1. You are given the following molecules:
PCl3, PCl5, HOF
(a) For each of these molecules, draw its three-dimensional structure.
(b) Name the three-dimensional structures in (a).
All of these are wrong answers!!!
(3 marks)
(3 marks)
PCl3 – 3bond pairs
--1 lone pair
Note P.7
1. You are given the following molecules:
PCl3, PCl5, HOF
(a) For each of these molecules, draw its three-dimensional structure.
(b) Name the three-dimensional structures in (a).
HOF– 2bond pairs
--2 lone pairs
PCl3 – 3bond pairs
--1 lone pair
PCl5 – 5bond pairs
--0 lone pair
(3 marks)
(3 marks)
1. You are given the following molecules:
PCl3, PCl5, HOF
(a) For each of these molecules, draw its three-dimensional structure.
(b) Name the three-dimensional structures in (a).
PCl3 – 3bond pairs
--1 lone pair
Electron clouds
distribution
(3 marks)
(3 marks)
PCl5 – 5bond pairs
--0 lone pair
HOF– 2bond pairs
--2 lone pairs
tetrahedral
Trigonal
bipyramidal
tetrahedral
Trigonal pyramidal
Trigonal
bipyramidal
V-shaped
3D structures
Name of the 3D
structures
Draw a 3D structure of CSH2
(c)(i) The polarity of a molecule can be explained by the concept of electronegativity.
What is meant by “electronegativity”?
(1 mark)
(c)(ii) Describe the trend in electronegativity of elements in period 2 of the Periodic
Table.
(1 mark)
P.12
(d) Which one, PCl3 or PCl5, is a polar molecule? Explain your answer.
(2 marks)
PCl3 is a polar molecule which consists of 3 polar P-Cl bonds arranged in trigonal
pyramidal shape. [1] The polarities of the polar bonds cannot cancel out each
other. [1]
X Polarities of the polar bonds in PCl3 are not identical
(They are identical!!!)
X 3 polar P-Cl bonds are arranged asymmetrically/symmetrically.
(Don’t mention this when it is a polar molecule)
But this explanation is necessary for a non-polar molecule
e.g1 . PCl5 is a non-polar molecule which consists of 5 polar P-Cl bonds arranged in
trigonal bipyramidal shape symmetrically. [1] The polarities of the polar bonds
cancel out each other. [1]
e.g2 . COS is a non-polar molecule….
2. A student studied the effect of electric field on a jet of propanone by using a positively
charged rod. The structure of propanone is shown below.
(a) What would be observed if a positively charged rod was placed near the jet of
propanone? Explain your observation.
(1 mark)
The propanone liquid is deflected. [0.5]The negative ends of the polar molecules
are attracted to the positively charged rod. [0.5]
X The jet / solution is deflected.
X The negative ions of the molecules are attracted to the positively charged rod.
X The molecule is polar.
(b) Describe and draw a diagram to show how a propanone molecule forms hydrogen
bonds with water molecules.
(2 marks)
P.24
The electrostatic attractions exists between the partial positively charged H atom of
water and the lone pair of electrons on O atom of propanone. [1]
(c)Two reagent bottles labelled A and B are left unattended on the bench. It is known that
one bottle contains pure propanone while the other contains a mixture of propanone and
water. Bottle B gives a stronger smell of propanone when the caps of the two bottles are
opened.
(i) Which bottle contains a mixture of propanone and water?
(1 mark)
(ii)Explain your answer in (c)(i) in terms of intermolecular forces.
(2 marks)
Book 2 Ch 17 E15
2.(c)(i) Bottle B. [1]
2.(c)(ii) When propanone dissolves in water, [1] propanone forms hydrogen bonds
with water molecules. [ 1} it gives the strong smell.
The table below shows three carbon compounds. Arrange the three compounds in order
of increasing boiling point. Explain your answer.
(4 marks)
Butane
VDW forces
Propan-1-ol
VDW forces
H-Bond (2H-bonds)/ size larger
/chain
Propanone
VDW forces
H-Bond (1 H-bond) / size smaller
/ branched
All of these are wrong answers!!!
The table below shows three carbon compounds. Arrange the three compounds in order
of increasing boiling point. Explain your answer.
(4 marks)
Butane
VDW forces
branched /size smaller
Propan-1-ol
VDW forces
Propanone
VDW forces
/ size larger
chain
H-Bond
All of these are wrong answers!!!
The table below shows three carbon compounds. Arrange the three compounds in order
of increasing boiling point. Explain your answer.
(4 marks)
Butane<Propanone<Propan-1-ol[0.5]
For all the 3 compounds, there are van der Waal’s forces between molecules. [0.5]
Propan-1-ol has the highest boiling point because it can form hydrogen bonding
between its molecules. [1]
The strength of hydrogen bonding is stronger than van der Waal’s forces. [0.5]
Propanone is polar and butane is non-polar, [1]
van der Waal’s forces between a polar molecules is higher than that between non-polar
molecules, [0.5]so propanone has a higher boiling point than butane.
3. Chlorine can be made by reacting concentrated hydrochloric acid with potassium
permanganate solution.
2KMnO4(aq) +16HCl(aq)  2MnCl2(aq) + 2KCl(aq) + 5Cl2(g) + 8H2O(l)
(a) Is the above reaction a redox reaction? Explain your answers in terms of the
oxidation number.
(2 marks)
(b) Deduce which species is an oxidizing agent.
(1 mark)
3.(a)Yes. The oxidation number of Mn in KMnO4 decreases from +7 to +2. KMnO4
is reduced. [1]
The oxidation number of Cl in HCl increases from -1 to 0. HCl is oxidized. [1]
3.(b) KMnO4. / MnO4-[1]
P.1
4. Magnesium oxide is insoluble in water, so it is difficult to titrate directly. Its purity can be
determined by performing the following steps.
Step 1: 4.06 g of impure magnesium oxide was completely dissolved in 100 cm3 of 2.00M
hydrochloric acid in excess.
Step 2: The solution formed is then diluted to 250.0 cm3 by distilled water.
Step 3: 25.0 cm3 of the dilute solution is titrated 0.02M sodium hydroxide solution. 19.7 cm3
of sodium hydroxide is required for the neutralization.
(a) Write a chemical equation with state symbols for the reaction in Step 1.
(1 mark)
(b) Describe how you can perform dilution in Step 2 by using suitable apparatus. (3 marks)
(c) Suggest a suitable indicator for the titration in Step 3, and give the colour change at the
end-point.
(1 mark)
4.(a) MgO(s) + 2HCl(aq)  MgCl2(aq) + H2O (l) [1/0]
4.(b) X Use a pipette to transfer the solution to the 250.0 cm3 volumetric flask
√(good) wash the 250.0cm3 volumetric flask with distilled water first.
Pour all the solution obtained from Step 1 to a (250.0 cm3) volumetric flask.
[1]
Rinse all the solution left in the beaker by distilled water and transfer the washing to the
volumetric flask. [1]
Add distilled water to the graduation mark of the volumetric flask [0.5]and shake the
volumetric flask thoroughly. [0.5]
4.(c) methyl orange[0.5], red to orange[0.5]/
phenolphthalein, colourless to pink.
(HCl add to NaOH)
P.28
P.30
Describe how can we dilute 1M HCl to 0.04M
P.26
4. Magnesium oxide is insoluble in water, so it is difficult to titrate directly. Its purity can be
determined by performing the following steps.
Step 1: 4.06 g of impure magnesium oxide was completely dissolved in 100 cm3 of 2.00M
hydrochloric acid in excess.
Step 2: The solution formed is then diluted to 250.0 cm3 by distilled water.
Step 3: 25.0 cm3 of the dilute solution is titrated 0.02M sodium hydroxide solution. 19.7 cm3 of
sodium hydroxide is required for the neutralization.
(d) Calculate the number of moles of hydrochloric acid added to the magnesium oxide. (0.5 mark)
(e) Calculate the number of moles of hydrochloric acid reacting with the magnesium oxide. (1.5
marks)
(f) Calculate the percentage purity of the magnesium oxide. (2 marks)
4.(d) 2 x 0.1 =0.2 mol [0.5]
4.(e) NaOH(aq) + HCl(aq)  NaCl(aq) + H2O (l)
no. of moles of NaOH in titration = 0.02 x 19.7/1000 = 0.000394
no. of moles of HCl in titration = 0.000394
no. of moles of 250 cm3 HCl = 0.000394 x10= 0.00394 [0.5]
no. of moles of HCl reacting with MgO = 0.2 -0.00394 =0.19606 [1]
4.(f) MgO(s) + 2HCl(aq)  MgCl2(aq) + H2O (l)
no. of moles of MgO = 0.19606 /2 =0.09803
mass of MgO = 0.09803 x 40.3 =3.95g [1]
% purity = 3.95/4.06 x100% =97.3% [1]
(g) What compounds could be present in the magnesium oxide that could lead to a false
value of its purity? Explain.
(2 marks)
Mg(OH)2 from MgO + H2O OR MgCO3 from the original mineral source, (1)
both of these compounds react with acid (1)and would lead to a false
titration value.
X MgO formed
X Mg
5 (a)(i)Write a chemical equation with state symbols for the reaction between nickel and
concentrated sulphuric acid.
(1 mark)
(ii)What property of concentrated sulphuric acid is responsible for the reaction in test tube A?
(0.5 mark)
(iii)State an observable change in test tube A.
(1 mark)
5.(a)(i) Ni (s) + 2H2SO4 (l) NiSO4 (s) + SO2(g) +2H2O (l) [1]
5.(a)(ii) oxidizing. [0.5]
5.(a)(iii) colourless gas bubbles can be seen. [1]
X Ni dissolves, solution turns from colourless to green.
(b)(i)State an observable change in test tube B.
(ii)Write an ionic equation for the reaction involved in test tube B.
(0.5 mark)
(1 mark)
(c) Reagent X in test tube C was used to absorb any excess sulphur dioxide. Suggest what X
could be.
(1 mark)
5.(b)(i) solution turns from yellow to green. [0.5]
5.(b)(ii) 2H2O (l) +SO2 (g) + 2Fe2+(aq) SO32- (aq) + 2Fe3+ (aq) + 4H+ (aq) [1]
5.(c) sodium hydroxide solution. [1] (acid base: SO2 + 2NaOH  Na2SO3 + H2O)
Limewater/ Ca(OH)2 solution [1] ] (acid base: SO2 + 2Ca(OH)2  CaSO3 + H2O)
/ Acidified KMnO4 solution (redox :SO2  SO42- and MnO4-  Mn2+ )
X anhydrous CaCl2
X water
X 2Cl- Cl2 + 2eX 7H2O + 2Cr3+ Cr2O72- + 14H+ + 6e-
voltmeter
(2e + Fe2+  Fe)
chromium rod
iron rod
salt bridge
chromium(III) chloride solution
iron(II) sulphate solution
6(a) As times goes by, the colour of iron(II) sulphate solution in the beaker gradually fades
out. State, with explanation, the direction of electron flow in the external circuit.
(2 marks)
Electron flow from chromium to iron in the external circuit. [1]
Fe2+ is reduced to form Fe, [0.5] so the concentration of Fe2+ decreases [0.5], so the colour
fades out.
X Cr is more reactive than Fe
(b) Write the half equation for the change occurring at the chromium rod. (1 mark)
(c) Write the half equation for the change occurring at the iron rod.
(1 mark)
(d) Which metal rod is an anode? Explain.
(1 mark)
(e) State two functions of the use of salt bridge.
(2 marks)
(f) Suggest why calcium nitrate solution cannot be used for making the salt bridge.
(1 mark)
(g)Suggest an appropriate chemical for making the salt bridge.
(1 mark)
6.(b) Cr(s) Cr3+ (aq)+ 3e-[1]
6.(c) Fe2+(aq) +2e- Fe(s) [1]
6.(d) Chromium rod. Oxidation occurs at chromium rod. [1]
6.(e) To complete the circuit by allowing ions to pass from one half cell to another. [1]
To reduce excess charges building up in the half cells by providing ions to it. [1]
6.(f) It can react with sulphate ions and form insoluble CaSO4 which blocks the salt
bridge. [0.5] , the voltage of the cell drops to 0 quickly. [0.5]
6.(g) saturated KNO3 / NaCl/ NaNO3 solution. [1] X Ca2+
P.22
Electrons flow from X to Cu
Reactivity: X>>Cu
Electrons flow from Y to Cu
Reactivity: Y>Cu
7(a) What is the function of the lemons in these cells?
(1 mark)
XXX act as a salt bridge
Act as an electrolyte.
7(b) Arrange metal X, metal Y, and copper in increasing order of reducing power. (power of
reducing other chemicals )
(1 mark)
Cu > Y > X
P.16
Use of salt bridge : when there are 2
electrolytes (solutions) which need to complete
the circuit
P.11
Electrons flow from X to Cu
Reactivity: X>>Cu
7(c) For Cell 1, write the half equation for the change that occurs at copper
strip.
(1 mark)
X Cu2+ + 2e- Cu
2H+(aq) +2e- H2(g)
7(d) Suggest a reason to explain why the voltages of both Cell 1 and Cell 2 drop quickly.
(1 mark)
X and Y can react directly with acid in lemon and produce hydrogen gas. Less electrons
passes to the wire, which lower the efficiency of the cell.[1] / gas bubbles built up at Cu
strip, which hinder the reaction of H+ to gain electron and the increases to resistance of
the cell.
X X and Y can react directly with acid in lemon, gas bubbles built up at X and Y.
X lemon will use up.
X direct reaction of Cu and acid in lemon.
X X and Y dissolves
X H2 formed block the ions move from X  Cu
8. Hydrogen peroxide (H2O2) is another oxide of hydrogen.
(a) What is the oxidation number of oxygen in hydrogen peroxide?
(1 mark)
b) In the presence of a dilute sodium hydroxide solution, hydrogen peroxide reduces
iron(III) ions and it is oxidized to oxygen. Write the half equation for the oxidization of
hydrogen peroxide.
(1 mark)
State the expected observation and write a chemical equation for the reaction involved.
(2 marks)
Redox Quiz (WS. P4)
b) In the presence of a dilute sodium hydroxide solution, hydrogen peroxide reduces
iron(III) ions and it is oxidized to oxygen. Write the half equation for the oxidization of
hydrogen peroxide.
(1 mark)
State the expected observation and write a chemical equation for the reaction involved.
(2 marks)
8.(b)(i)
H2O2  O2
H2O2  O2 +2H+ + 2e2OH- + H2O2  O2 +2H+ + 2e- + 2OH2OH- (aq) +H2O2 (aq)  O2 (g) +2H2O (l) + 2e- [1]
8.(b)(ii) colourless gas bubbles can be seen. / solution turns form yellow to
green [1]
2Fe3+ +2OH- +H2O2  2Fe2+ +O2 +2H2O [1]
X H2O2 dissolves
9. (a) Describe the structure of buckminsterfullerene.
(2 marks)
P.32
9.(a) It consists of 60 carbon atoms [0.5] bonded with covalent bonds. [0.5] in a
nearly spherical configuration [0.5] with each 5-membered ring is surrounded by
five 6-membered rings. [0.5]
X open-caged
X wrong spellings
X Carbon molecules.
(b) Compare the differences of two physical properties between graphite and
buckminsterfullerene.
(2 marks)
9.(b) Graphite has a higher melting point than C60. [1]Graphite is insoluble in
aqueous or non-aqueous solvent, but C60 is soluble in organic solvent. [1]
No need explanation!
c) State the differences of the structures and bondings between graphite and
buckminsterfullerene,
(2 marks)
9.(c) Graphite has a giant covalent structure with strong covalent bonds between C
atoms [1], with van der Waal’s forces between layers.
and C60 has a simple molecular structure with weak van der Waal’s forces between
C6o molecules. [1]
P.34
2.Compare the difference in properties between diamond , graphite and C60.
Explain differences in terms of their structure and bonding.
10. Four iron-made objects are placed separately in gel with rust indicator solution
containing potassium hexacyanoferrate(III), and allowed to stand in air for some time. Write
down the observation and giving the relevant explanation for each of the cases on the
answer book.
Prevent Fe to form Fe2+
X 1. No blue colour
X Zn rust
X Fe does not rust
X Colourless gas bubbles because Zn reacts with the hot gel
(Mg can react with the hot water in the gel not Zn )
P.249, 304
11. Lead (Pb) is an element in Group IV of the Periodic Table.
An oxide of lead, X, contains 90.6% of lead by mass. Calculate the empirical
formula of X. (3 marks)
Let the mass of lead oxide be 100g
Pb
O
Mass in gram 90.6
9.4
90.6
9.4
No. of moles 207.2
[1]
16
0.4373 0.5875
1
1.3
mole ratio
3
4
[1]
Empirical formula of X is Pb3O4
[1]
X 1.3 cannot round up to 1
X multiply by 10
Divided both answers by the
smaller answer
Muitlply by 2/3/4/…. Until you
can get a whole number
12. An astronaut exhales 1.1 kg of carbon dioxide per day in a spaceship. This carbon dioxide
can be absorbed by solid lithium hydroxide.
(a) Write a balanced full equation for the reaction between lithium hydroxide and carbon dioxide.
(1 mark)
(b) Calculate the mass of lithium hydroxide required per day for a crew of 3 astronauts. (3 marks)
(c) An alternative to lithium hydroxide for the removal of carbon dioxide in the spaceship is
sodium peroxide according to the following equation:
2Na2O2(s) + 2CO2(g)  2Na2CO3(s) + O2 (g)
Suggest ONE advantage of using sodium peroxide rather than lithium hydroxide. (1 mark)
(a) LiOH (s) + CO2 (g)
base+ acid  salt + water
2LiOH (s) + CO2 (g)  Li2CO3 (s) + H2O (l)
Or 2LiOH + H2CO3  Li2CO3 + 2 H2O
[1]
b) total mass of CO2 produced by 3 astronauts = 3 x 1.1 x 1000g
3300g [1]
No. of moles of CO2 produced = 3300/(12+32) = 75 [1]
no. of moles of LiOH required = 2 x 75 x (6.9+ 16+1) = 3585 g
(c) Advantage: It produces oxygen for the astronauts to breathe.
[1]
[1]
1. Which of the following combinations is INCORRECT?
Chemical
Method of storage
A. Calcium
Under water
B. Potassium
Under paraffin oil
C. Ethanol (alcohol)
In a cool place
D. Potassium permanganate solution
In a brown bottle
---Ca reacts with water
--- Potassium permanganate solution, AgNO3 and HNO3 can be
decompose under sunlight
2. Calcium carbonate can be obtained from quicklime through two processes as
shown below.
Process 1
Process 2
Quicklime  Limewater  Calcium carbonate
CaO
Ca(OH)2
CaCO3
Which of the following combinations is correct?
Process 1
Process 2
A. Adding water
Adding Na2CO3(aq)
B. Adding Na2CO3(aq)
Adding water
C. Adding water
Heating
D. Heating
Adding water
CaO + H2OCa(OH)2
Ca(OH)2 (aq) + Na2CO3(aq) 
CaCO3(s) +NaOH (aq)
A
Test
A
B
Procedure
Concentrated hydrochloric acid is used to wash
the platinum wire which is used to transfer Y
into non-luminous flame
Acidified silver nitrate solution is added to the
solution of sample Y
Observation
Golden yellow
flame
White precipitate
3. A student has performed the following tests on solid sample Y:
Y is
A. Calcium carbonate.
B. Calcium chloride.
C. Sodium chloride.
D. Potassium sulphate.
C
4. X is an element. It can form a cation X2+ which has an electronic arrangement
2, 8, 8. Which of the following statements concerning X are correct?
(1) X is a strong oxidizing agent.
(2) X is in Period 4 of the Period Table.
(3) X burns in oxygen with a brick red flame.
A.
B.
C.
D.
(1) and (2) only
(2) and (3) only
(1) and (3) only
(1),(2) and (3)
X--- 2,8,8,2
B
***5. When a neutron is added to the nucleus of an atom,
A. the atomic number of the atom increases.
B. the attraction between the nucleus and the electrons increases.
C. the new atom is an isotope of the original atom.
D. the atomic mass does not change.
A ---Proton increases
B. --- neutron carries no charge
D --- mass increased by 1
C --- isotope– atoms of same elements with same no. of P and diff
no. of n.
***6. How many moles of C2H6 contain y hydrogen atoms? (L represents
the Avogadro’s constant.)
A. y/L
B. L/y
C. y/6L
D. 6y/L
I mole C2H6 contains 6L hydrogen atoms
X moles C2H6 contains y hydrogen atoms
X/1 = y/6L
C
7. What is the number of moles of Fe3+ ions in 0.1dm3 of 0.5 M Fe2(SO4)3
solution?
A. 0.1 × 0.5
B. 2 × 0.1 × 0.5
C. 0.1 × 0.5 × 6.02 × 1023
D. 2 × 0.1 × 0.5 × 6.02 × 1023
No. of moles of Fe2(SO4)3 = 0.1 x 0.5
No of moles of Fe3+ = 0.1 x 0.5 x 2
B
8. Which of the following metal oxides CANNOT be
reduced to form metal when heated with carbon?
A. Copper(II) oxide
B. Iron(III) oxide
C. Zinc oxide
D. Magnesium oxide
9. X and Y are elements. The melting points of their chlorides are given below:
Melting point / oC
Chloride of X
772
Chloride of Y
-68
Which of the following statements is correct?
A. Both X and Y are metals.
B. The chloride of Y is a solid at room temperature.
C. The chloride of X conducts electricity in the solid state.
D. The chloride of Y is a covalent compound.
B- at 25oC, higher than m.p. liquid/gas
D– simple molecular structure , with low m.p
10. Hydrochloric acid is prepared in the laboratory by bubbling hydrogen chloride gas
into water using an inverted funnel as shown below.
The main purpose of using an inverted funnel is to
A. obtain a more concentrated solution.
B. increase the rate at which the gas dissolves in water.
C. increase the solubility of the gas in water.
D. prevent sucking back of water.
(air pressure in the delivery tube lower than outside, water pushes up.)
 With inverted funnel,  increases surface area of absorption of HCl gas
***11. Which of the following substances have the giant structure shown?
A.Potassium chloride
B.Sodium oxide
C.Hydrogen chloride
D.Calcium sulphide
A. (1) and (4) only
B. (2) and (3) only
C. (1) and (2) only
D. (1), (2) and (4) only
6 anions with 6 cations  1:1
KCl  1:1
Na2O  2:1
HCl covalent
CaS  1:1
A
12. What kind of bondings/attraction forces exist in ammonium chloride?
(1) Ionic bond
(2) Covalent bond
(3) Dative bond
(4) Van der Waal’s forces
A.
B.
C.
D.
(2) and (3) only
(1) and (2) only
(1), (2) and (4) only
(1), (2) and (3) only
NH4Cl
NH4+ Cl-  ionic compound  ionic bond
NH4+  N-H covalent bond , one NH bond is a dative covalent bond
C
13. X,Y and Z are metals. Y can displace X from a solution of the nitrate of
X. Oxides of X and Y can reduced by hydrogen but not the oxide of Z.
Which of the following arrangements represents the correct
descending order of reactivity of the metals?
A. Z > Y > X
B. X > Y > Z
C. Z > X > Y
D. X > Z > Y
Y can displace X from a solution of the nitrate of X.
 Y>X
Oxides of X and Y can reduced by hydrogen but not the oxide of Z.
oxide of Z can only use electrolysis
 Z > X,Y
A
14. 25 cm3 of a sodium hydrogencarbonate solution requires 26.25 cm3 of
0.5M sulphuric acid for complete neutralization. The mass of sodium
hydrogenarbonate in 250 cm3 of the above solution is
A. 1.10g.
B. 5.51g.
C. 11.03g.
D. 22.05g.
2NaHCO3 + H2SO4  Na2SO4 + 2CO2 + H2O
Mole of H2SO4 = 0.5 x 26.25/1000 = 0.013125
Mole of 25 cm3 NaHCO3 = 0.013125 x2 =0.02625
Mole of 250 cm3 NaHCO3 =0.02625x 10 = 0.2625
Mass of NaHCO3 = 0.2625 x 84 = 22.05g
D
15. A hydrated sample of sodium carbonate (Na2CO3•nH2O) is titrated
against dilute hydrochloric acid. It is found that 1.24 g of the hydrated
sample react with 13.34 cm3 of 1.5 M hydrochloric acid. What is the value
of n?
A. 1
B. 3
C. 5
D. 7
Na2CO3 + HCl  2NaCl + CO2 + H2O
Moles of HCl = 1.5 x13.34/1000 = 0.02
Moles of Na2CO3= 0.02
1.24/ molar mass of Na2CO3•nH2O = 0.02
molar mass of Na2CO3•nH2O = 62
23x2 + 12+ 16x3 + 18n = 62
n=1
A
16. Sewage discharged from electroplating factories contains heavy metal ions
like Ni2+, Cr3+ etc. Which of the following substances is suitable for eliminating
these harmful ions?
A.
Calcium nitrate solution
B.
Sodium chloride solution
C.
Potassium sulphate solution
D. Sodium hydroxide solution
Alkali reacts with metal ions formed ppt (metal hydroxides ) which can filter
off.
D (note P.8 ,P9 reaction of alkalis)
17. Which of the following hydroxides is INSOLUBLE in both excess
dilute sodium hydroxide solution and excess dilute aqueous
ammonia?
A. Cu(OH)2
B. Fe(OH)3
C. Pb(OH)2
D. Zn(OH)2
B
**18. When calcium granules are added to water, colourless gas bubbles are
formed. The mixture is then filtrated to obtain a clear solution. Which of the
following is correct if excess dilute hydrochloric acid is added to the clear
solution?
A. Gas bubbles are formed.
B. There is no visible change.
C. A white precipitate is formed.
D. The clear solution turns brick red.
Ca + 2H2O  Ca(OH)2 + H2
 milky suspension, colourless gaas bubbles
Ca(OH)2 + HCl
Base + Acid  salt + water
B
****19. 25.0 cm3 of 0.252M sulphuric acid sample was diluted to 250
cm3. What is the pH of the diluted acid solution?
A. 0.50
B. 1.30
C. 1.60
D. 1.90
Molarity of H+ in 0.252M sulphuric acid = 0.252 x2 = 0.504
Molarity of H+ in 0.0252M sulphuric acid= 0.0504
pH= -log (0.0504)
B
20. Which of the following reactions can be used to prepare lead(II)
sulphate?
(1)
Reaction of lead with dilute sulphuric acid
(2)
Reaction of lead(II) oxide with dilute sulphuric acid
(3)
Reaction of lead(II) nitrate solution with sodium sulphate solution
A. (3) only
B. (1) and (2) only
C. (1) and (3) only
D. (2) and (3) only
PbSO4 is insoluble
 2 soluble salts mix together to form ppt PbSO4
A
****21. Which of the following substances can be used to distinguish between
concentrated nitric acid and concentrated sulphuric acid?
(1) Sodium carbonate powder
(2) Copper turnings
(3) Cane sugar
A.
B.
C.
D.
(1) and (2) only
(1) and (3) only
(2) and (3) only
(1), (2) and (3)
(2) Brown gas NO2 (oxidizing power)/ colourless gas bubbles SO2 (oxidizing power)
(3) No change/ turns black C(dehydrating power)
(1) colourless gas bubbles / colourless gas bubbles (acidic property – CO2)
B, D
22. Consider the following chemical equation :
x VO2+ (aq) + y H+ (aq) + 2I- (aq)  x VO2+ (aq) + z H2O (l) + I2 (aq)
(V is the symbol for the element vanadium.)
Which of the following combinations is correct?
x y z
A. 1 2 1
B. 1 4 2
C. 2 4 2
D. 3 6 3
C
Which of the following statements concerning the above chemical cell is/are correct?
(1)
The solution near electrode Y changes from colourless to purple.
(2)
Electrons flow from electrode X to electrode Y through the external circuit.
(3)
Sodium chloride solution can be used to make the salt bridge.
A.
(1) only
B.
(2) only
C.
(2) and (3) only
D.
(1) and (3) only