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GCSE Questions
and Answers
Calculations
6 consecutive GCSE Chemistry papers: 2002-7
1
I
1
H
1
Relative Atomic Masses you’ll need.
II
 Atomic Number at the top
 Relative Atomic Mass at the bottom
Ammonium
3 4
Li Be
7 9
11 12
Na Mg
23 24
III
IV
V
VI
VII VIII
2
He
4
NH4+
Hydroxide
Nitrate
Hydrogencarbonate
Hydrogen sulphate
OHNO3HCO3HSO4-
Carbonate
Sulphate
CO32SO42-
5 6 7 8 9 10
B C N O F Ne
11 12 14 16 19 20
13 14 15 16 17 18
Al Si P S Cl Ar
27 28 21 32 35.5 40
19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36
K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr
39 40 45 48 51 52 55 56 59 59 64 65 70 73 75 79 80 84
2
Covalent elements and Compounds
Covalent Elements: Covalent Compounds
Hydrogen
Chlorine
Bromine
Iodine
Oxygen
Nitrogen
Helium
Neon
Argon
H2
Cl2
Br2
I2
O2
N2
He
Ne
Ar
Water
H2O
Carbon dioxide
CO2
Carbon monoxide
CO
Sulphur dioxide
SO2
Sulphur trioxide
SO3
Ammonia
NH3
Hydrogen peroxide H2O2
Nitrogen monoxide NO
Nitrogen dioxide
NO2
Sulphur
Phosphorus
S8
P4
3
2002, Paper 1
To obtain full marks in this question, you must show
your working out
5
When washing soda crystals,
Na2CO3.10H2O, are left exposed to the
atmosphere they lose water of
crystallisation. The longer they are left,
the more water is lost. The amount of
water of crystallisation remaining can be
found in two ways: either by heating to
remove all the remaining water or by
titration.
5a) 2.675g of a sample of crystals were
heated to constant mass. The mass of
the residue was 1.325g.
(i)
Why was the sample heated to constant
mass?
______________________________ [1]
(ii) Calculate the number of moles of
anhydrous sodium carbonate in the
residue.
______________________________ [2]
Consequential marking applies throughout
5a)
(i)
(ii)
2.675g of a sample of crystals were heated to
constant mass. The mass of the residue was
1.325g.
Why was the sample heated to constant mass?
To ensure that all water [1] (of crystallisation)
was lost.
Calculate the number of moles of anhydrous
sodium carbonate in the residue.
Na2CO3 = 1.325 = 0.0125 [1]
106 [1]
(iii) Calculate the mass of water lost and from
this calculate the number of moles of
water lost.
_________________________________
______________________________ [3]
(iv) From your answers to part (a)(ii) and (iii)
above, calculate the value of x in the
formula Na2CO3.xH2O.
_________________________________
______________________________ [2]
(iii) Calculate the mass of water lost and from
this calculate the number of moles of water lost.
Mass of water lost = (2.765-1.325) g
= 1.35g [1]
Number of moles lost = 1.35 = 0.075 [1]
18 [1]
(iv) From your answers to part (a)(ii) and (iii) above,
calculate the value of x in the formula
Na2CO3.xH2O.
Ratio of moles Na2CO3 : H2O
0.0125 : 0.075 [1]
1
: 6 [1]
b)
1.775g of a different sample of washing soda was
dissolved in distilled water and made up of a total
volume of 250cm3. 25.0cm3 of this solution were
titrated with 0.08 mol/dm3 (moles per litre) nitric
acid. 31.25cm3 of acid were required. The
equation for the reaction is:
Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2
(i)
Calculate the number of moles of nitric acid used
in the titration.
______________________________________
___________________________________ [2]
b)
1.775g of a different sample of washing soda was
dissolved in distilled water and made up of a total
volume of 250cm3. 25.0cm3 of this solution were
titrated with 0.08 mol/dm3 (moles per litre) nitric
acid. 31.25cm3 of acid were required. The
equation for the reaction is:
Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2
(i)
Calculate the number of moles of nitric acid used
in the titration.
Number of moles HNO3 = 31.25 x 0.08
1000[1]
= 2.5 x 10-3
(ii)
Calculate the number of moles of sodium
carbonate present in the 25.0cm3 sample.
_______________________________________
____________________________________[2]
(iii) Calculate the number of moles of sodium
carbonate present in 250cm3 of solution.
_______________________________________
____________________________________[2]
(ii)
Calculate the number of moles of sodium
carbonate present in the 25.0cm3 sample.
Na2CO3 + 2HNO3 → 2NaNO3 + H2O + CO2
mole ratio 1 : 2 [1]
Number of moles Na2CO3 = 2.5 x 10-3
2
= 1.25 x 10-3
(iii) Calculate the number of moles of sodium
carbonate present in 250cm3 of solution.
Number of moles Na2CO3 in 250cm3
= 1.25 x 10-3 x 10 [1]
= 1.25 x 10-2 [1]
(iv) Using your answer to part (b)(iii) and the
mass of sodium carbonate crystals
weighed out find the value of x in the
formula Na2CO3.xH2O.
_________________________________
______________________________[2]
(iv) Using your answer to part (b)(iii) and the
mass of sodium carbonate crystals
weighed out find the value of x in the
formula Na2CO3.xH2O.
Number of moles = moles
rfm
rfm = mass =
1.775
no. of moles
1.25 x 10-2
= 142 [1]
rfm Na2CO3 = 106
rfm x H2O = 142 – 106 =36 [1]
x = 36 = 2 [1]
18
c)
Sodium hydrogencarbonate decomposes when it
is heated into sodium carbonate according to the
equation: 2NaHCO3 → NaCO3 + H2O + CO2
1.68g of sodium hydrogencarbonate were placed
in a test tube and heated in a Bunsen flame for
some time.
(i)
Calculate the number of moles of sodium
hydrogencarbonate used.
_______________________________________
____________________________________ [2]
c)
Sodium hydrogencarbonate decomposes when it
is heated into sodium carbonate according to the
equation: 2NaHCO3 → NaCO3 + H2O + CO2
1.68g of sodium hydrogencarbonate were placed
in a test tube and heated in a Bunsen flame for
some time.
(i)
Calculate the number of moles of sodium
hydrogencarbonate used.
Number of moles NaHCO3 = 1.68
84 [1]
= 0.02 [1]
(ii) Calculate the number of moles of sodium
carbonate formed.
___________________________________
________________________________ [2]
(iii) Calculate the mass of sodium carbonate
expected to be formed.
___________________________________
________________________________ [2]
(ii) Calculate the number of moles of sodium
carbonate formed.
Mole ratio NaHCO3 : Na2CO3
2
:
1 [1]
number of moles Na2CO3 formed
= 0.02 = 0.01 [1]
2
(iii) Calculate the mass of sodium carbonate
expected to be formed.
Mass Na2CO3 expected = 0.01 x 106 [1]
= 1.06g [1]
(iv)
Calculate the volume of carbon dioxide
produced in this experiment.
(1 mole of gas
occupies 24dm3 at room temperature and
pressure)
_____________________________________
__________________________________ [2]
(iv)
Calculate the volume of carbon dioxide
produced in this experiment.
(1 mole of gas occupies 24dm3 at room
temperature and pressure)
Mole ratio NaHCO3 : CO2
2
: 1
[1]
Number of moles CO2 expected
= 0.02 = 0.01 [1]
Volume of CO2 expected
= 24 x 0.01
= 0.24dm3 [1]
or 240cm3
2003, Paper 1
4
To obtain full marks in this question, all steps in the
calculation must be shown.
a)
Copper carbonate, CuCO3 decomposes on heating
according to the following equation:
CuCO3 → CuO + CO2
(Relative atomic masses: C = 12; O = 16; Cu = 64)
The following results were obtained in an experiment in
which a sample of copper carbonate was to be heated.
Mass of empty crucible = 24.21g
Mass of crucible and sample of copper carbonate = 27.31g
(i)
What colour change would you observe on
heating the sample of copper carbonate?
_______________________________ [2]
(ii) Calculate the mass of copper carbonate in
the crucible.
_______________________________ [1]
(i)
What colour change would you observe on
heating the sample of copper carbonate?
Green [1] to black [1]
[2]
(ii) Calculate the mass of copper carbonate in
the crucible.
27.31 – 24.21 = 3.1g
[1]
(iii) Calculate the mass of solid copper
oxide, CuO, which you would expect
to obtain from the complete
decomposition of this sample of
copper carbonate.
(iii) Calculate the mass of solid copper oxide,
CuO, which you would expect to obtain
from the complete decomposition of this
sample of copper carbonate.
3.1 = 0.025 [1] moles of CuCO3
124[1]
CuCO3 : CuO 1:1 [1]
0.025 [1] moles of CuO
0.025 x 80 [1] = 2.0 [1]g
(iv) How would you ensure all the copper
carbonate had decomposed?
_______________________________ [1]
(iv) How would you ensure all the copper
carbonate had decomposed?
Heat to constant mass
[1]
b)
The experiment was carried out, heating
the sample for 3 minutes. Not all of the
copper carbonate had decomposed in this
time. The results obtained are shown
below:
Mass of empty crucible
= 24.21g
Mass of crucible and sample of copper carbonate
= 27.31g
Mass of crucible and solid after heating for 3 minutes = 26.43g
(i)
What mass of solid remained after heating
for 3 minutes?
_______________________________ [2]
(ii) Calculate the mass of carbon dioxide gas
lost in this experiment.
_______________________________ [2]
(i)
What mass of solid remained after heating
for 3 minutes?
26.43 – 24.21 [1] = 2.22 [1]g
(ii) Calculate the mass of carbon dioxide gas
lost in this experiment.
27.31 – 26.43 [1] = 0.88 [1] g
(iii) Calculate the number of moles of carbon
dioxide gas lost in this experiment.
_______________________________ [2]
(iv) Calculate the mass of copper carbonate
which must have been decomposed in this
experiment.
__________________________________
_______________________________ [ ]
(iii) Calculate the number of moles of carbon
dioxide gas lost in this experiment.
0.88
44[1]
= 0.02 [1] moles of CO2
(iv) Calculate the mass of copper carbonate
which must have been decomposed in this
experiment.
CO2 : CuCO3 = 1 : 1
0.02 [1] moles of CuCO3 decomposed
0.02 x 124 = 2.48 [1]g of CuCO3
decomposed
v)
Using your answer to parts (a)(ii) and
(b)(iv), calculate the percentage of copper
carbonate in the original sample which
must have been decomposed in this
experiment.
[2]
v)
Using your answer to parts (a)(ii) and
(b)(iv), calculate the percentage of copper
carbonate in the original sample which
must have been decomposed in this
experiment.
2.48
3. 1
x 100 [1] = 80%
[2]
c)
Copper carbonate reacts with sulphuric
acid and crystals of hydrated copper
sulphate, CuSO4.5H2O, can be obtained.
Calculate the percentage of water of
crystallisation in hydrated copper
sulphate, CuSO4.5H2O.
(Relative atomic masses: H=1; O=16; Cu=64)
[3]
c)
Copper carbonate reacts with sulphuric acid and
crystals of hydrated copper sulphate,
CuSO4.5H2O, can be obtained.
Calculate the percentage of water of
crystallisation in hydrated copper sulphate,
CuSO4.5H2O.
(Relative atomic masses: H=1; O=16; Cu=64) [3]
CuSO4.5H2O RFM = 250 [1]
Mass of water = 18 x 5 = 90 [1]
Percentage water = 90 x 100 =36% [1]
250
d)
Carbon dioxide reacts with excess sodium
hydroxide solution according to the following
equation:
CO2 + 2NaOH → Na2CO3 + H2O
(1 mole of any gas at room temperature and
pressure occupies 24 000cm3)
120cm3 of carbon dioxide gas is passed into
150cm3 of 0.1mol/dm3 moles per litre.mol/l
sodium hydroxide solution.
(i)
Calculate the number of moles in 120cm3 of
carbon dioxide gas.
_____________________________________
__________________________________ [2]
(i)
Calculate the number of moles in 120cm3 of
carbon dioxide gas.
120
24000
[1] = 0.005 [1] moles CO2
(ii)
Calculate the number of moles of sodium
hydroxide needed to react with this amount of
carbon dioxide.
_____________________________________
__________________________________ [2]
(ii) Calculate the number of moles of sodium
hydroxide needed to react with this
amount of carbon dioxide.
CO2 : NaOH [1]
0.005 moles x 2 = 0.01 moles NaOH
needed [1]
(iii) Calculate the number of moles in 150cm3 of
0.1mol/dm3 sodium hydroxide solution.
_____________________________________
__________________________________ [2]
(iii) Calculate the number of moles in 150cm3
of 0.1mol/dm3 sodium hydroxide solution.
150x0.1
1000
[1] = 0.015 [1] moles NaOH
[2]
(iv) How many moles of sodium hydroxide are left
at the end of the reaction?
_____________________________________
__________________________________ [2]
(iv) How many moles of sodium hydroxide are
left at the end of the reaction?
Moles NaOH needed to react with CO2 =
2 x 0.005 = 0.01
Moles NaOH remaining = 0.015 – 0.01
[1] = 0.005 [1]
e)
Sodium hydroxide can be neutralised by
hydrochloric acid according to the equation.
NaOH + HCl → NaCl + H2O
Using your answer to part (d)(iv) calculate the
number of 0.5 mol/dm3 hydrochloric acid
which would be required to neutralise the
sodium hydroxide left at the end of the
reaction.
_____________________________________
__________________________________ [4]
e)
Sodium hydroxide can be neutralised by
hydrochloric acid according to the equation.
NaOH + HCl → NaCl + H2O
Using your answer to part (d)(iv) calculate the
number of 0.5 mol/dm3 hydrochloric acid which
would be required to neutralise the sodium
hydroxide left at the end of the reaction.
NaOH : HCl = 1.1 [1]
moles HCl required = 0.005 [1]
volume HCl required =
0.005x1000
0.5
[1] = 10 [1] cm3
2004, Paper 1
5
To obtain full marks in this question, all
steps in the calculation must be shown.
a)
In 1908 a German chemist called Fritz
Haber succeeded in combining nitrogen
with hydrogen to form ammonia.
N2 + 3H2 → 2NH3
Calculate the volume of nitrogen gas,
measured at room temperature and
pressure, needed to produce 10dm3 of
ammonia.
Calculate the volume of nitrogen gas,
measured at room temperature and
pressure, needed to produce 10dm3 of
ammonia.
1 volume of nitrogen needs 2 volumes of
ammonia [1] hence 10dm3 needs
5dm3/4.99 [1] dm3
b)
A concentrated solution of ammonia can be used
as a fertiliser. To determine the concentration of
the ammonia it was first diluted by measuring
10.0cm3 and making the volume up to 1dm3
(1000cm3).
A 25.0cm3 sample of this dilute ammonia solution
was then titrated against 0.05 mol/dm3 (moles per
litre) sulphuric acid. The 25.0 cm3 of diluted
ammonia required 12.5cm3 of the acid for
neutralisation.
The equation for the titration is
2NH3 + H2SO4 → (NH4)2SO4
(i) Calculate the number of moles of
sulphuric acid used in the titration.
(ii) Calculate the number of moles of
ammonia in the 25.0cm3 sample
which reacted with the acid.
(i)
Calculate the number of moles of
sulphuric acid used in the titration.
Number of moles of sulphuric acid =
12.5 x0.05
1000 = 0.000625
[2]
(ii) Calculate the number of moles of
ammonia in the 25.0cm3 sample which
reacted with the acid.
From equation (1:2) 1 mole sulphuric acid reacts
with 2 moles of ammonia [1]
0.000625 moles acid reacts with 0.00125 moles
ammonia [1]
(iii) Calculate the concentration of the
dilute ammonia solution in mol/dm3
(moles per litre).
(iv) Calculate the concentration of the
original concentrated ammonia
solution in mol/dm3 (moles per litre).
(iii) Calculate the concentration of the dilute ammonia
solution in mol/dm3 (moles per litre).
Concentration of ammonia =
0.00125x1000
25
= 0.05 [1] mol/dm3
(iv) Calculate the concentration of the original
concentrated ammonia solution in mol/dm3
(moles per litre).
Diluted 100 times [1]
original conc = 0.05 x 100 = 5 [1] mol/dm3
(v) Calculate the concentration of the
original concentrated ammonia
solution in g/dm3
(v) Calculate the concentration of the
original concentrated ammonia
solution in g/dm3
RFM of NH3 = 17[1]
conc g/dm3 = 5 x 17[1]
= 85[1] g/dm3
c)
Solid fertilisers are easier to store,
hence fertilisers like solid ammonium
chloride are preferred over ammonia
solution. To produce ammonium
chloride, ammonia is reacted with
hydrochloric acid, according to the
equation below.
NH3 + HCl → NH4Cl
What mass of ammonium chloride is
formed when 73g of hydrochloric acid are
completely neutralised by ammonia?
(Relative atomic masses: H=1, N=14,
Cl=35.5
What mass of ammonium chloride is formed when 73g
of hydrochloric acid are completely neutralised by
ammonia?
(Relative atomic masses: H=1, N=14, Cl=35.5
RFM of HCl = 36.5 [1]
moles of HCl = 73/36.5 = 2[1]
(ratio 1:1 hence) moles NH4Cl = 2[1]
RFM of NH4Cl = 53.5[1]
2 x 53.5 = 107g[1]
d)
Another important fertiliser made from
ammonia is urea. It contains 20.00%
carbon, 6.66% hydrogen, 46.67% nitrogen
and 26.67% oxygen. Calculate the
formula of urea.
(Relative atomic masses: H=1, C=12,
N=14, O=16)
d)
Another important fertiliser made from
ammonia is urea. It contains 20.00% carbon,
6.66% hydrogen, 46.67% nitrogen and 26.67%
oxygen. Calculate the formula of urea.
(Relative atomic masses: H=1, C=12, N=14, O=16)
In 100g of compound there are:
20/12 moles C = 1.67 [1]
6.67/1 moles H = 6.67 [1]
46.67/14 moles N = 3.33[1]
26.67/16 moles O = 1.67[1]
formula is CH4N2O [1]
accept any correct whole number multiple.
e)
Car exhaust fumes contain harmful
nitrogen monoxide gas. Research has
shown that when a stream of ammonia gas
is injected into the hot exhaust a reaction
occurs which converts the harmful
nitrogen monoxide, NO, to nitrogen gas
according to the equation below.
6NO + 4NH3 → 5N2 + 6H2O
(i) How many moles of ammonia would
be needed to react with 0.6 moles of
nitrogen monoxide, NO?
___________________________ [1]
(i) How many moles of ammonia would
be needed to react with 0.6 moles of
nitrogen monoxide, NO?
0.4 moles
[1]
(ii) The average car emits 0.033 moles of
nitrogen monoxide per km. How
many moles of ammonia would be
needed to convert this to N2 gas?
___________________________ [2]
(ii) The average car emits 0.033 moles of
nitrogen monoxide per km. How
many moles of ammonia would be
needed to convert this to N2 gas?
6 moles NO: 4 moles
NH3/0.033moles NO: 0.033/6x4 [1]
= 0.022 [1] moles per km
(iii) Using your answer to (e)(ii) calculate
the mass of ammonia needed to
convert 0.033 moles of NO to N2 gas.
___________________________ [2]
(iii) Using your answer to (e)(ii) calculate
the mass of ammonia needed to
convert 0.033 moles of NO to N2 gas.
0.022 x 17[1] = 0.374g [1]
2005, Paper 2
2
Lead is extracted from the ore galena, PbS.
a)
The ore is roasted in air to produce lead(II) oxide,
PbO.
2PbS(s) + 3O2(g) → 2PbO(s) + 2SO2(g)
(Relative Atomic Masses: Pb=207, S=32, O=16)
(i)
Calculate the mass of lead(II) oxide, PbO,
produced from 2390g of galena, PbS. (Show all
steps in your calculations.
[5]
(i)
Calculate the mass of lead(II) oxide, PbO,
produced from 2390g of galena, PbS. (Show all
steps in your calculations.
[5]
RFM PbS = 207+32 = 239 [1]
2390
Moles PbS = 239 = 10 [1]
Moles PbS = 10 [1]
RFM PbO = 270+16 = 223 [1]
Mass PbO = 10x223 = 2230g or 2.23kg [1]
The lead(II) oxide is reduced to lead by heating it with
carbon in a blast furnace.
PbO(s) + C(s) → Pb(l) + CO(g)
The molten lead is tapped off from the bottom of the
furnace.
(ii)
Using your answer to part (a)(i), calculate the
mass of lead that would eventually be produced.
(ii)
Using your answer to part (a)(i), calculate the
mass of lead that would eventually be produced.
PbO: Pb = 1:1 [1]
Moles Pb = 10 [1]
Mass Pb = 10 x 270 = 2070g or 2.07kg [1]
b)
Lead metal forms several oxides. The formula of
lead oxide may be represented as PbxOy.
In an experiment to find the formula of a sample
of lead oxide, a porcelain dish was weighed and
the mass recorded. The porcelain dish was then
filled with the lead oxide and reweighed. The
mass was again recorded.
The dish was placed in a hard-glass tube and heated in a
stream of hydrogen gas. The hydrogen reduced all of
the lead oxide to a bead of silvery lead metal. The
apparatus was allowed to cool and the dish and its
contents were reweighed.
(i)
Calculate the mass of lead metal produced.
_____________________________________ [1]
(ii)
Calculate the mass of oxygen present in the lead
oxide.
_____________________________________ [1]
(i)
(ii)
Calculate the mass of lead metal produced.
27.56 – 21.35 = 6.21g
[1]
Calculate the mass of oxygen present in the lead
oxide.
28.20 – 27.56 = 0.64g
[1]
(iii) Using your answers to (i) and (ii),
calculate the formula of the sample of
lead oxide.
(Relative atomic masses: Pb=207, O=16)
(iii) Using your answers to (i) and (ii),
calculate the formula of the sample of
lead oxide.
(Relative atomic masses: Pb=207, O=16)
Moles Pb =
= 0.03
[1]
Moles O =
= 0.04
[1]
Ratio 3:4 so formula is Pb3O4
[1]
c)
Titration is a
technique used by
chemists to find
the concentration
of a solution. The
apparatus used in a
titration is shown
opposite.
(i)
Identify the pieces of apparatus A and B.
A is a ________________________________ [1]
B is a ________________________________ [1]
(ii)
Describe in detail, stating precautions to ensure
safety and accuracy, how you would transfer
25.0cm3 of an alkali into the conical flask using
the piece of apparatus A.
_____________________________________
_____________________________________
_____________________________________
_____________________________________ [3]
(i)
(ii)
Identify the pieces of apparatus A and B.
A is a pipette
B is a burette
[1]
[1]
Describe in detail, stating precautions to ensure
safety and accuracy, how you would transfer
25.0cm3 of an alkali into the conical flask using
the piece of apparatus A.
Rinse with deionised water [1] rinse with alkali
[1] use safety pipette filler/safety goggles [1] to
draw up liquid until bottom of meniscus on line
[1] release [1] into conical flask touch tip of
pipette to surface of alkai [1] (Max [3])
(iii) Describe in detail, stating precautions to
ensure accuracy, the steps you would take
to prepare the piece of apparatus B for use
in a titration.
_______________________________
_______________________________
_______________________________
_______________________________ [4]
(iii) Describe in detail, stating precautions to
ensure accuracy, the steps you would take
to prepare the piece of apparatus B for use
in a titration.
Rinse with deionised water [1] rinse with
solution [1] fill burette with solution [1]
use funnel [1] ensure jet is filled [1] ensure
no air bubbles [1] (Max [4])
d)
Limewater is calcium hydroxide solution. In a
titration to find the concentration of calcium
hydroxide in limewater, 25.0cm3 of limewater
required 16.4cm3 of hydrochloric acid of
concentration 0.040 mol/dm3 for neutralisation.
(Relative atomic masses: Ca=40, O=16, H=1)
Ca(OH)2 + 2HCl → CaCl2 + 2H2O
(i)
Calculate the concentration of the calcium
hydroxide in mol/dm3 (mol per litre).
Answer __________________ mol/dm3 [4]
(ii)
Calculate the concentration of the calcium
hydroxide in g/dm3 (grams per litre).
Answer __________________ mol/dm3 [4]
(i)
Calculate the concentration of the calcium
hydroxide in mol/dm3 (mol per litre).
Answer
16.4 x0.04
1000
= 0.000656 [1]
Moles Ca(OH)2 =
0.000328
25
(ii)
0.000656
2
0.000328 [1]
x 1000 [1] = 0.01312 mol/dm3
Calculate the concentration of the calcium
hydroxide in g/dm3 (grams per litre).
Answer Mass = mol x RFM
= 0.01312 x 74 [1]
= 0.971 [1] g/dm3
2006, Paper 2
7a) Oxygen forms ozone gas, O3, in the upper
atmosphere according to the equation:
3O2(g) → 2O3(g)
150m3 of oxygen reacts completely to
form ozone.
(i)
State Avogadro’s Law
_______________________________
_______________________________
_______________________________ [3]
(i)
State Avogadro’s Law
Equal volumes of gas [1] under the same
conditions of temperature and pressure [1]
contain the same number of particles [1] or
moles of ozone = 4.2 moles
[3]
(ii) Using Avogadro’s Law or otherwise,
calculate the volume of ozone gas
produced.
____________________________m3 [2]
(ii) Using Avogadro’s Law or otherwise,
calculate the volume of ozone gas
produced.
150 x2[1] 100[1]m3
3
[2]
b)
In the laboratory ozone gas can be produced by
passing an electrical discharge through dry air.
450cm3 of ozone gas is produced when the
temperature is 300K and the pressure is 1
atmosphere.
The ozone gas is compressed using a pressure of
8 atmospheres and the temperature is decreased to
200K. Calculate the volume of ozone gas under
these new conditions.
_______________________________cm3 [4]
b)
Calculate the volume of ozone gas under these
new conditions.
P1V1 P2V2

[1]
T1
T2
8
xV
1x450 
2 [1]
300
200
1x450 x200
V2
[1]
300 x8
3
V
2

37
.
5
[
1
]
cm
_______________________________
[4]
c)
Ozone is used in very small amounts in
underground railway stations to remove
compounds which cause stations to be
stuffy. One of the compounds which is
formed is formaldehyde CH2O.
Calculate the percentage by mass of
carbon in CH2O.
____________________________% [3]
c)
Calculate the percentage by mass of
carbon in CH2O.
RFM (CH2O) = 30 [1]
% carbon =
12
x100
30
[1] = 40% [1]
[3]
d)
1.92g of sulphur dioxide SO2, reacts
completely with ozone to form 2.40g of
sulphur trioxide, SO3.
(i)
Calculate the number of moles of sulphur
dioxide used.
_______________________________ [2]
d)
1.92g of sulphur dioxide SO2, reacts
completely with ozone to form 2.40g of
sulphur trioxide, SO3.
(i)
Calculate the number of moles of sulphur
dioxide used.
RFM (SO2) = 64 [1]
Moles =
1.92
64
= 0.03 [1]
[2]
(ii) Calculate the mass of ozone which reacts.
_______________________________ [1]
(iii) Calculate the number of moles of ozone
which reacts.
_______________________________ [2]
(ii) Calculate the mass of ozone which reacts.
2.4 – 1.92 = 0.48g
[1]
(iii) Calculate the number of moles of ozone
which reacts.
RFM (O3) = 48[1]
Moles =
0.48
48
= 0.01 [1]
[2]
(iv) Calculate the number of moles of sulphur
trioxide formed.
_______________________________ [2]
(iv) Calculate the number of moles of sulphur
trioxide formed.
RFM (SO3) = 80[1]
2.4
Moles =
80
= 0.03[1]
[2]
(v) Using your answers to (i), (iii) and (iv) or
otherwise, balance the symbol equation for
the reaction.
Equation: SO2 + O3 → SO3
[1]
(v) Using your answers to (i), (iii) and (iv) or
otherwise, balance the symbol equation for
the reaction.
Equation: SO2 + O3 → SO3
[1]
Ratio: SO2 : O3 : SO3 = 0.03 : 0.01 : 0.03 = 3 : 1 : 3
Equation: 3SO + O3 → 3SO2 balancing numbers = [1]
e)
Oxygen gas is prepared by the
decomposition of hydrogen peroxide
solution using the catalyst manganese(IV)
oxygen.
2H2O2 → 2H2O + O2
A solution of hydrogen peroxide is
labelled 0.1mol/dm3 (moles per litre).
25.0cm3 of this solution is decomposed
completely using manganese(IV) oxide.
(i)
What is meant by the term catalyst?
_______________________________
_______________________________ [3]
(i)
What is meant by the term catalyst?
Substance which speeds up/increases the
rate of [1] a (chemical) reaction [1]
without being used up/chemically
unchanged at the end [1]
[3]
(ii) Calculate the volume of oxygen gas
produced in this decomposition. State the
units.
(1 mole of any gas occupies a volume of 24dm3)
__________________________________ [7]
(ii) Calculate the volume of oxygen gas
produced in this decomposition. State the
units.
(1 mole of any gas occupies a volume of 24dm3)
25.0 x 0.1
Moles of H2O2 = 1000  0.0025[1]
Ratio: H2O2 : O2 = 2 : 1 [1]
0.0025
[1]  0.00125[1]
Moles of O2 =
2
Volume of oxygen = 0.00125 x 24 [1] = 0.03 [1]dm3
[1]
(or volume of oxygen = 0.00125 x 24000 [1] = 30[1]cm3[1]
2007, Paper 1
5
Borax is a salt which is hydrated and is used in
cleaning agents. The formula may be represented
by Na2B4O7.xH2O. Borax dissolves in water to
give a solution which acts as a weak alkali.
a)
4.775g of Borax were weighed out and made up
to a volume of 250cm3 with deionised water.
25.0cm3 portions of this solution were titrated
against nitric acid of concentration 0.094 mol/dm3
(moles per litre). The results were recorded in the
table below.
Rough Titration
1st Accurate Titration
2nd Accurate Titration
Initial
burette
reading
(cm3)
0.0
0.0
0.0
Final
burette
reading
(cm3)
26.9
26.7
26.5
Volume of
nitric acid
used (titre)
(cm3)
26.9
26.7
26.5
Rough Titration
1st Accurate Titration
2nd Accurate Titration
(i)
Initial
burette
reading
(cm3)
0.0
0.0
0.0
Final
burette
reading
(cm3)
26.9
26.7
26.5
Volume of
nitric acid
used (titre)
(cm3)
26.9
26.7
26.5
Calculate the average titre.
_______________________________ [2]
(i)
Calculate the average titre.
26.2 [2]. If rough used average = 26.7
award [1]
[2]
(ii) The indicator used was methyl orange.
State the colour change of the indicator in
this titration.
From ____________ to ____________ [2]
(ii) The indicator used was methyl orange.
State the colour change of the indicator in
this titration.
From orange or yellow [1] to pink or red
[1] (wrong way round) = [1]
[2]
(iii) Calculate the number of moles of nitric
acid used in this titrations.
_______________________________ [2]
(iii) Calculate the number of moles of nitric
acid used in this titrations.
Moles =
26.6 x 0.094
1000
[1] = 0.0025 [1] (2.5 x 10-3)
The equation for the reaction is:
Na2B4O7 + 2HNO3 + 5H2O → 2NaNO3 + 4H3BO3
(iv) Use the equation to deduce the number of moles
of Borax which reacted with the nitric acid.
__________________________________
_____________________________________ [2]
The equation for the reaction is:
Na2B4O7 + 2HNO3 + 5H2O → 2NaNO3 + 4H3BO3
(iv) Use the equation to deduce the number of moles
of Borax which reacted with the nitric acid.
0.0025
1 mole borax : 2 moles nitric acid or 2
= 0.00125 [1] (1.25 x 10-3) [2]
(v) Calculate the concentration of the Borax in
mol/dm3 (moles per litre)
_______________________________
_______________________________ [2]
(v) Calculate the concentration of the Borax in
mol/dm3 (moles per litre)
conc =
0.00125x1000
25
[1] = 0.05 [1] mol/dm3 [2]
(vi) From the mass of Borax used, calculate the
concentration of Borax in g/dm3
_______________________________ [1]
(vi) From the mass of Borax used, calculate the
concentration of Borax in g/dm3
4.775g in 250cm3 → 4.775 x 4 = 19.1 [1]
g/dm3
[1]
(vii) Using your answers to parts (v) and (vi)
find the formula mass of the Borax,
Na2B4O7.xH2O, and hence find the value
of x.
(Relative atomic masses: H = 1; B = 11; O = 16; Na = 23)
_______________________________
_______________________________
_______________________________ [3]
(vii) Using your answers to parts (v) and (vi)
find the formula mass of the Borax,
Na2B4O7.xH2O, and hence find the value
of x.
(Relative atomic masses: H = 1; B = 11; O = 16; Na = 23)
19.1
0.05 = 382 [1]
382 = 202 [1] + 18x
18x = 180
x = 10 [1]
[3]
b)
When Borax crystals are left in air they
lose some of their water of crystallisation.
To find the value of x in a sample of
hydrated Borax Na2B4O7.xH2O, which had
been left in air for a month, the sample
was heated to constant mass. 7.28g of
hydrated Borax produced 4.04g of
anhydrous Borax.
(Relative atomic masses: H = 1; B = 11; O = 16; Na = 23)
(i)
What is meant by “heated to constant
mass”?
_______________________________ [2]
(ii) Calculate the mass of water lost.
_______________________________ [1]
(i)
What is meant by “heated to constant
mass”?
heating and weighing [1] repeat until 2
readings the same [1]
[2]
(ii) Calculate the mass of water lost.
3.25g, 7.28 – 4.04 = 3.24g
[1]
(iii) Calculate the number of moles of water
lost.
_______________________________
_______________________________ [2]
(iv) Calculate the number of moles of
anhydrous Borax.
_______________________________ [1]
(iii) Calculate the number of moles of water
lost.
moles =
3.24
18
[1] = 0.18 [1]
[2]
(iv) Calculate the number of moles of
anhydrous Borax.
Moles =
4.04
202
= 0.02
[1]
(v) Using your answers to (iii) and (iv)
determine the value of x in
Na2B4O7.xH2O.
_______________________________
_______________________________ [2]
(v) Using your answers to (iii) and (iv)
determine the value of x in
Na2B4O7.xH2O.
borax : water
0.02 : 0.18 [1]
1
:
9
x = 9[1]
[2]
c)
When anhydrous Borax is heated it
decomposes according to the equation:
Anhydrous Borax → Sodium metaborate + Boric Oxide
Na2B4O7
→ 2NaBO2
+ B2O3
Calculate the mass of sodium metaborate
which is produced when 5.05g of
anhydrous Borax is heated.
(Relative atomic masses: B = 11; O = 16; Na = 23)
Calculate the mass of sodium metaborate
which is produced when 5.05g of
anhydrous Borax is heated.
(Relative atomic masses: B = 11; O = 16; Na = 23)
Moles of anhydrous borax =
= 0.025 [1]
Ratio 1 borax
:
2 sodium metaborate
0.025
:
0.050[1]
0.05x66[1] :
= 3.3 [1]g
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136