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Chemistry
WAEC / WASSCE 2007
Stoichiometry, Mole Concept & Volumetric Analysis
The following substances are heavy chemicals except
A
sodium trioxocarbonate (IV)
B
tetraoxosulphate (VI) acid
C
lead (lV) tetraethyl
D
sodium hydroxide
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option C.
Source: Myschool Past Questions Archive (Record #37252)
Chemistry
WAEC / WASSCE 2007
Metals, Transition Elements & Extraction
Pig-iron is brittle because it contains
A
a high percentage of carbon as impurity
B
calcium trioxosilicate (IV)
C
unreacted haematite
D
undecomposed limestone
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #37255)
Chemistry
WAEC / WASSCE 2007
Kinetic Theory of Matter & Gas Laws
What volume will 0.5 g of H2 occupy at s.t.p? [H = 1; 1 mole of a gas occupy 22.4 dm3 at s.t.p.]
A
2.24 dm3
B
5.60dm3
C
11.20dm3
D
44.80dm3
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option B.
Source: Myschool Past Questions Archive (Record #37262)
Chemistry
WAEC / WASSCE 2007
Stoichiometry, Mole Concept & Volumetric Analysis
Which of the these metals will not liberate hydrogen from dilute HCI?
A
Copper
B
Iron
C
Magnesium
D
Zinc
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #37263)
Chemistry
WAEC / WASSCE 2007
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
All your burette readings (initial and final) as well as the size size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
A is a solution containing 1.04 g HCl per 500 cm\(^3\) of solution. B was prepared by diluting 50.0 cm\(^3\) of a saturated solution of Na\(2\)CO\(_3\) at room temperature to 1000 cm\(^3\)
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^{3}\) portions of B using methyl orange as indicator. Repeat the titration to obtain consistent titres. Tabulate your results and calculate the average volume of acid used.
9b) From your results and information provided above, calculate the;
(i) concentration of A in moldm\(^{-3}\)
(ii) concentration of B in mol dm\(^{-3}\)
(iii) solubility of Na\(_2\)CO\(_3\) in mol dm\(^{-3}\)
(iv) volume of CO\(_2\) that would be liberated from 1 dm\(^3\) of B if the titration were carried out at s.t.p.
The equation for the reaction is Na\(_2\)CO\(_{3(aq)}\) + 2HCl\(_{(aq)}\) \(\to\) 2NaCl\(_{(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
[H = 1; C = 12; O = 16; Na = 23; Cl = 35.5; Molar volume of gas at s.t.p = 22.4 dm\(^3\)]
A
Copper
B
Iron
C
Magnesium
D
Zinc
medium Difficulty
Verified Correct Answer: Option A
Chemical Principles & Mechanism
(a)
Burette Reading
Rough Titration (cm\(^3\))
1st Titration (cm\(^3\))
2nd Titration (cm\(^3\))
3rd Titration (cm\(^3\))
Final
25,00
24.90
34.90
24.90
Initial
0.00
0.00
10.00
0.00
Volume of acid used
25.00
24.90
24.90
24.90
Average volume of acid used
= 24.90 + \(\frac{24.90}{3}\) + 24.90 = \(\frac{74.7}{3}\)
= 24.90 cm\(^3\)
(b) Concentration of A in g/dm\(^3\) = \(\frac{1.04 \times 1000}{500}\)
= 2.08 g/dm \(^3\)
Molar mass of HCl = 1 + 35.5 = 36.5 g/mol
Concentration of A in mol/dm\(^3\) = \(\frac{\text{concentration in g/dm}^3}{\text{molar mass}}\)
= \(\frac{2.08}{36.5}\)
= 0.057 mol/dm\(^3\)
(ii) \(\frac{C_AV_A}{C_BV_B} = \frac{N_A}{N_B}\)
\(\frac{0.057 \times 24.90}{C_B \times 25} = \frac{2}{1}\)
C\(_B\) = \(\frac{0.057 \times 24.90}{25 \times 2}\) = \(\frac{1.4193}{50}\)
= 0.02839
= 0.0284 mol/dm\(^3\)
(iii) To calculate the solubility of Na\(_2\)CO\(_3\) in mol/dm\(^3\). The dilution factor of B = 50 to 1000
= 1 : 20
The solubility = 0.0284 x 20
= 0.568 mol/dm\(^3\)
(iv) To calculate the volume of CO\(_{2(g)}\) liberated from 1 dm\(^3\) at s.t.p. from the equation of the reaction, one mole of Na\(_2\) CO\(_3\) liberate 22.4 x 0.0284
= 0.64 dm\(^3\) of CO\(_2\)
Source: Official Exam / Past Question Archive
Chemistry
WAEC / WASSCE 2007
Water, Solutions & Solubility
Solution
(a) Define the following terms:(i) Saturated solution; (ii) Solubility.
(b) In an experiment to determine the solubility of a given salt Y, the following data were provided:
Mass of dry empty dish = 7.16 g
Mass of dish + saturated solution of salt Y = 17.85 g
Mass of dish + salt Y = 9.30 g Temperature of solution = ^\circC
Molar mass of salt Y = 100
Density of solution Y = 1.00 g cm\(^{-3}\) Calculate the solubility of salt Y in
(i) g dm\(^{-3}\) of solution, (ii) mol dm\(^{-3}\) of solution.
(c) State the type of bond broken on melting each of the following substances: (i) NaCI\(_{(s)}\) (ii) CO\(_{2(s)}\) (iii) SiO\(_{2(s)}\) (iv) Al\(_{(s)}\)
(d) Explain the following observations: (i) the chemical reactivity of alkali metals increases down the group; (ii) Mg has higher melting point than Na; (iii) K is a better reducing agent than Na.
(e)(i) What are isotopes? (ii) Lithium exists as \(^6_3\)Li and \(^7_3\)Li in the ratio 2:25. Calculate the relative atomic mass of the lithium.
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
(a)(i) A saturated solution is one that will dissolve no more of the solute at a particular temperatuie in the presence of excess solute.
(ii) Solubility is the,maximum amount of solute (number of moles/mass) that dissolves in 1dm\(^3\) of solvent at a particular temperature.
(b) Mass of saturated solution = 17.85 — 7.16 = 10.69
Mass of dry salt = 9.30 — 7.16 = 2.14
Volume of solution = 10.69cm\(^3\)
Since volume of solution = \(\frac{mass}{density}\)
and density = 1.0g/dm\(^3\)
(i) Solubility in g/dm\(^3\)
10.69cm\(^3\) of solution Y contains 2.14g of salt Y
1cm\(^3\) of solution Y contains \(\frac{2.14 }{10.69}\)
1000cm\(^3\) = \(\frac{2.14 \times 1000}{10.69}\) = 200.19g/dm\(^3\)
(ii) Solubility in mol/dm\(^3\) = \(\frac{\text{solubility in g/dm}^3}{\text{molar mass of salt Y}}\)
= \(\frac{200.19}{100}\)
= 2.00 mol/dm\(^3\)
(c) NaCI — Ionic bond of electrovalent bond
C\(O_2\) — van der Waals forces
SiO — Covalent bond Al\(_{(s)}\) — Metallic bond.
(d)(i) Reactivity is due to loss of electrons, i.e the ease of electron loss, hence ionization energy decreases down the group.
(ii) Mg has more valence electrons contributing to the stronger binding energy compared to Na.
(iii) Potassium loses electron more readily than sodium, has lower ionization energy than sodium.
(e)(i) Isotopes are atoms of the same element with the same atomic number but different mass number.
(ii) Li = \(\frac{2 \times 6}{27}\) + \(\frac{25 \times 7}{27}\)
= 6.925
Source: Myschool Past Questions Archive (Record #52386)
Chemistry
WAEC / WASSCE 2007
Acids, Bases, pH & Salts
Solution
a) Define the following in term:. of electron transfer: (i) oxidation; (ii) reduction.
(b)(i) Determina the oxidation stale of phosphorus in each of the following structures: I. POCI\(_3\) II. PH\(_3\).
(ii) State with reasons whether the following compounds will form acidic, neutral or basic aqueous solutions: I. NaNO\(_3\) II. Na\(_2\)H\(_4\)CI; Ill. Na\(_2\)CO\(_3\).
(c) Consider the set-up
(i) What is the gas produced in the experiment illustrated by the set-up above?
(ii) Name the method of collection of gas
(iii) Give a reason for your answer in (c)(ii) above
(iv) State the function of the concentrated H\(_2\)SO\(_4\) in the conical flask
(v) Give of collection of the gas one I. physical property; II. chemical property of the gas
(vi) State one chemical test to identify the gas.
(d) A 4.3 g hydrated sodium tetraoxosulphate (VI) (Na\(_2\)SO\(_4\).xH\(_2\)O) was heated to remove the water of crystallization. The remaining anhydrous salt had a mass of 2.12 g. Calculate the value of x in the t I hydrated salt. [H = 1; O = 16; Na = 23; S = 32 ]
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
(a)(i) Oxidation is the process of electron loss. Reduction is the process of electron gain
(b)(i) POCI\(_3\)
O.N of P + O.N of oxygen + (O.N of CI) x 3 = 0
P + —2 + —1 x 3 = 0
—P — 2 — 3 = 0
P = — 5
(ii) I. NaNO\(_3\) is neutral because it is a salt of a strong acid and a strong base, its ions do not undergo hydrolysis.
II. NH\(_4\)CI is acidic because it is a salt of a strong acid and a weak base. The NH\(^+_4\) undergoes hydrolysis.
III. N\(a_4\)CO\(_3\) is basic because it is a salt of weak acid and ntrong base. The CO\(^{2-}_3\) undergoes hydrolysis.
(c)(i) Hydrogen chloride gas
(ii) downward delivery or upward displacement of air.
(iii) The gas is denser than air
(iv) drying the gas.
(v) Physical properties
— denser than air — very soluble in water — colourless gas — fumes in moist air
Chemical Properties
— gives cloud or white fumes in contact with ammonia vapour
(vi) Pass the gas into AgNO\(_{3(aq)}\) a white precipitate of AgCl is formed wihich is soluble in aqueous ammonia but insoluble in HNO\(_3\)
(d) Mass of water = 4.30 — 2.12 = 2.18g
Mass of Na\(_2\)SO\(_4\) = 23 x 2 + 32 + 16 x 4 = 142
Molar mass of H\(_2\)O = 2 + 16 = 18
2.12g of anhydrous salt contain 2.18g of water
142g of anhydrous salt contain \(\frac{2.18 \times 142}{ 2.12}\)
= 146.018
X(H\(_2\)O) = 146
18x = 146
x = \(\frac{146}{18}\)
= 8
Source: Myschool Past Questions Archive (Record #52387)
Chemistry
WAEC / WASSCE 2007
Organic Chemistry I: Hydrocarbons & Crude Oil
Solution
(a)(i) What is 2 homologous series?
(ii) Give two homologous series present in petroleum
(iii) Give one example of a compound belonging to each of the homologous series in (a)(ii).
(iv) Name two fractions obtained from the fractional distillation of petroleum
(v) Why is there a gradual change in the physical properties of petroleum fractions?
b) Write a two-step balanced chemical equation for the reaction of (i) ethanol with excess concentration traoxosulphate (VI) acid at high temperature
(ii) excess ethanol with concentrated tetraoxosulphate (VI) acid at lower temperature.
(c) An organic compound of relative molecular mass 46, on analysis vas ound to contain 52.0% carbon, 13.3% hydrogen and 34.7% oxygen.
(i) Determine its I. empirical formula, II. molecular formula.
(ii) Draw two possible structures of the compound and name one of them [O = 16; C = 12; H = 1 ]
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
(a)(i) Homologous series is a series are family of organic compounds with a general molecular formula, similarity in property and differing from successive member by CH\(_2\) of 14 a.m.u.
(ii) Alkanes, alkenes, cycloalkar les and aromatics.
(iii) Alkane — methane; ethane, etc.
Alkene — ethene, propene, butene, etc
Cycloalkanes — cyclohexane,
cyclopropane Aromatic — benzene.
(iv) Gas Petroleum ether Gasoline (petrol) Gas oil/diesel oil Lubricating oil Biturhen Kerosine.
(v) Because of a steady change (increase or decrease) in molecular mass of tho constituents.
(b)(i)
I. C\(_2\)H\(_5\)OH\(_{(l)}\) + H\(_2\)SO\(_4\)\(_{(l)}\) \(\to\) C\(_2\)H\(_5\)HSO\(_4\)\(_{(l)}\) + H\(_2\)O\(_{(l)}\)
II. C\(_2\)H\(_5\)HSO\(_4\)\(_{(l)}\) \(\to\) C\(_2\)H\(_4\)\(_{(g)}\) + H\(_2\)SO\(_4\)\(_{(l)}\)
(ii) I. C\(_2\)H\(_5\)OH\(_{(l)}\) + H\(_2\)SO\(_4\)\(_{(l)}\) ---> C\(_2\)H\(_5\)HSO\(_4\)\(_{(l)}\) + H\(_2\)O\(_{(l)}\)
II. C\(_2\)H\(_5\)HSO\(_4\)\(_{(l)}\) \(\to\) C\(_2\)H\(_5\)OC\(_2\)H\(_5\)\(_{(l)}\) + H\(_2\)SO\(_4\)
(c)(i)
(I) C H O
52.0 13.3 34.7
12 1 16
4.33 13.3 2.17
2.17 2.17 2.17
1 6 1
Emperica formula = C\(_2\)H\(_6\)O
(II) (C\(_2\)H\(_6\)O)\(_n\) = 46
(12 x 2 + 6 x 1 + 16)\(_n\) = 46
46\(_n\) = 46
n = \(\frac{46}{46}\) = 1
molecular formula = C\(_2\)H\(_6\)O
(ii)
Source: Myschool Past Questions Archive (Record #52388)
Chemistry
WAEC / WASSCE 2007
Electrolysis & Electrochemical Cells
Solution
(a) Write an equation in each case to represent the (i) \(\beta\)-decay of \(^{24}_{11}Na\) to give Mg.
ii) reaction of sodium with cold water.
b)(i) State two differences between reaction (a)(i) and (ii)
ii) State two applications of the type of reaction represented in(a)(i)
(c) Consider the reaction represented by the equation:
Mg\(_{(s)}\) + 2HCI\(_{(aq)}\) \(\to\) MgCl\(_{(aq)}\) + H\(_{(q)}\)
(i) Name the type of reaction involved
(ii) Give two ways by which the reaction could be made faster.
(iii) What volume of hydrogen gas would be produced from 6.0 g of the magnesium? [ H = 1; 1 mole of gas occupy 22.4 dm\(^3\) at s.t.p. ]
(d) What is (i) an electrolyte?; (ii) electrolysis?
(e)(i) Give one metal that is extracted using electrolytic process.
(ii) Name the ore of the metal.
(iii) What is the substance discharged at each electrode when dilute NaCI is electrolysed using graphite electrodes?
(iv) Why would aqueous NaCI conduct electricity but solid NaCI would not?
(v) Give one industrial use of NaCI.
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
(a) (i)
\(^{24}_{11}Na\) \(\to\) \(^{24}_{12}Mg\) + \(_1\beta\)
(ii) 2Na\(_{(s)}\) + 2H\(_2\)O\(_{(l)}\) \(\to\) 2NaOH\(_{(aq)}\) + H\(_{2(g)}\)
(b)
Reaction (i)
Reaction(i)
Takes place in the nucleus or nucleus reaction
Takes place in the valence shell or chemical reaction
Large amount of heat is released
Small amount of heat is released
Involves proton, neutrons and electrons.
Involves valence electrons
New nuclides are formed
Newxompounds are formed
Not affected by external factors e.g temperature, pressure, catalyst.
Usually affected by external factors.
(ii) Sterilizing instruments
-- treatment of cancer
- carbon dating
- pest control, etc.
(c)(i) Displacement reaction.
(ii) using Mg powder i.e increasing the surface area of Mg.
- using higher concentration of HCI
- increase temperature of the reaction mixture.
(iii) 24g Mg = 22.4 dm\(^3\) of H\(_2\) gas
6g Mg = \(\frac{6g \times 22.4 dm^3}{24g}\)
= 5.6 dm\(^3\).
(d)(i) An electrolyte is a compound which when molten or in aqueous solution will conduct electricity.
(ii) Electrolysis is the decomposition of an electrolyte by the passage of an electric current.
(e)(i) Aluminium
(ii) Bauxite
(iii) Cathode
- Hydrogen Anoda
- Oxygen
(iv) In the liquid state the Na+ and CI\(^-\) ions are mobile while in the solid state the ions ere in fixed position and cannot conduct electricity.
(v) Industrial uses of NaCI In the manufacture of CI\(_{2(g)}\)
— Salting out in soap manufacture
— Production of NaOH
— Solvay process
Source: Myschool Past Questions Archive (Record #52389)
Chemistry
WAEC / WASSCE 2007
Acids, Bases, pH & Salts
Solution
i) What is the name of the process used for the industrial preparation of tetraoxosulphate (VI) acid?
(ii) State the catalyst used in (a)(i)
(iii) Show by means of bdanced chemical equations only, the industrial preparation of tetraoxosulphate (VI) acid from sulphur (IV) oxide.
(b)(i) Distinguish between dehydration and drying
(ii) Explain why concentrated tetraoxosulphate (VI) acid cannot be used to dry ammonia
(iii) What is the drying agent for ammonia?
(c)(i) Give one example of I. a chloride which is soluble in hot water, II. a trioxocarbonate (IV) which does not decompose on heating, Ill. an amphoteric oxide
(ii) List three methods for the preparation of salts
(iii) State one method for the recovery of salt from its solution.
(d)(i) State Gay Lussac's law of combining volumes
(ii) Consider the reaction represented by the following equation: C\(_2\)H\(_{4(g)}\) + 3O\(_{2(g)}\) \(\to\) 2H\(_2\)O\(_{(g)}\) + 2CO\(_{2(g)}\). What is the volume of oxygen required for the complete combustion of 12.5 cm\(^{3}\) of ethene?
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
(a)(i) Contract process
(ii) Vanadium (v) oxide (V\(_2\)O\(_5\))
(iii) 2SO\(_{2(g)}\) + O\(_{2(g)}\) \(\rightleftharpoons\) 2SO\(_{3(g)}\)
SO\(_{3(g)}\) + H\(_2\)SO\(_4\)\(_{(/)}\) \(\to\) H\(_2\)S\(_2\)O\(_7\)
H\(_2\)S\(_2\)O\(_7\)\(_{(/)}\) + H\(_2\))\(_{(/)}\) \(\to\) 2H\(_2\)SO\(_{4(aq)}\)
(b)(i) Dehydration is the removeal of the element of water from a compound with change in the chemical composition of the compcu while drying is the removal of associated water from a substance without affecting it chemical composition.
(ii) Concentrated H\(_2\)SO\(_4\) reacts with ammonia to form ammonium tetraoxosulphate (VI)
(iii) CaO or silica gel.
(c)(i) I. lead chloride (PbCl\(_2\))
II. K\(_2\)CO\(_3\) or Na\(_2\)CO\(_3\)
III. ZnO or Al\(_2\)O\(_3\)
(ii) I - Reaction of an acid with base (neutralization reaction)
- Reactiction of an acid with a metal
- Reaction of an acid with trioxocarbonate (IV)
- Double decomposition
- Direct combination of its constituent element
- A metal with halogens
(iii) - Evaporation
- Crystallization
(d)(i) Gay lussac's Law of coming volumes states that when gases react, they do so in volumes which are in simple ratio to one another and to the volumes of the products if gaseous, provided that the temperature and pressure remains constant.
(ii) C\(_2\)H\(_{4(g)}\) + 3O\(_{2(g)}\) \(\to\) 2H\(_2\)O + 2Cl\(_{2(g)}\)
1 cm\(^3\) of ethene requires 3cm\(^3\) of oxygen
12.5cm\(^3\) of ethene will require 3 x 12.5cm\(^3\) of oxygen = 37.5cm\(^3\)
Source: Myschool Past Questions Archive (Record #52390)
Chemistry
WAEC / WASSCE 2007
Reaction Rates & Chemical Equilibrium
Solution
(i) Draw and label an energy profile diagram of an endothermic reaction. Indicate on your diagram the I. activation energy II. heat change, \(\Delta\)H.
(ii) Explain how the rate of reaction is affected by I. addition of a catalyst, II. increase in temperature.
(b)(i) Write an equation for the thermal decomposition of calcium trioxocarbonate (IV)
(ii) Determine the volume of carbon (IV) oxide measured at s.t.p. that would he produced by the thermal decomposition of 10g calcium trioxocarbonate (IV). [ Ca = 40; O = 16; C = 12 ]
(c)(i) Give one use cf each of the following forms of carbon: I. coal; II. wood charcoal; Ill. carbon (IV) oxide.
(ii) Write a balanced chemical equation to show what happens when each of the following compounds is heated strongly: I. NaNO\(_{3(s)}\) II. MgCO\(_{3(s)}\).
(d) Consider the following compounds: CaO, CaCO\(_3\), Ca(OH)\(_2\), NaOH Which of them is
(i) used in the manufacture of cement;
(ii) used to detect the presence of carbon (IV) oxide;
(iii) used to liberate carbon (IV) oxide when dilute acid is added;
(iv) hygroscopic;
(v) deliquescent?
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
(1) (a)
(ii) I. Catalyst increases the rate of the reaction cy cowering activation energy.
II. The rate of reaction increases with increase in temperature because there will be more effective collisions.
(b)(i) CaCO\(_{3(S)}\) \(\to\) CaO\(_{(s)}\) + CO\(_{2(g)}\)
(ii) Molar mass of CaCO\(_2\) = 40 + 12 + 16 x 3 =100
From the equation in b(i) above,
100g of CaCO\(_{(3)}\) produces 22.4 dm\(_3\) of CO\(_2\) at s.t.p
10g of CaCO\(_3\) will produce \(\frac{10 \times 22.4}{100}\)
= 2.24dm\(^3\) of CO\(_2\)
(c)(i) I. Coal is used as fuel
II. Wood charcoal is used for absorbing poisonous gases
Ill. CO\(_2\) is used in the following:
— in fire extinguishers
—manufacture of washing soda by solvay process
— in carbonated drinks.
(ii) I. 2NaNO\(_{3(s)}\) ---> 2NaNO\(_{2(s)}\) + O\(_2\)
II. MgCO\(_{3(s)}\) \(\to\) MgO\(_{(s)}\) + CO\(_{2(g)}\)
(d)(i) CaCO\(_3\) — cement
Ca(HO)\(_2\) — detection of CO\(_2\)
CaCO\(_3\) — liberates C\(O_2\) with addition of acid
CaO — hygroscopic
NaOH — deliquescent.
Source: Myschool Past Questions Archive (Record #52391)
Chemistry
WAEC / WASSCE 2007
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
(a) State an indicator suitable for the titration of;
(i) dilute HCl and NaOH\(_{3(aq)}\)
(ii) dilute CH\(_3\)COOH and KOH\(_{(aq)}\)
(iii) dilute HCl and NH\(_{3(aq)}\).
Give a reason for your answer in each case.
(b) Calculate the volume of water that would be added to 50 cm\(^3\) of 0.10 mol dm\(^{-3}\) of HCI to dilute it to 0.010 mol dm \(^{-3}\)
(c) Name one gas that could be used to demonstrate the fountain experiment.
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
(a)(i) Phenolphthalein or methyl orange or methyl red. Titration of strong acid and strong base.
(ii) Phenolphthalein reasons; Titration of a weak acid and strong base
(iii) Methyl orange or Methyl red reason: Are strong acid and weak base
(b) Using dilution law;
C\(_1\)V\(_1\) = C\(_2\)V\(_2\)
0.1 x 50 = 0.010 x V\(_2\)
V\(_2\) = \(\frac{0.1 \times 50}{0.010}\) = 500 cm\(^3\)
Volume of distilled water to be added to the original solution to make 0.01 mol/dm\(^3\)
= 500 - 50
= 450 cm\(^{3}\)
(c) Hydrogen chloride gas or ammonia or sulphur (IV) oxide
Source: Myschool Past Questions Archive (Record #60044)
Chemistry
WAEC / WASSCE 2007
Nature of Matter & Separation Techniques
Solution
Credit will be given for strict adherence to the instructions, for observations precisely recorded, and for accurate inferences. All tests, Observations, and inferences must be clearly entered in your answer book, in ink, at the time they are made.
C is a mixture of two inorganic salts. Carry out the following exercises on C. Record your observations and identify any gas(es) evolved. State the conclusion you draw from the result of each test.
(a) Put all of C in a test tube and add about 10 cm\(^3\) of distilled water. Stir, filter, and keep the filtrate and the residue.
(b) Put the residue into a test tube and add about 5 cm\(^3\) of dilute HCl. Shake to dissolve
(i) To about 2 cm\(^3\) of the solution, add NaOH\(_{(aq)}\) in drops and then in excess
(ii) To another 2 cm\(^3\) portion of the solution, add NH\(_{3(aq)}\) in drops and then in excess.
(c) To about 2 cm\(^3\) portion of the filtrate, add few drops of dilute HNO\(_3\), and then AgNO\(_{3(aq)}\) a followed by aqueous NH\(_3\) in excess.
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
test
Observation
Inference
(a) C + water and filter
White residue partially dissolved
(b) Residue +
dil. HCl + shake
Effervescence occur,
Colorless gas evolved.
Gas turns lime water milky
Gas is co\(_2^-\)
CO\(_3^{2-}\) or HCO\(_3\) present
(i) Solution in (b) + NaOH in drops
Then in excess
White gelatinous
ppt. ppt soluble in excess NaOH
Zn\(^{2+}\), Al\(^{3+}\)
or Pb\(^{2+}\) present
(ii) Solution in (b) +
NH\(_3\) solution in drops
then in excess
White gelatinous
ppt.ppt soluble
in excess NH\(_{3(aq)}\)
Zn\(^{2+}\)
confirmed
(c) Filtrate + dil. HNO\(_3\)
+ AgNO\(_{3(aq)}\) +
NH\(_3\) solution in excess
No visible reaction
White ppt formed.
ppt dissolved in excess NH\(_{3(aq)}\)
Cl\(^{-}\) present
Cl\(^{-}\) confirmed
Source: Myschool Past Questions Archive (Record #60045)
Chemistry
WAEC / WASSCE 2007
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
All your burette readings (initial and final) as well as the size size of your pipette, must be recorded but no account of experimental procedure is required. All calculations must be done in your answer book.
A is a solution containing 1.04 g HCl per 500 cm\(^3\) of solution. B was prepared by diluting 50.0 cm\(^3\) of a saturated solution of Na\(2\)CO\(_3\) at room temperature to 1000 cm\(^3\)
(a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0 cm\(^{3}\) portions of B using methyl orange as indicator. Repeat the titration to obtain consistent titres. Tabulate your results and calculate the average volume of acid used.
9b) From your results and information provided above, calculate the;
(i) concentration of A in moldm\(^{-3}\)
(ii) concentration of B in mol dm\(^{-3}\)
(iii) solubility of Na\(_2\)CO\(_3\) in mol dm\(^{-3}\)
(iv) volume of CO\(_2\) that would be liberated from 1 dm\(^3\) of B if the titration were carried out at s.t.p.
The equation for the reaction is Na\(_2\)CO\(_{3(aq)}\) + 2HCl\(_{(aq)}\) \(\to\) 2NaCl\(_{(aq)}\) + H\(_2\)O\(_{(l)}\) + CO\(_{2(g)}\)
[H = 1; C = 12; O = 16; Na = 23; Cl = 35.5; Molar volume of gas at s.t.p = 22.4 dm\(^3\)]
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
(a)
Burette Reading
Rough Titration (cm\(^3\))
1st Titration (cm\(^3\))
2nd Titration (cm\(^3\))
3rd Titration (cm\(^3\))
Final
25,00
24.90
34.90
24.90
Initial
0.00
0.00
10.00
0.00
Volume of acid used
25.00
24.90
24.90
24.90
Average volume of acid used
= 24.90 + \(\frac{24.90}{3}\) + 24.90 = \(\frac{74.7}{3}\)
= 24.90 cm\(^3\)
(b) Concentration of A in g/dm\(^3\) = \(\frac{1.04 \times 1000}{500}\)
= 2.08 g/dm \(^3\)
Molar mass of HCl = 1 + 35.5 = 36.5 g/mol
Concentration of A in mol/dm\(^3\) = \(\frac{\text{concentration in g/dm}^3}{\text{molar mass}}\)
= \(\frac{2.08}{36.5}\)
= 0.057 mol/dm\(^3\)
(ii) \(\frac{C_AV_A}{C_BV_B} = \frac{N_A}{N_B}\)
\(\frac{0.057 \times 24.90}{C_B \times 25} = \frac{2}{1}\)
C\(_B\) = \(\frac{0.057 \times 24.90}{25 \times 2}\) = \(\frac{1.4193}{50}\)
= 0.02839
= 0.0284 mol/dm\(^3\)
(iii) To calculate the solubility of Na\(_2\)CO\(_3\) in mol/dm\(^3\). The dilution factor of B = 50 to 1000
= 1 : 20
The solubility = 0.0284 x 20
= 0.568 mol/dm\(^3\)
(iv) To calculate the volume of CO\(_{2(g)}\) liberated from 1 dm\(^3\) at s.t.p. from the equation of the reaction, one mole of Na\(_2\) CO\(_3\) liberate 22.4 x 0.0284
= 0.64 dm\(^3\) of CO\(_2\)
Source: Myschool Past Questions Archive (Record #60046)
Chemistry
UTME / JAMB 2007
Organic Chemistry II: Functional Groups, Polymers & Biomolecules
Diagram
Solution
Which of the compounds above is a tertiary alkanol?
Attached Figure / Diagram
A
D
B
C
C
A
D
B
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
In order to identify the different types of alcohol
Identify the Carbon with the OH Group: Find the carbon atom that has the hydroxyl group (OH) attached to it.
Count the Carbon Bonds: Count how many carbon atoms are directly bonded to that carbon atom.
Classify the Alcohol:
If the carbon with the OH group is bonded to one other carbon, it's a primary alcohol.
If the carbon with the OH group is bonded to two other carbons, it's a secondary alcohol.
If the carbon with the OH group is bonded to three other carbons, it's a tertiary alcohol.
Source: Myschool Past Questions Archive (Record #72244)