); (b) OH–
ion in (b); (c) Cl– ion in (c)
|
| H6. |
How
many grams of solute are there in solutions (a) and (c) of Exercise H4?
|
| H7. |
(a) |
How many moles of
lead iodide are formed when 25 mL 0.266 mol/L potassium iodide is added to
a solution containing excess lead nitrate? |
|
(b) |
If a small excess
of magnesium chloride solution is added to 50.0 mL 0.415 mol/L solution of
sodium hydroxide how many moles of magnesium hydroxide precipitate out? |
|
(c) |
What mass of (i)
lead iodide (ii) magnesium hydroxide is formed in (a) and (b)
respectively?
|
| H8. |
Sodium
sulfate solution was added to 100.0 mL of a solution containing lead ions
until all the lead had been precipitated as lead sulfate. 0.855 g lead
sulfate was formed. What was the concentration in moles per litre of lead
ions in the original solution?
|
| H9. |
25.0 mL
of sea water was accurately diluted to 250 mL. 50.0 mL of this diluted
solution required 26.7 mL 0.114 mol/L silver nitrate solution to
precipitate out all the chloride as silver chloride. Calculate the
molarity of chloride in the sea water. Assuming that all the chloride had
been present as sodium chloride, calculate the percentage (w/v) of sodium
chloride in the sea water.
|
| H10. |
To determine the
concentration of cadmium ion in the waste water from an electroplating
factory, an analyst took 250.0 mL of the waste water and slowly added
sodium hydroxide solution until no more cadmium hydroxide precipitated.
The precipitate was filtered off, washed and dried and weighed; it had a
mass of 0.642 g. Calculate the concentration of cadmium ion in the
original waste water in mol/L, g/100 mL and in ppm (assuming a density of
1.00 g/mL). |
For page 224
| J1. |
What
quantity of heat is required to heat |
|
(a) |
45 g ethanol from
10oC to 55oC |
|
(b) |
32 g ethyl acetate
through 25 K?
|
| J2. |
If you spill 10 mL
water at 60oC on your hand it burns (stings, hurts)
significantly more than if it had
been 10 mL ethyl acetate at 60oC. Use Table 8.4 on page 237 to explain
why.
|
| J3. |
If 100
g water at 50oC is added with stirring to 250 g water at 20oC
in a well insulated container of negligible heat capacity, what will the
final temperature be? (Hint: heat flows from the hot water to the cold
water: the heat lost by the hot water equals the heat gained by the cold
water. Let the final temperature be x; set up an equation for heat
lost equals heat gained and solve for x.)
|
| J4. |
If 100
g ethanol at 50oC is added to 250 g water at 20oC
as in Exercise J3, what will be the final temperature? The hint in J3
applies here also.
|
For page 226
For
Exercises K1 and K2 take the density and specific heat capacity of the
final solutions as 1.00 g/mL and 4.2 J K–1 g–1
respectively and assume negligible heat losses.
|
| K1. |
When
8.5 g sodium hydroxide was dissolved in 200 mL water at 18.2oC,
the temperature rose to 28.1oC. Calculate the molar heat of
solution of sodium hydroxide. Write a chemical equation for the process this
refers to.
|
| K2. |
The
molar heat of solution of sulfuric acid is –90 kJ/mol. 10 mL (=18 g)
concentrated sulfuric acid (take as 100%) was added to 100 mL water at 20oC.
What was the final temperature?
|
| K3. |
A
convenient form of heat pack sometimes used by athletes consists of a
plastic bag containing a supersaturated solution of sodium thiosulfate. To
use it the athlete shakes the bag vigorously to cause the excess
thiosulfate to crystallise out of solution. Using information in Exercise
34(b) on page 241, explain how this causes the bag to become hot. What
would you do in order to be able to re-use this heat pack?
There is another type of heat pack that consists of a pouch of calcium
chloride crystals inside a plastic bag of water. To use this the athlete
ruptures the pouch containing the calcium chloride which causes the bag to
get hot. Using information in Exercise 34(a) on page 241 explain why this
bag of water becomes hot. Is this type of heat pack re-usable? Explain.
|
| K4. |
(a) |
If a power station
discharges 5 x 104 L of used cooling water at 60oC
into a lake containing 2 x 106 L of water at 15oC,
what will be the final temperature of the lake, assuming complete mixing
and no heat losses? (See the hint for Exercise J3.) |
|
(b) |
The assumptions in
(a) are unrealistic. Describe what would be the real effect of discharging
this hot water into the lake.
|
Answers
to Further exercises
| A1. |
The
water at the surface of the lake cools; this causes its density to
increase (data in Exercise 2 on page 190) and so it sinks and
causes higher temperature water to come to the surface so it in
turn cools and sinks. This is normal convection. When the
temperature at the surface reaches 4oC, convectional
mixing stops because as the water cools below 4oC its
density decreases and so it remains on the surface. (In the
table in Exercise 2 it looks like 5oC, but more
extensive data shows that the maximum density occurs at 4oC.)
Hence the surface water cools to 0oC while water below
the surface remains at 4oC. Ice therefore forms on the
surface of the lake. Ice has a lower density than water at 0oC
or 4oC so it stays on the surface (floats). Water below
the ice is very slowly cooled by conduction (water is a poor
conductor) and so the thickness of the ice gradually increases. In
other words ice forms from the top down.
The significance of this for fish is that there is generally a
layer of liquid water (at about 4oC) under the ice of a
frozen lake: species of fish common to areas where lakes freeze
can survive quite well at such low temperatures. Only under
extreme conditions does a lake freeze completely to the bottom. If
the lake completely froze, the fish would die, first because they
could not move or feed but also because there would be
insufficient oxygen dissolved in the ice.
|
| A2. |
(b) |
1.022 g/mL
(c) 1.7% |
|
(d) |
rises up
(because of the greater density, less water has to be displaced to
support the weight of the ship)
|
| A3. |
(b) |
(i)
–0.9oC (ii) –11.0oC (iii)
18.5% (iv) 23.2% |
|
(c) |
–28oC.
Not very, because it involves an extrapolation beyond the range of
the experimental points. In fact the freezing point versus
composition curve has a decided break at 23%: the freezing point
then starts to increase as composition increases. There is a
freezing point minimum of –21.2oC at 23% NaCl. A 25%
solution has a freezing point considerably above 21oC
though there is no way of telling this from the data given. |
|
(d) |
Adding salt
to ice lowers the freezing point to a value below ambient air
temperature and so the ice melts.
|
| A4. |
As
the sea water is boiled it loses water so the concentration of
salt in the remaining water increases. Dissolved substances
increase the boiling point as was shown by Exercise 3 on Page 190
(they also decrease the freezing point – that exercise and
Exercise A3). Hence as concentration increases, the boiling point
increases. When pure water is boiled there is no change in
concentration, so boiling point remains constant at 100oC.
|
| A5. |
Yes.
Waste water flows to rivers and oceans where as part of the
natural water cycle pure water evaporates to form clouds that
produce rain that runs back into rivers and reservoirs to be used
again.
Primarily Sydney Water wants to delay as long as possible the need
to build new dams to meet the demands of a growing population
(because a new dam would cause considerable detrimental
environmental effects in the Sydney region). Another factor is to
reduce the cost of treating and piping increasing quantities of
water to households.
|
| B1. |
The structural formulae are:
 |
| B2. |
(b)
and (d);
(b) because O–Cl bonds are polar and the molecule is bent so the
two polar bonds 'add' up to give a net dipole.
(d) because As–F bonds are polar and the molecule is pyramidal
so the three polar bonds 'add' up to produce a net dipole
In (a) the bond is non-polar (same atoms on each end) so the
molecule is non-polar. In (c) the four bonds are non-polar and so
the molecule is non-polar. Even if the bonds were polar they would
cancel out to make a non-polar molecule because SiH4
has a symmetrical tetrahedral shape.
|
| B3. |
Cl2
< CHCl3 < NaCl
NaCl is an ionic lattice and so strong electrostatic attractions
between ions prevent ion pairs separating and forming a vapour so
boiling point is very high. Both chlorine and trichloromethane are
small covalent molecules so it is relatively easy to separate
molecules from one another (only weak intermolecular forces
holding molecules to one another). Chlorine is non-polar so there
are only weak dispersion forces present; trichloromethane is polar
(three polar C–Cl bonds adding up to give a net dipole) so there
are dipole–dipole forces as well as dispersion forces, so CHCl3
has a higher boiling point than chlorine.
|
| B4. |
BCl3
must have a symmetrical structure so that the three B–Cl dipoles
cancel out and make the molecule non-polar. PCl3 must
be non-symmetrical so that the three P–Cl dipoles do not cancel
out, leaving the molecule polar.
|
| C1. |
There
is hydrogen bonding in HF; this is the strongest type of
intermolecular force so it causes HF to have a high boiling point.
In neon there are only weak dispersion forces so its boiling point
is very low.
|
| C2. |
Methanol,
because there are hydrogen bonds between pairs of molecules. In
methanal there are only dipole–dipole interactions plus
dispersion forces which are much weaker.
|
| C3. |
It
is intermolecular forces that determine boiling points, not the
strength of chemical bonds within individual molecules. Hydrazine
has the higher boiling point because of strong hydrogen bonds that
are not present in phosphine (which has only weak dispersion
forces). (Actually the N–N bond in hydrazine is much weaker than
the P–H bonds in phosphine.)
|
| C4. |
There
are strong hydrogen bonds between pairs of H2O2
molecules but only weak dispersion forces in O2 so H2O2
has a much higher boiling point. In aqueous solution H2O2
forms hydrogen bonds with water and so it has a high solubility. O2
forms only weak dispersion forces with water so its solubility is
low.
|
| C5. |
(a) |
SCl2
is a small covalent molecule with relatively weak intermolecular
forces and so a low boiling point; MgCl2 is an ionic
lattice and so has a very high boiling point (no discrete
molecules in the solid). |
|
(b) |
There are
strong intermolecular forces in methanol because hydrogen bonding
occurs (H attached to O) but only weak intermolecular forces in CH3Br
(no hydrogen bonding) and so methanol has a much higher boiling
point.
|
| C6. |
(a) |
water >
ethylene glycol > ethanol = methanol > hexane > pentane |
|
(b) |
It appears
that provided molecular weights are not too different, surface
tension increases as intermolecular forces increase. Water,
ethylene glycol, ethanol and methanol with strong hydrogen bonding
have higher surface tensions than hexane and pentane (very
weak intermolecular forces).
For an explanation for how intermolecular forces give rise to
surface tension, see pages 192–3 and Figure 7.8.
|
| D1. |
Your
diagrams should show: |
|
(a) |
Mg2+
and SO42– ions scattered randomly through
the solution: no molecules. |
|
(b) |
neutral
molecules of urea scattered randomly through the solution: no
ions.
|
| D2. |
Soluble:
CaBr2 (ionic), ammonia (strong hydrogen bonds with
water or reacts with water), nitrous acid (strong hydrogen bonds
with water or reacts with water), glucose (strong hydrogen bonds
with water)
Insoluble: CF4 (covalent molecule), nickel
(metal: the only metals that dissolve in water are those that
react with it such as Na and Li), carborundum (covalent lattice),
neon (non-polar atom or molecule because in this case the atom is
a molecule)
|
| D3. |
NiCl2.6H2O,
nickel chloride hexahydrate; 2.63 g
|
| D4. |
MgCl2.6H2O
(1.24 g is 0.0130 mol MgCl2; 1.40 g H2O is
0.0778 mol, so 1 : 6)
|
| D5. |
(a) |
Water does
not wet steel so the surface tension of water pulls the needle out
of the water; this 'pull' is sufficient to overcome the downward
gravitational force resulting from the greater density of the
needle. |
|
(b) |
With
detergent in the water the surface tension is much less and so
cannot stop the needle sinking.
|
| D6. |
Ethanol
forms hydrogen bonds with water whereas dimethyl ether cannot.
(Remember that to form hydrogen bonds the hydrogen must be
attached to an O, N or F atom.)
|
| E1. |
(a) |
silver
iodide, AgI |
(f) |
no
precipitate |
|
(b) |
aluminium
hydroxide, Al(OH)3 |
(g) |
iron(II)
carbonate, FeCO3 |
|
(c) |
no
precipitate |
(h) |
silver
chloride, AgCl |
|
(d) |
lead
sulfide, PbS |
(i) |
zinc
hydroxide, Zn(OH)2 |
|
(e) |
barium
sulfate BaSO4 |
(j) |
iron(III)
sulfide, Fe2S3
|
| E2. |
(a) |
Ag+(aq)
+ I–(aq) ®
AgI(s)
The (aq)s in this and the following equations are not
completely necessary, but the (s) is. |
|
(b) |
Al3+(aq)
+ 3OH–(aq) ®
Al(OH)3(s) |
|
(c) |
no reaction |
|
(d) |
Pb2+(aq)
+ S2–(aq) ®
PbS(s) |
|
(e) |
Ba2+(aq)
+ SO42–(aq) ®
BaSO4(s) |
|
(f) |
no reaction |
|
(g) |
CO32–(aq)
+ Fe2+(aq) ®
FeCO3(s) |
|
(h) |
Ag+(aq)
+ Cl–(aq) ®
AgCl(s) |
|
(i) |
Zn2+(aq)
+ 2OH–(aq) ®
Zn(OH)2(s) |
|
(j) |
3S2–(aq)
+ 2Fe3+(aq) ®
Fe2S3(s)
|
| E3. |
(a) |
Ag+
+ NO3– + K+ + I–
®
AgI(s) + NO3– + K+
Pb2+ + 2NO3– + 2Na+
+ S2– ®
PbS(s) + 2NO3– + 2Na+
2Na+ + CO32– + Fe2+
+ SO42– ®
FeCO3(s) + 2Na+ + SO42–
You could put (aq)s on the reactant ions in these equations
but they are not completely necessary: however the (s)s
are. |
|
(b) |
AlCl3(aq)
+ 3NaOH(aq) ®
Al(OH)3(s) + 3NaCl(aq)
BaCl2(aq) + H2SO4(aq)
®
BaSO4(s) + 2HCl(aq)
3(NH4)2S(aq) + Fe2(SO4)3(aq)
®
Fe2S3(s) + 3(NH4)2SO4(aq)
|
| E4. |
Mix
solutions of |
|
(a) |
sodium (or
potassium or ammonium) sulfate and lead nitrate |
|
(b) |
magnesium
nitrate (or chloride or sulfate) and sodium (or potassium)
carbonate |
|
(c) |
copper
sulfate (or nitrate or chloride) and sodium (or ammonium) sulfide
and then filter off the precipitate and wash and dry it.
|
| F1. |
39.4
g/100 g
|
| F2. |
(a) |
no; only
18.5 g would dissolve: 3.5 g would remain as solid |
|
(b) |
yes; in 200
g water we could dissolve 74 g
|
| F3. |
(a) |
|
|
(b) |
(i)
(6 ± 0.5) g/100 g (ii) (39.8 ± 0.5) g/100 g |
|
(c) |
(i)
(43 ± 1)oC (ii) (96 ± 1) oC (±
2oC is acceptable) |
|
(d) |
yes,
because solubility at 90oC is 42 g/100 g or 84 g/200 g.
On cooling solid KClO3 would crystallise out; the
amount of solid formed would be 60 g, because the solubility at 10oC
is only 5 g/100 g or 10 g in 200 g.
|
| F4. |
|
0.016 mol.
In both experiments, mixing solutions to form a precipitate and
dissolving solid to form a saturated solution, an equilibrium is
set up
Ca2+(aq) + 2OH–(aq)  Ca(OH)2(s)
The equilibrium position is the same, regardless of the direction
from which it is approached. In the precipitation experiment we
have 0.016 mol Ca2+ (and so 0.032 mol OH–)
in equilibrium with solid Ca(OH)2, so if we start with
solid Ca(OH)2 and 100 g pure water we will end up with
the same 0.016 mol Ca2+ in solution at equilibrium.
It takes a long time for a solid to reach equilibrium with a
solution to make it saturated. It is only at equilibrium that we
can make any predictions.
|
| F5. |
(a) |
A, C, D, F
are electrolytes; D and E are non-electrolytes. A and C are strong
electrolytes, D and F are weak. |
|
(b) |
A and C
present as ions; D and F mostly as molecules but with a few ions,
B and E as molecules only.
|
| F6. |
CH3CO2H
+ H2O
H3O+ + CH3CO2–
NH3 + H2O NH4+
+ OH–
HNO3 + H2O ®
H3O+ + NO3–
The reversible arrow shows that the first two reactions do not go
to completion: that is, only some of the molecules react to form
ions. The forward arrow shows that the last reaction goes to
completion.
The first two reactions are equilibrium reactions.
|
| F7. |
HNO2
+ NaOH ®
H2O + NaNO2 or
HNO2 + OH– ®
H2O + NO2–
(a) is weak, (b) is strong; because 90 is not much less than
150 (conductivity of salts is usually lower than that of acids or
alkalis) whereas 12 is very much less than 150, indicating that
only a few of the molecules must have ionised (reacted with
water).
In nitrous acid there are mainly HNO2 molecules with
only a few ions H+ and NO2–
ions . In sodium nitrite there are only Na+ and NO2–
ions present.
|
| G1. |
26.5% |
| G2. |
(a) |
360 g/kg,
26.5%(w/w) |
|
(b) |
0.220 g/g,
18.0%(w/w) |
|
(c) |
For (a) 360
g/L; for (b) 220 g/L |
| G3. |
(a) |
500 g |
(b) |
500 g |
(c) |
310 g/L |
| G4. |
(a) |
10 g |
(b) |
3.8
g |
| G5. |
(a) |
4.92 g/mL;
49.2 g/L |
(b) |
0.98
g/L; 980 ppm |
| G6. |
(a) |
0.024 g/L |
(b) |
0.13
g |
| G7. |
0.096%;
5.0 mL
|
|
|
| H1. |
(a) |
0.024 mol/L |
(c) |
4.96
mol/L |
|
(b) |
0.856 mol/L |
(d) |
0.0240
mol/L |
| H2. |
(a) |
0.125 mol |
(b) |
0.0355 mol |
(c) |
5.00 g;
0.604 g |
| H3. |
(a) |
5.65 g |
(b) |
1.54
g |
| H4. |
(a) |
7.76 x 10–3
mol |
(c) |
5.83
x 10–5 mol |
|
(b) |
4.11 x 10–4
mol |
(d) |
0.375
mol |
| H5. |
(a) |
1.55 x10–2
mol |
(b) |
8.22 x 10–4
mol |
(c) |
1.75 x 10–4
mol |
| H6. |
In
(a) 0.823 g; in (c) 7.77 x 10–3 mol |
| H7. |
(a) |
3.33 x 10–3
mol |
(c) |
(i)
1.54 g (ii) 0.606 g |
|
(b) |
0.0104 mol |
|
|
| H8. |
0.0281
mol |
|
|
| H9. |
0.609
mol/L; 3.56%(w/v) |
| H10. |
0.0175
mol/L; 0.197 g/100 mL; 1970 ppm
|
| J1. |
(a) |
4.9 x 103
J |
(b) |
1.6 x 103
J |
| J2. |
Water
has a much higher specific heat capacity than ethanol (4.14
compared with 1.94) so the 10 mL water gives out much more heat in
cooling from 60oC to body temperature than does ethanol
and so there is a greater burning sensation. |
| J3. |
28.6oC |
| J4. |
25.7oC
|
| K1. |
–39
kJ/mol; NaOH(s) ®
NaOH(aq) |
| K2. |
59.4oC |
| K3. |
Exercise
34(b) on page 241 showed that dissolving sodium thiosulfate was
endothermic. This means that when sodium thiosulfate crystallises
out of solution the process is exothermic (heat released). Hence
breaking the supersaturated solution (causing thiosulfate to
crystallise out) releases heat and so the pack becomes hot.
Put the pack into hot water (80 to 90oC) for some time
until the crystals re-dissolve then cool the pack slowly to room
temperature.
Exercise 34(a) showed that dissolution of calcium chloride is
exothermic so mixing the solid with water causes dissolution to
occur and so heat is released. This pack is not re-usable because
there is no easy of separating the solid from the solution.
|
| K4. |
(a)
16.1oC |
|
(b) |
Mixing
would not be very efficient. The water near the outlet would get
much hotter than 16.1oC while that at considerable
distance away would warm up only slightly. Much of the heat
transferred to the lake would be lost to the atmosphere especially
if there were regular breezes across the surface of the lake.
The significant rise in temperature near the outlet would have
considerable environmental effects: it would accelerate growth of
water plants and other aquatic organisms and could damage the
breeding cycles of fish, some of which are very sensitive to
temperature. |
|