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Chemistry
WAEC / WASSCE 2017
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.
State what would be observed if the following reactions are carried out in the laboratory:
(i) methyl orange is dropped into a solution of lime juice:
(ii) hydrogen sulphide gas is bubbled through Iron (III) chloride solution:
(iii) sulphur (IV) oxide gas is bubbled into acidified solution of KMnO\(_4\):
(iv) ethanoic acid is added to a solution of Ka\(_2\)CO\(_3\)
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
(a)(i) Solution of lime juice turns to pink or red,
(ii) The FeCl\(_3\) solution changes colour from brown to green and a yellow deposit.
(iii) The purple colour of KmnO\(_4\) solution turns colourless or decolourised.
(iv) The solution of K\(_2\)CO\(_3\) reacts with the addition of ethanoic acid to evolve a colourless, odourless gas with bubbling or effervescence
Source: Myschool Past Questions Archive (Record #59979)
Chemistry
WAEC / WASSCE 2017
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.
F is 2 mixture of two inorganic salts. Carry out the following exercises on F. Record your observations and identify any gas(es) evolved. State the conclusions you draw from the result of each test.
(a) Put all of F in a beaker and add about 10 cm\(^3\) of distilled water. Stir well and filter. Keep the filtrate and the residue.
(b)(i) To about 2cm\(^3\) of the filtrate. add NaOH\(_{(aq)}\) in drops and then in excess.
(ii) To another 2cm\(^3\) portion of the solution, add a few drops of NH3\(_{(aq)}\) in drops and then in excess.
(c) To about 2cm\(^3\) of the solution, add a few drops of HNO\(_{3(aq)}\) followed by few drops of the drops of AgNO\(_{3(aq)}\)
(d)(i) Put all the residue into a clean test-tube and add HNO\(_{3(aq)}\)
(ii) To a portion of the solution from (d)(i)) add NaOH\(_{(aq)}\) in drops and then in excess.
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
TEST
OBSERVATION
INTERFERENCE
(a)
F + distil water + stir + filter
Effervescence (bubbling) occurs. Colourless and odourless gas evolved after filtration pale blue/colourless filtrate and green residue
(b)(i)
Filtrate + NaOH\(_{(aq)}\) in drops
Then in excess
White precipitate formed
Precipitate insoluble in excess NaOH\(_{(aq)}\)
Ca\(^{2+}\)/pb\(^{2+}\)
Ca\(^{2+}\) present
(b)(ii)
Filtrate + NH\(_3\) \(_{(aq)}\) in drops
Then in excess
No visible reaction
No visible reaction
Ca\(^{2+}\) present
Ca\(^{2+}\) confirmed
(b)(i)
Filtrate + NaOH\(_{(aq)}\) in drops
Then in excess
Blue gelatinous precipitate formed
Precipitate insoluble in excess
Cu\(^{2+}\) is present
(b)(ii)
Filtrate + NH\(_3\) \(_{(aq)}\) in drops
Then in excess
Pale (light) blue gelatinous precipitate formed
Precipitate dissolves to form a deep blue solution
Cu\(^{2+}\)
Cu\(^{2+}\) confirmed
(c)
Filtrate + NHO\(_3\) \(_{(aq)}\) + AgNO\(_3\)\(_{(aq)}\)
No visible reaction
No gas involved
White precipitate formed
Cl\(^{-}\) present
(d)(i)
Residue + HNO\(_{3(aq)}\)
Effervescence occurs, colourless and odourless gas evolves that turn blue litmus paper to red and turns lime water milky
CO\(_{3(g)}\)
CO\(_{3^{2-}}\) present
(d)(ii)
Solution from (d)(i) + NaOH\(_{(aq)}\) in drops
Then in excess
Blue precipitate formed
present Precipitate insoluble in excess NaOH\(_{(aq)}\)
Cu\(^{2+}\)
Cu\(^{2+}\) present
Source: Myschool Past Questions Archive (Record #59980)
Chemistry
WAEC / WASSCE 2017
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 is 0.200 moldm\(^3\) of HCl. C is a solution containing 14.3g of Na2C\(O_3\).xH\(_2\)O in 500 cm\(^3\) of solution.
a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0cm\(^3\) portions of C using methyl orange as indicator. Repeat the titration to obtain Consistent titre values. Tabulate vour results and calculate the average volume of A used. The equation for the reaction is;
Na\(_2\)CO\(_3\) . xH\(_2\)O + 2HCl\(_{(aq)}\) \(\to\) 2NaCl\(_{(aq)}\) + CO\(_{2(g)}\) + (x + 1)\(_3\)H\(_2\)O\(_{(l)}\)
(b) From your results and the information provided. calculate the:
(i) concentration of C in moldm\(^{-3}\)
(ii) concentration of C in gdm\(^{-3}\)
(iii) molar mass of Na\(_2\)CO\(_3\) . xH\(_2\)O
(iv) the value of x in Na\(_2\)CO\(_3\) . xH\(_2\)O. [H =1.0; C = 12.0; O = 16.0; Na = 23.0]
Credit will be given for strict adherence to the instruction, for observations precisely recorded and for accurale references. All tests. obsenations and influences must be cleary entered in the booklet in ink at the same time they are made.
medium Difficulty
Verified Correct Answer: Option
Chemical Principles & Mechanism
Indicator = Methyl Orange
Volume of the base used = 25.00cm\(^3\)
Titration
Rough Titre
1st Titre
2nd Titre
3rd Titre
Final burette readings cm\(^3\)
24.70
24.80
24.70
24.90
Initial burette reading cm\(^3\)
0.00
0.00
0.00
0.00
Volume of acid used cm\(^3\)
24.70
24.80
24.70
24.90
Average Titre = \(\frac{1st + 2nd + 3rd}{3}\)
= \(\frac{24.80 + 24.70 + 24.90}{3}\)
= 24.80cm\(^3\)
Alternatively 2 concordant titres can be used to calculate average titre
Equation of thereaction; Na\(_2\)CO\(_3\).XH\(_2\)O + 2HCl\(_{(aq)}\) \(\to\) 2Nacl\(_{(aq)}\) \(\to\) 2Nacl\(_{(aq)}\) + CO\(_{2(aq)}\) + (X + 1)H\(_2\)O\(_{(l)}\)
\(\frac{C_AV_A}{C_BV_B} = \frac{n_A}{n_B}\)
C\(_A\) = Molar concentration of HCl\(_{(aq)}\) in moldm\(^{3}\)
V\(_A\) = Volume of acid used in cm\(^3\) = 24.80cm\(^3\)
C\(_B\) = Molar concentration of Na\(_2\)Cu\(_3\) . xH\(_2\)O in molddm\(^3\)
n\(_A\) = 2
n\(_B\) = 1
(b)(i) Concentration of C in moldm\(^{-3}\) = ?
From the equation reaction;
\(\frac{C_AV_A}{C_BV_B} = \frac{n_A}{n_B}\)
C\(_A\) = 0.200 molddm\(^{-3}\)
C\(_B\) = ?
V\(_A\) = 24.80 cm
V\(_B\) = 25.00 cm\(^3\)
Substitution of known values
\(\frac{0.200 \times 24.80}{C_B \times 25.00} = \frac{2}{1}\)
C\(_B\) = \(\frac{1 \times 0.200 \times 24.80}{2 \times 25.00}\)
C\(_B\) = 0.0992 moldm\(^{-3}\)
(ii) Concentration of C in gdm\(^3\) = ?
500 cm\(^2\) \(\to\) 14.3 g
1000cm\(^3\) \(\to\) \(\frac{14.3}{500}\) x 1000g
= 28.6 dm\(^{-3}\)
(iii) Molar mass of Na\(_2\)CO\(_3\) . XH\(_2\)O
Molar conc in moldm\(^{-3}\) = \(\frac{\text{conc. in gdm}^{-3}} {\text{molar mass} }\)
Molar mass gmol\(^{-1}\) = \(\frac{\text{conc in g dm}^{-3}}{\text{molar conc.in moldm}^{-3}}\)
\(\frac{28.6 gdm^{-3}}{0.0992 moldm^{-3}}\)
= 288.3065
= 288 gmol\(^{-1}\)
(iv) Value of x in Na\(_2\)CO\(_3\) . xH\(_2\)O?
[H = 1.0, C = 12.0, O = 16.0, Na = 23.0]
N\(_2\)CO\(_3\) . xH\(_2\)O = 288
2(23) + 12 + 3 (16) + x(2(1) + 16) = 288
46 + 12 +48 +18x = 288
106 + 18x = 288
18x = 288 - 106
18x = 182
x = \(\frac{182}{18}\)
x = 10.11
x = 10
Source: Myschool Past Questions Archive (Record #59981)
Chemistry
UTME / JAMB 2017
Stoichiometry, Mole Concept & Volumetric Analysis
Solution
The IUPAC nomenclature of the structure above is
A
3-methybut-3-ene
B
2-methylbut-1-ene
C
2-ethylprop-1-ene
D
2-methylbut-2-ene
medium Difficulty
Verified Correct Answer: Option B
Chemical Principles & Mechanism
CH2CH2 - CCH3CH2
Start the numbering from the terminal carbon.
Source: Official Exam / Past Question Archive
Chemistry
WAEC / WASSCE 2017
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 is 0.200 moldm\(^3\) of HCl. C is a solution containing 14.3g of Na2C\(O_3\).xH\(_2\)O in 500 cm\(^3\) of solution.
a) Put A into the burette and titrate it against 20.0 cm\(^3\) or 25.0cm\(^3\) portions of C using methyl orange as indicator. Repeat the titration to obtain Consistent titre values. Tabulate vour results and calculate the average volume of A used. The equation for the reaction is;
Na\(_2\)CO\(_3\) . xH\(_2\)O + 2HCl\(_{(aq)}\) \(\to\) 2NaCl\(_{(aq)}\) + CO\(_{2(g)}\) + (x + 1)\(_3\)H\(_2\)O\(_{(l)}\)
(b) From your results and the information provided. calculate the:
(i) concentration of C in moldm\(^{-3}\)
(ii) concentration of C in gdm\(^{-3}\)
(iii) molar mass of Na\(_2\)CO\(_3\) . xH\(_2\)O
(iv) the value of x in Na\(_2\)CO\(_3\) . xH\(_2\)O. [H =1.0; C = 12.0; O = 16.0; Na = 23.0]
Credit will be given for strict adherence to the instruction, for observations precisely recorded and for accurale references. All tests. obsenations and influences must be cleary entered in the booklet in ink at the same time they are made.
A
is formed naturally
B
is cheap
C
can be recycled after use
D
cannot be regenerated once used up
medium Difficulty
Verified Correct Answer: Option D
Chemical Principles & Mechanism
Indicator = Methyl Orange
Volume of the base used = 25.00cm\(^3\)
Titration
Rough Titre
1st Titre
2nd Titre
3rd Titre
Final burette readings cm\(^3\)
24.70
24.80
24.70
24.90
Initial burette reading cm\(^3\)
0.00
0.00
0.00
0.00
Volume of acid used cm\(^3\)
24.70
24.80
24.70
24.90
Average Titre = \(\frac{1st + 2nd + 3rd}{3}\)
= \(\frac{24.80 + 24.70 + 24.90}{3}\)
= 24.80cm\(^3\)
Alternatively 2 concordant titres can be used to calculate average titre
Equation of thereaction; Na\(_2\)CO\(_3\).XH\(_2\)O + 2HCl\(_{(aq)}\) \(\to\) 2Nacl\(_{(aq)}\) \(\to\) 2Nacl\(_{(aq)}\) + CO\(_{2(aq)}\) + (X + 1)H\(_2\)O\(_{(l)}\)
\(\frac{C_AV_A}{C_BV_B} = \frac{n_A}{n_B}\)
C\(_A\) = Molar concentration of HCl\(_{(aq)}\) in moldm\(^{3}\)
V\(_A\) = Volume of acid used in cm\(^3\) = 24.80cm\(^3\)
C\(_B\) = Molar concentration of Na\(_2\)Cu\(_3\) . xH\(_2\)O in molddm\(^3\)
n\(_A\) = 2
n\(_B\) = 1
(b)(i) Concentration of C in moldm\(^{-3}\) = ?
From the equation reaction;
\(\frac{C_AV_A}{C_BV_B} = \frac{n_A}{n_B}\)
C\(_A\) = 0.200 molddm\(^{-3}\)
C\(_B\) = ?
V\(_A\) = 24.80 cm
V\(_B\) = 25.00 cm\(^3\)
Substitution of known values
\(\frac{0.200 \times 24.80}{C_B \times 25.00} = \frac{2}{1}\)
C\(_B\) = \(\frac{1 \times 0.200 \times 24.80}{2 \times 25.00}\)
C\(_B\) = 0.0992 moldm\(^{-3}\)
(ii) Concentration of C in gdm\(^3\) = ?
500 cm\(^2\) \(\to\) 14.3 g
1000cm\(^3\) \(\to\) \(\frac{14.3}{500}\) x 1000g
= 28.6 dm\(^{-3}\)
(iii) Molar mass of Na\(_2\)CO\(_3\) . XH\(_2\)O
Molar conc in moldm\(^{-3}\) = \(\frac{\text{conc. in gdm}^{-3}} {\text{molar mass} }\)
Molar mass gmol\(^{-1}\) = \(\frac{\text{conc in g dm}^{-3}}{\text{molar conc.in moldm}^{-3}}\)
\(\frac{28.6 gdm^{-3}}{0.0992 moldm^{-3}}\)
= 288.3065
= 288 gmol\(^{-1}\)
(iv) Value of x in Na\(_2\)CO\(_3\) . xH\(_2\)O?
[H = 1.0, C = 12.0, O = 16.0, Na = 23.0]
N\(_2\)CO\(_3\) . xH\(_2\)O = 288
2(23) + 12 + 3 (16) + x(2(1) + 16) = 288
46 + 12 +48 +18x = 288
106 + 18x = 288
18x = 288 - 106
18x = 182
x = \(\frac{182}{18}\)
x = 10.11
x = 10
Source: Official Exam / Past Question Archive
English Language
UTME / JAMB 2017
Antonyms (Opposite in Meaning)
Solution
Choose the word that is most nearly opposite in the meaning to the underlined word or group of words that will fill in the sentence.
At first, we found life in the town very exciting but soon it became rather _______
A
tiring
B
disturbing
C
burdensome
D
boring
medium Difficulty
Verified Correct Answer: Option D
Grammar & Contextual Explanation
exciting: to bring out, stimulate or arouse interest
boring: causing boredom or unable to engage one's interest
Source: Myschool Past Questions Archive (Record #46468)
English Language
UTME / JAMB 2017
Concord (Subject-Verb Agreement)
Solution
Select the option that best explain the information conveyed in the sentence
The man's Achilles' heel is women ________
A
His weakness is women
B
His strength is women
C
He fights women
D
He respects women
medium Difficulty
Verified Correct Answer: Option A
Grammar & Contextual Explanation
The phase 'Achilles' heel' means a small problem or weakness in a person or system that can result in failure. The phrase has it origin traced to the legend of Achilles, which tells the talk of Achilles, how he was dipped into the river by his mother, That is in order to make him invulnerable. His heel wasn't covered by the water and he was later killed by a narrow wound to his heel.
Source: Myschool Past Questions Archive (Record #46470)
English Language
UTME / JAMB 2017
Oral English: Stress, Rhymes & Intonation
Solution
Choose the most appropriate stress pattern from the options below
Constitution
A
CON—sti—tu—tion
B
con—STI—tu—tion
C
con—sti—TU—tion
D
con—sti—tu—TION
medium Difficulty
Verified Correct Answer: Option C
Grammar & Contextual Explanation
'Constitution' is stressed on the third syllable as seen in option C
Source: Myschool Past Questions Archive (Record #46472)
English Language
UTME / JAMB 2017
Lexis, Structure & Register (Cloze & Vocabulary)
Solution
Choose the option that has the same consonant as the one represented by the letter(s) underlined
clash
A
sharp
B
cheap
C
leisure
D
school
medium Difficulty
Verified Correct Answer: Option A
Grammar & Contextual Explanation
Option A sharp has the same consonant at the beginning of the word
Source: Myschool Past Questions Archive (Record #46473)
English Language
UTME / JAMB 2017
Oral English: Vowel & Consonant Sounds
Solution
Choose the option that has the same consonant as the one represented by the letter(s) underlined
who
A
vehicle
B
khaki
C
rhyme
D
hammer
medium Difficulty
Verified Correct Answer: Option D
Grammar & Contextual Explanation
Option D has the same consonant sound at the beginning of the word 'hammer'
Source: Myschool Past Questions Archive (Record #46474)
English Language
UTME / JAMB 2017
Oral English: Stress, Rhymes & Intonation
Solution
Choose the option to which the given sentence relates
My Boss has an amiable DISPOSITION
A
Does your boss have an amiable disposition?
B
Does my boss have an unfriendly disposition?
C
Does my colleague have an amiable disposition?
D
Does my boss have an amiable complexion?
medium Difficulty
Verified Correct Answer: Option D
Grammar & Contextual Explanation
Still going by the rule when answering questions on emphatic stress, the option that doesn't have the emphasized word. In this case 'DISPOSITION' is the emphasized word on the only option without it is option D
Source: Myschool Past Questions Archive (Record #46475)
English Language
UTME / JAMB 2017
Oral English: Stress, Rhymes & Intonation
Solution
Choose the most appropriate stress pattern from the options below
palatable
A
PAL—a—ta—ble
B
pal—A—ta—ble
C
pal—a—TA—ble
D
pal—a—ta—BLE
medium Difficulty
Verified Correct Answer: Option A
Grammar & Contextual Explanation
Some words are made up of a root and a suffix:
dangerous = root –> danger + ous –< suffix
commercial = root –> commerc(e) + al –< suffix
In some words with suffixes, the stress stays on the same syllable as in the root. Compare:
‘danger and: ‘dangerous
In other words, the suffix changes the stressed syllable. Compare:
‘commerce and: com‘mercial
Suffixes which don’t usually change the stress pattern in the root word include (however they maybe some exceptions.) -able, -age, -al, -er, -ful, -less, -ness, -ous and -fy. For example:
comfort – ‘comfortable
per‘cent – per’centage
e‘lectric – e‘lectrical
‘amplify –‘amplifier
re‘gret – re‘gretful
re‘gard – re‘gardless
‘foolish – ‘foolishness
di‘saster – di‘sastrous
‘beauty – ‘beautify
(However they maybe some exceptions.) Exceptions with -able and -al include:
ad‘mire – ‘admirable
pre‘fer – ‘preferable
‘medicine – me‘dicinal
‘agriculture – agri‘cultural
Source: Myschool Past Questions Archive (Record #46476)
English Language
UTME / JAMB 2017
Idioms, Figurative Expressions & Sentence Interpretation
Solution
Choose the interpretation that you consider most appropriate for each sentence.
Despite their newness on the job, the workers were remunerated rather handsomely. This means that _________?
A
the salaries were poor
B
those paid were handsome in appearance
C
the salaries were attractive
D
workers were paid by handsome accountants
medium Difficulty
Verified Correct Answer: Option C
Grammar & Contextual Explanation
The word 'remunerate' as used in the sentence means paid for services rendered or work done, while the word 'handsomely' means to a large degree, substantially and also connotes attractiveness
Source: Myschool Past Questions Archive (Record #46477)
English Language
UTME / JAMB 2017
Antonyms (Opposite in Meaning)
Solution
Choose the option opposite in meaning to the word or phrase in italics
The chairman is Parsimonious
A
Avaricious
B
thrifty
C
generous
D
accustomed
medium Difficulty
Verified Correct Answer: Option C
Grammar & Contextual Explanation
The word Parsimonious means very unwilling to spend money or use resources. On the other hand 'Generous is the opposite', meaning to give more of something especially money
From the above meaning of each words of option, generous best describes the opposite of parsimonious, as every other option is synonymous to parsimonious
Source: Myschool Past Questions Archive (Record #46478)
English Language
UTME / JAMB 2017
Question Tags, Modals & Pronouns
Solution
Choose the interpretation that you consider most appropriate for each sentence.
Now that Michael has become rich, Nancy has begun to make much of him. This means __________
A
now values Michael
B
now gets a lot of money from Michael
C
only recently married Michael
D
no longer wants to leave Michael
medium Difficulty
Verified Correct Answer: Option A
Grammar & Contextual Explanation
Make much of means to give much attention to someone or something or to treat very well
Source: Myschool Past Questions Archive (Record #46479)