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Chemistry
WAEC / WASSCE 2016
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
(a) (i) Draw the structures of the isomers of the alkene with molecular formurat C\(_4\)H\(_8\)
(ii) State the class of alkanols to which each of the following compounds belongs:
I. CH\(_3\)C(CH\(_3\))\(_2\)OH;
II. CH\(_3\)CH(CH\(_3\))CH\(_2\)OH;
III. CH\(_3\)CH\(_2\)CH(CH\(_3\))OH.
(b) (i) Write the formulae of the products formed in the following reactions:
I. CH\(_3\)CH\(_2\)COOH \(\frac{K_{(s)}}{}\)
II. CH\(_3\)CH\(_2\)COOH. \(\frac{C_4H_6OH, heat}{Conc.H_2SO_4}\)
III. CH\(_3\)CH\(_2\)CH\(_2\)CH\(_2\)OH \(\frac{H^+/KMnO_4}{(excess)}\)
(ii) Name the major product(s) of each of the reactions in (b)(i).
(c) A gaseous hydrocarbon R of mass 7.0 g occupies a volume of 2.24 dm\(^3\) at s. t.p. If the percentage composition by mass of hydrogen is 14.3, determine its:
(i) empirical formula;
(ii) molecular formula. [ H = 1.00, C = 12.0, Molar volume of gas at s.t.p, = 22.4 dm\(^3\) ]
(d) Define structural isomerism.
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
(a)(i) Structural Isomers of C\(_4\)H\(_4\)
(ii) (I)Tertiary(3^\circ)alkanol.
(II) Primary (1^\circ)alkanol.
(III) Secondary (2^\circ) alkanol
(b)(i)
(I) CH\(_3\)CH\(_2\)COOK, H\(_2\)
(II) CH\(_3\)CH\(_2\)COOC\(_4\)H\(_9\) H\(_2\)O
(III) CH\(_3\)CH\(_2\)CH\(_2\)COOH
(ii) Major products in (b)(i) (1) — Potassuim propanoate (II) — Butyl propanoate (III) — Butanoic acid
(C)(i) % carbon by mass = 100 – 14.3 (l) = 85.7%
Ratio of atoms = \(\frac{C}{\frac{85.7}{12.0}}\) = 7.1415; \(\frac{H}{\frac{14.3}{1.0}}\) = 14.3
Divide by smaller number = \(\frac{7.1415}{7.1415}\) = 1; \(\frac{14.3015}{7.1415}\) = 2
Empirical Formular = CH\(_2\)
(i) 2.24dm\(^3\) = 7.0g of R
22.4dm\(^3\) = \(\frac{7.0}{2.24}\) x 22.4 = 70gmol\(^1\)
OR
Molar volume of gas = 22.4 dm\(^3\) at s.t.p 22.4
22.4 dm\(^3\) gas = 1 mole
22.4 dm\(^3\) = \(\frac{2.24}{22.4} \times 1\) = 0.1 mol
But numbers of moles = 0.1
Mr = \(\frac{7}{0.1}\) = 70gmol\(^1\)
Molecular formular = C\(_5\)H\(_{10}\)
(CH_2)n = 70
14n = 70
n = \(\frac{70}{4}\) = 5
(d) The occurence of / existence of / condition whereby compounds with the same molecular formula have different structural formulae.
Source: Myschool Past Questions Archive (Record #51013)
Chemistry
WAEC / WASSCE 2016
Metals, Transition Elements & Extraction
Solution
(a)(i) Name a suitable drying agent for the preparation of carbon (IV) oxide in the laboratory.
(ii) Using one chemical test, distinguish between carbon (II) oxide and carbon (IV) oxide.
(b)(i) Describe briefly how oxygen and nitrogen could be obtained separately from air on an industrial scale
(ii) State how a lighted splint can be used to distinguish between samples of oxygen and nitrogen.
(c)(i) Give one reason why bauxite is usually preferred as the ore for the extraction of aluminium.
(ii). List two main impurities. usually present in bauxite.
(iii) State the function of sodium hydroxide solution in the extraction of aluminium from its ore.
(iv) Explain briefly why it is difficult to extract aluminium by chemical reduction of aluminium oxide
(v) Write an equation for the reaction of aluminium oxide with aqueous sodium hydroxide.
(d) (i) The melting and boiling points of sodium chloride are 801 ^\circC and 141.3 ^\circC respectively. Explain briefly why sodium chloride does not conduct electricity at 25^\circC but does so between 801 ^\circC and 1413 ^\circC.
(ii) State the reason why sodium metal is stored under paraffin oil in the laboratory.
(e)(i) State what would be observed when aqueous sodium trioxocarbonate(IV) is added to a solution containing iron (III) ions
(ii) Write a balanced equation for the reaction in (e)(i).
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
(a) (i) Concentrated tetraoxosulphate (vi) acid fused calcuim chloride.
(ii) Pass or bubble each gas into lime water, if lime water turns milky then CO, is present if no reaction then CO is present.
(b) (i) –Atmospheric air is compressed and cooled to liquefy it. – The liquefied air undergoes fractional distillation. – Nitrogen boils out first at – 196^\circC followed by oxygen at – 183^\circC.
(ii) – Splint burns brighter / continues to burn in oxygen. Splint is extinguished in nitrogen
(c) (i) – Bauxite is more abundant than any other ore. – Bauxite has less percentage of silicates as impurity.
(ii) . – iron (III) oxide. – Trioxosilicates(IV) / silica / sand.
(iii) To dissolve the aluminium oxide so that the impurities can be filtered off
(iv) Aluminium is very reactive to be reduced by the eommon reducing agenfleoaction.
(v) Al\(_2\)O\(_3\) + 2NaOH \(\to\) 2NaAlO\(_2\) +H\(_2\)O or Al\(_2\)O\(_2\) + 2NaOH + 3H\(_2\)O --> 2NaAl(OH)\(_4\)
(d) (i) The ions are not free at 25^\circC since sodium chloride is in the solid state but between 801^\circC and 1413^\circC the ions are free / mobile since it is now in the molten / liquid state.
(ii) Sodium metal is very reactive with air/ it is easily oxidised in air / tarnishes easily.
(e) (i) Brown precipitate / reddish brown / orange solid formed
(ii) 3Na\(_2\)CO\(_{3(aq)}\) + 2Fe\(^{3+}\) --> F\(e_2\) (CO\(_3\))\(_{3(s)}\) + 6Na\(^{+}\)
Source: Myschool Past Questions Archive (Record #51015)
Chemistry
WAEC / WASSCE 2016
Non-Metals & Their Compounds
Solution
(a)(i) Determine the oxidation number of sulphur in Na\(_2\)S\(_2\)O\(_3\)
(ii) Name the allotropes of sulphur.
(iii) State two ways in which the structure of graphite and diamond are similar.
(b)(i) Name two green-house gases.
(ii) State one effect of an increased level of green-house gases on the environment.
(iii) State one source from which nitrogen (I) oxide is released into the environment.
(iv) Write a chemical equation to show the effect of heat on each of the following compounds: I. KNO\(_{3(s)}\) II. AgNO\(_{3(s)}\)
(C)(i)Describe briefly how pure crystals of calcium chloride could be obtained from a solution of calcium chloride
ii) Explain briefly each of the following observations:
I. ammonia gas is highly soluble in water;
II. boiling ploint of chlorine is lower than that of iodine
(d) Consider the reaction represented by the following equation:- 2NaCl + H\(_2\)SO\(_{4(s)}\) \(\to\) Na\(_2\)SO\(_{4(s)}\) + 2HCl\(_{(g)}\)
Calculate the volume of HCl gas that can be obtained at s.t.p. from 5.85 g of sodium. chloride. [ Na = 23.0, 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) (i) let x = oxidation number of sulphur 2(1) + 2 + 3(-2) = 0
x = +2
(ii) Allotropes of sulphur are: – rhombic / alpha sulphur. – monoclinic / Beta / prismatic sulphur. – amorphous / delta sulphur. – plastic sulphur
(iii) – Macromolecule / giant molecule /3 dimensional. – carbon atoms are covalently bonded. – crystalline in nature.
(b) (i) Carbon(IV)oxide. – methane. – nitrogen(I)oxide. –water vapour – ozone – chlorofluorocarbons.
(ii) – Global warming / climate change.
(iii) – Ammonium trioxonitrate (V) /aerosol (accept ammonium nitrate). – Combustion of fuel at high temperature.
(iv)
(I) 2KNO\(_3\) --> 2KNO\(_2\) + O\(_2\)
(II) 2AgNO\(_{_3(s)}\) ---> 2Ag\(_s\) + 2NO\(_{2(g)}\) + O\(_2\)
(c) (i) – Heat the solution to concentrate it or leave in a warm place.
- partially evaporate solution and allow to cool filter off crystals.
- pick out crystals
- Dry crystals with filter paper.
(ii)(1) Ammonia readily associates with water molecules through hydrogen bonding.
(II) Chlorine molecules are smaller or chlorine molecules have fewer number of electrons and the strength of van der waals forces between molecules are weaker as compared to that of iodine OR iodine molecules are higher and the wander waal's forces between molecules are stronger as compared to that of chlorine.
(d) Mr(NaCl) = 23.0 + 35.5 = 58.5g/mol\(^1\)
Mass of NaCl = 5.85g
Molar mass of NaCl= 58.5g/mol\(^1\)
Amount of NaCl in 85g = \(\frac{5.85g}{58.5gmol^{-1}}\) = 0.1mol
1 mol of NaCI = 22.4dm\(^3\) of HCI
:. 0.1 mol = \(\frac{0.1}{1}\) x 22.4 = 2.24dm\(^3\)
OR
Molar mass NaCl = 23.0 + 35.5 = 58.5gg/mol\(^1\)
58.5g of NaCl = 22.4dm\(^3\) HCl
58.5g = \(\frac{22.4}{58.5}\) x 5.85
= 2.24dm\(^3\)
Source: Myschool Past Questions Archive (Record #51036)
Chemistry
UTME / JAMB 2016
Kinetic Theory of Matter & Gas Laws
Solution
From the diagram above, an ideal gas can be represented by
A
K
B
M
C
L
D
N
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
see diagram in textbook
Source: Official Exam / Past Question Archive
Chemistry
WAEC / WASSCE 2016
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet
A solution of 0.050 moldm\(^3\) H\(_2\)C\(_2\)O\(_4\) (ethanedioic acid). B is a solution of KMnO\(_4\), (potassium tetraoxomanganate (VII), of unknown concentration.
(a) Put B into the burette. Pipette 20.0 cm\(^3\) or 25.0 cm\(^3\) of A into a Conical flask and add about 10.0 cm\(^3\) of dilute H\(_2\)SO\(_4\), Heat the mixture to about 40^\circC - 50^\circC and titrate it while still hot with B. Repeat the titration to obtain consistent titre values. Tabulate your results and calculate the average volume of B used. The equation of reaction is;
2MnO\(_{4(aq)}^-\) + 5C\(_2\)O\(_{4(aq)}^{2-}\) + 16H\(^+_{(aq)}\) \(\to\) 2MnH\(^{2+}_{(aq)}\) + 8H\(_2\)O\(_{(l)}\) + 10CO\(_{2(g)}\)
(b) From your results and the information provided, calculate the:
(i) concentration of MnO\(_2^-\) in B in moldm\(^{-1}\)
(ii) concentration of KMnO\(_4^-\) in B in gdm\(^{-3}\)
(iii) volume of CO\(_2\) evolved at s.t.p when 25.0 cm\(^3\) of H\(_2\)C\(_2\)O\(_4\) reacted completely. [0 = 16.0, K= 39.0, Mn = 55.0, Molar volume of gas at s.t.p.= 22.4 dm\(^3\) mol\(^{-1}\)]
Credit will be given for strict adherence to the instructions. for observations precisely recorded and jor accurate inferences. All tests, observations and inferences must be clearly entered in your answer book in ink, at the time they are made
A
Dihydric Alkanol
B
Tertiary Alkanol
C
Trihydric Alkanol
D
Monohydric Alkanol
medium Difficulty
Verified Correct Answer: Option C
Chemical Principles & Mechanism
Burette Reading (cm\(^3\))
Rough
1st Titration
2nd Titration
3rd Titration
Final Burette Readings (cm\(^3\))
36.50
30.40
41.70
26.80
Initial Burette readings (cm\(^3\))
12.30
05.90
17.10
01.90
Vol. of acid used (cm\(^3\))
24.20
24.50
24.60
24.90
Titre value = \(\frac{24.50 + 24.60}{2}\) = 24.55cm\(^3\)
(b)(i) Molar concentration of KMNO\(_4\)
\(\frac{C(MnO_4^-) \times V(MnO_4^-)}{C(C_2O_4^{2-}) \times V(C_2O_4^{2-})} = \frac{n(MnO_4^-)}{n(C_2O_4^{2-})}\)
\(\frac{C(MnO_4^-) \times 24.55}{0.05 \times 25} = \frac{2}{5}\)
C(MnO\(_4^-\)) = \(\frac{0.05 \times 25 \times 2}{24.55 \times 5}\)
= 0.0204 mol/dm
(ii) Mass concentration of KmnO\(_4\)
Molar mass of KmnO\(_4\) = 39 + 55 + (16 x 4) = 158g/mol
Mass conc. KmnO\(_4\) = molar conc. x molar mass
= 0.0204 x 158 = 3.22g/dm\(^3\)
(iii) Volume of CO\(_2\) (gas evolved)
\(\frac{n(CO_2)}{n (H_2C_2O_4)}\) = \(\frac{10}{5}\)
n(CO\(_2\)) = \(\frac{10}{5}\) x 0.05
= 2 x 0.05 = 0.100 mol
V(CO\(_2\)) at s.t.p = n x Vm
25cm\(^3\) of H\(_2\)C\(_2\)O\(_4\) = 0.100 x 22.4dm\(^3\)mol\(^{-1}\) x \(\frac{25}{1000}\)
= 0.056dm\(^3\) or 56.0 cm\(^3\)
Alternative method
Volume of CO\(_2\) gas evolved
Amount of H\(_2\)C\(_2\)O\(_4\) used = 0.050 x \(\frac{25}{1000}\) = 0.00125 mol
From the equation of the reaction
5 mol of C\(_2\)O\(_4^{2-}\) = 10 mol of CO\(_2\)
0.00125 mol C\(_2\)O\(_4^{2-}\) = 0.00125 mol x \(\frac{10}{5}\)
= 0.0025 mol CO\(_2\)
@S.T.P
1 mol of CO\(_2\) = 22.4dm\(^3\)
0.00250 mol = 22.4 x 0.0025
= 0.056 dm\(_3\) or 56.0 cm\(^3\)
Source: Official Exam / Past Question Archive
Chemistry
WAEC / WASSCE 2016
Nature of Matter & Separation Techniques
Solution
All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet
(a) What difference in physical properties enable the separation of mixtures by:
(i) simple distillation,
(ii) paper chromatography;
(iii) fractional distillation.
(b) Give a reason for each of the following practices during titration in the laboratory.
(i) White tile is placed under the conical flask.
(ii) Burette readings are always recorded to two decimal places.
(iii) Calculate the volume of 2.5 moldm\(^{-3}\) stock HČI required to prepare 500 cm\(^3\) of 0.20 moldm HCI.
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
(a)(i) Wide difference in boiling points of components (for miscible liquids)/ difference in physical state (for liquid - solid solution)
(ii) Difference in absorption/retention rate of components
(ii) Close difference in boiling points of components.
(b)(i) To make the endpoint colour change colour easily visible/ in order not to overshoot the endpoint.
(ii) To ensure (high) accuracy of the titre value
(iii) Using Dilution formula
M\(_1\)V\(_1\) = M\(_2\)V\(_2\)
M\(_1\) = 2.5, V\(_1\) = ?, M\(_2\) = 0.20, V\(_2\) = 500
\(\frac{M_1V_1}{M_1}\) = \(\frac{M_2V_2}{M_1}\)
Make V\(_1\) the subject
V\(_1\) = \(\frac{M_2V_2}{m_1}\)
= \(\frac{0.20 \times 500}{2.5}\)
= 40.0 cm\(^3\)
Source: Myschool Past Questions Archive (Record #59983)
Chemistry
WAEC / WASSCE 2016
Nature of Matter & Separation Techniques
Solution
All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet
C is a mıxture of two salts, containing one cation and two anions. 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 beaker and add about 10 cm\(^3\) of distilled water. Stir well and filter. Keep the filtrate and the residue.
(b) To about 2 cm\(^3\) of the filtrate, add few drops of AgNO\(_{(aq)}\), followed by HNO\(_{3(g)}\). Add excess NH\(_{3(aq)}\) to the resulting mixture.
(c)(i) Put all of the residue into a clean test tube and add about 5 cm of HNO\(_{3(aq)}\)
(ii) To about 2 cm\(^3\) of the solution from 2(c)(i), add NaOH\(_{(aq)}\) in drops and then in excess
(iii) To another 2 cm\(^3\) of the solution from 2(c)(i), add NH\(_{3(aq)}\) in drops and then in excess.
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
Test
Observation
Inference
(a)
C\(_{(s)}\) + distilled water + filter
Party dissolved/soluble colourless filtrate obtained, water residue
C contains soluble and insoluble salts
(b)
Filtrate C + AgNO\(_{3(aq)}\)
+ HNO\(_{3(aq)}\)
+ NH\(_{3(aq)}\)
White precipitate
Precipitate insoluble
Precipitate soluble/dissolved
Cl\(^{-}\) or CO\(_3^{2-}\) or SO\(_3^{2-}\)
Cl\(^-\)
Cl\(^-\) present
(c)(i)
Residue + HNO\(_{3(aq)}\)
Effervescence, colourless,
Odourless gas evolved
Gas turned lime water milky
CO\(_2\) gas from CO\(_3^{2-}\) evolved
(ii)
Solution from C(i) +
NaOH\(_{3(aq)}\) drops
Then in excess
White gelatinous precipitate
Precipitate dissolved/soluble
Zn\(^{2+}\) or Al\(^{3+}\) present
(iii)
Solution from C(i) + NH\(_{3(aq)}\) drops
Then in excess
White gelatinous precipitate
Precipitate dissolved/soluble
Zn\(^{2+}\) or Al\(^{3+}\) present
Zn\(^{2+}\) present
Source: Myschool Past Questions Archive (Record #59984)
Chemistry
WAEC / WASSCE 2016
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account of experimental procedure is required. All calculations must be done in your answer booklet
A solution of 0.050 moldm\(^3\) H\(_2\)C\(_2\)O\(_4\) (ethanedioic acid). B is a solution of KMnO\(_4\), (potassium tetraoxomanganate (VII), of unknown concentration.
(a) Put B into the burette. Pipette 20.0 cm\(^3\) or 25.0 cm\(^3\) of A into a Conical flask and add about 10.0 cm\(^3\) of dilute H\(_2\)SO\(_4\), Heat the mixture to about 40^\circC - 50^\circC and titrate it while still hot with B. Repeat the titration to obtain consistent titre values. Tabulate your results and calculate the average volume of B used. The equation of reaction is;
2MnO\(_{4(aq)}^-\) + 5C\(_2\)O\(_{4(aq)}^{2-}\) + 16H\(^+_{(aq)}\) \(\to\) 2MnH\(^{2+}_{(aq)}\) + 8H\(_2\)O\(_{(l)}\) + 10CO\(_{2(g)}\)
(b) From your results and the information provided, calculate the:
(i) concentration of MnO\(_2^-\) in B in moldm\(^{-1}\)
(ii) concentration of KMnO\(_4^-\) in B in gdm\(^{-3}\)
(iii) volume of CO\(_2\) evolved at s.t.p when 25.0 cm\(^3\) of H\(_2\)C\(_2\)O\(_4\) reacted completely. [0 = 16.0, K= 39.0, Mn = 55.0, Molar volume of gas at s.t.p.= 22.4 dm\(^3\) mol\(^{-1}\)]
Credit will be given for strict adherence to the instructions. for observations precisely recorded and jor accurate inferences. All tests, observations and inferences must be clearly entered in your answer book in ink, at the time they are made
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
Burette Reading (cm\(^3\))
Rough
1st Titration
2nd Titration
3rd Titration
Final Burette Readings (cm\(^3\))
36.50
30.40
41.70
26.80
Initial Burette readings (cm\(^3\))
12.30
05.90
17.10
01.90
Vol. of acid used (cm\(^3\))
24.20
24.50
24.60
24.90
Titre value = \(\frac{24.50 + 24.60}{2}\) = 24.55cm\(^3\)
(b)(i) Molar concentration of KMNO\(_4\)
\(\frac{C(MnO_4^-) \times V(MnO_4^-)}{C(C_2O_4^{2-}) \times V(C_2O_4^{2-})} = \frac{n(MnO_4^-)}{n(C_2O_4^{2-})}\)
\(\frac{C(MnO_4^-) \times 24.55}{0.05 \times 25} = \frac{2}{5}\)
C(MnO\(_4^-\)) = \(\frac{0.05 \times 25 \times 2}{24.55 \times 5}\)
= 0.0204 mol/dm
(ii) Mass concentration of KmnO\(_4\)
Molar mass of KmnO\(_4\) = 39 + 55 + (16 x 4) = 158g/mol
Mass conc. KmnO\(_4\) = molar conc. x molar mass
= 0.0204 x 158 = 3.22g/dm\(^3\)
(iii) Volume of CO\(_2\) (gas evolved)
\(\frac{n(CO_2)}{n (H_2C_2O_4)}\) = \(\frac{10}{5}\)
n(CO\(_2\)) = \(\frac{10}{5}\) x 0.05
= 2 x 0.05 = 0.100 mol
V(CO\(_2\)) at s.t.p = n x Vm
25cm\(^3\) of H\(_2\)C\(_2\)O\(_4\) = 0.100 x 22.4dm\(^3\)mol\(^{-1}\) x \(\frac{25}{1000}\)
= 0.056dm\(^3\) or 56.0 cm\(^3\)
Alternative method
Volume of CO\(_2\) gas evolved
Amount of H\(_2\)C\(_2\)O\(_4\) used = 0.050 x \(\frac{25}{1000}\) = 0.00125 mol
From the equation of the reaction
5 mol of C\(_2\)O\(_4^{2-}\) = 10 mol of CO\(_2\)
0.00125 mol C\(_2\)O\(_4^{2-}\) = 0.00125 mol x \(\frac{10}{5}\)
= 0.0025 mol CO\(_2\)
@S.T.P
1 mol of CO\(_2\) = 22.4dm\(^3\)
0.00250 mol = 22.4 x 0.0025
= 0.056 dm\(_3\) or 56.0 cm\(^3\)
Source: Myschool Past Questions Archive (Record #59985)
English Language
UTME / JAMB 2016
Prescribed Literature Texts & Literary Appreciation
These questions are based on A.H. Mohammed's The Last Days at Forcados High School.
At the end of term in Forcados High School. Nene was saddled with the responsibility of
A
leading her house in the race
B
packaging the gifts for presentation
C
arranging the school hall for dinner
D
organizing the carol night
medium Difficulty
Verified Correct Answer: Option D
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #44760)
English Language
UTME / JAMB 2016
Prescribed Literature Texts & Literary Appreciation
Efua was not in the good books of teachers at Forcados High School because
A
the students of Forcados did not like her
B
she was expelled from her previous school
C
she bribed the principal before she was admitted
D
her admission into senior class was unusual
medium Difficulty
Verified Correct Answer: Option D
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #44761)
English Language
UTME / JAMB 2016
Prescribed Literature Texts & Literary Appreciation
In the novel, Ansa looked around glumly when Jimi was engrossed in laughter and chatter because he was
A
neglected by Jimi
B
anxious to go home
C
distracted by a band playing on the field
D
given twelve strokes of the cane by the principal
medium Difficulty
Verified Correct Answer: Option A
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #44762)
English Language
UTME / JAMB 2016
Prescribed Literature Texts & Literary Appreciation
Mama Shilifat had an affection for Jimi because she always gave him
A
money to buy chewing gum
B
advice against bad students
C
the courage to do well in his class
D
the largest pieces of puff-puff
medium Difficulty
Verified Correct Answer: Option D
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #44763)
English Language
UTME / JAMB 2016
Prescribed Literature Texts & Literary Appreciation
Ansa licked his lips and said furtively, look this is between you and me. Jimi would be angry if i told you but his brother is ---- well, we suspect he sometimes sells drugs and...'
From the excerpt above, Ansa was talking to
A
Risikat and Nene
B
Novi
C
Caro
D
Efua and Nene
medium Difficulty
Verified Correct Answer: Option D
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #44764)
English Language
UTME / JAMB 2016
Prescribed Literature Texts & Literary Appreciation
The speculation amongst the students of Forcados High School was that Jimi was dating
A
Risikat
B
Joke
C
Caro
D
Efua
medium Difficulty
Verified Correct Answer: Option D
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option D.
Source: Myschool Past Questions Archive (Record #44765)
English Language
UTME / JAMB 2016
Prescribed Literature Texts & Literary Appreciation
Mr Solade sent Jimi to Ibadan for the holidays in order to
A
save Jimi from Wole's bad influence
B
heed Jimi's request to stay with his friends
C
allow Jimi spend some time with his friends
D
allow Jimi to know more about Ibadan
medium Difficulty
Verified Correct Answer: Option A
Grammar & Contextual Explanation
No expanded working has been attached for this question yet. The verified answer is Option A.
Source: Myschool Past Questions Archive (Record #44766)