Chemistry 2017 JAMB Past Questions

1. The general formula of alkanones is

A. RCHO
B. RCOR'
C. RCOOH
D. RCOOR

Correct Answer: Option B

2.The constituent common to duralumin and alnico is

A. Co
B. Mn
C. Al
D. Mg

Correct Answer: Option C

Explanation

Constituents of duralumin are: Al, Cu, Mg, Mn.
Constituents of Alnico are: Al, Ni and Co

In 1909, the alloy of duralumin was discovered by Alfred Wilon consisting of 94% Al, 4% Cu, 1% Mg and 1% Mn(Manganese)
Alnico is an acronym referring to a family of iron alloys which in addition to iron are composed primarily of Al, Ni and Co.

3.The shape of the S-orbital is

A. elliptical
B. spiral
C. circular
D. spherical

Correct Answer: Option D

Explanation

The shape of the S-orbital is spherical.

4.Aluminium hydroxide is used in the dyeing industry as a

A. dye
B. dispersant
C. salt
D. mordant

Correct Answer: Option D

5.The tincture of iodine means iodine dissolved in

A. ethanol
B. bromine chloride
C. chlorine water
D. water

Correct Answer: Option A

Explanation

It is also called weak iodine solution. Tincture solutions are characterized by the presence of alcohol.

6.Temporary hard water is formed when rain water containing dissolved carbon(IV) oxide flows over deposits of

A. CaCO₃
B. Na₂CO₃
C. Na₂SO₄
D. CaSO₄

Correct Answer: Option A

Explanation

Permanent hardness in water= Mg & Ca Sulphate.
Temporary hardness in water= Mg & Ca Carbonate.

Temporary hardness of water is caused by Magnesium and calcium hydrogencarbonate. It is formed when rainwater containing dissolved CO₂ flows over deposit of CaCO₃

i.e CaCO₃ + CO₂ + H₂O → Ca(HCO₃)₂

7.The acid anhydride that will produce weak acid in water is

A. SO₃
B. NO₂
C. SO₂
D. CO₂

Correct Answer: Option D

Explanation

H₂CO₃ is an example of a weak acid while H₂SO₄ and HNO₃ are examples of a strong acid.

CO₂ combines with water to give a weak trioxocarbonate (IV) acid.

CO₂ + H₂O → H₂CO₃

8.The process that occurs when two equivalent forms of a compound are in equilibrium is

A. Isotopy
B. Resonance
C. Isomerism
D. Reforming

Correct Answer: Option B

Explanation

Resonance involves two forms of a compound.
Isomerism involves two or more forms of an element.
Reforming involves the rearrangement of molecule.

This deals with a two forms of a molecule where the chemical connectivity is the same but the electrons are distributed differently around the structure.

9.In the laboratory preparation of ethyl ethanoate, the water present in the mixture is removed using a solution of

A. an hydrous CaCl₄
B. concentrated NaCO₄
C. dilute NaOH
D. concentrated H₂SO₄

10.The constituent of air necessary in the rusting process are

A. O₂ and H₂O
B. Ar and CO₂
C. CO₂ and H₂O
D. O₂ and CO₂

Correct Answer: Option A

Explanation

Rust is an iron oxide, a usually red oxide formed by the redox reaction of iron and oxygen in the presence of water or air moisture.

11.For a general equation of the nature xP + yQ → mR + nS, the expression for the equilibrium constant is

A. k [P]^x [Q]^y
B. $\frac{[P]^{x} [Q]^{y}}{[R]^{m} [S]^{n}}$

\frac{[R]^{m} [S]^{n}}{[P]^{x} [Q]^{y}}

\frac{m [R] n [S]}{x [P] y [Q]}

Correct Answer: Option C

Explanation

Expression for equilibrium constant

k =

\frac{concentration of product}{concentration of reactant}

12. A given mass of gas occupies 2dm³ at 300k. At what temperature will its volume be doubled, keeping the pressure constant?

A. 400k
B. 480k
C. 550k
D. 600k

Correct Answer: Option D

Explanation

At constant pressure connotes Charle's law

\frac{V_{1}}{T_{1}}

\frac{V_{2}}{T_{2}}

\frac{2 d m^{3}}{300 k}

= 2 x

\frac{2 d m^{3}}{T_{2}}

T₂= 600k

13.The oxidation number of iodine in KIO₃ is

A. +7
B. +3
C. +5
D. +6

Correct Answer: Option C

Explanation

If the rules are followed, it is non-negotiable that +5 is the suitable answer to the question.

14. An isomer of C₅H₁₂ is

A. 2–ethyl butane
B. butane
C. 2–methyl butane
D. 2–methyl propane

Correct Answer: Option C

Explanation

C₅H₁₂ has 5 carbons acid 12 hydrogens.

2–methyl butane

No of carbon=5

No of hydrogen=12

15.When few drops of concentrated trioxonitrate(V) acid is added to an unknown sample and wanned an intense yellow colouration is observed. The likely functional group present in the sample is

A. NH₃—C—C=O
B. CHO
C. CO
D. CNH₂
Correct Answer: Option A

Explanation
Xanthopreitic test for the presence of protein, when conc nitric acid is added to the drop, an intense yellow colouration is observed.

It contains all the functional group of protein which includes the amino, alkanol and the carboxylic group. Adding few drops of conc HNO₃ to a protein, gives an intense yellow colouration.

It is called Xanthopreitic test

16. A sample of orange juice is found to have a PH of 3.80. What is the concentration of the hydroxide ion in the juice?

A. 1.6 $\times$ 10^-4
B. 6.3 $\times$ 10^-11
C. 6.3 $\times$ 10^-4
D. 1.6 $\times$ 10^-11

Correct Answer: Option B

Explanation

P^H = - Log[ H⁺ ]

P^H + P^OH =14

P^OH = 14 - 3.8

P^OH = 10.2

P^OH = - Log[ OH^- ]

10.2 = - Log[ OH^- ]

10^-10.2 = [ OH^- ]

[ OH^- ] = 6.3 x 10^-11

17. Incomplete oxidation of ethanol yields

A. CH₃COOH
B. CH₃COCH₃
C. CH₃CH₂OCH₂CH₃
D. CH₂CHO

Correct Answer: Option D

Explanation

C2H5OH → CH3CHO → CH3COOH
Ethanol.....oxidation Ethanal........Ethanoic Acid
Primary alcohol oxidises to aldehyde and later to carboxylic acid.
Secondary alcohol oxidises to ketones.
Ethanol is an example of primary alcohol and primary alcohol can be oxidised to aldehyde and carboxylic acid. Wherein, incomplete oxidation of primary alcohol yields aldehyde also known as alkanal while complete oxidation of primary alcohol yields carboxylic acid.

18. The salt formed from a weak acid and a strong base hydrolyzes in water to form

A. A saturated solution
B. an acidic solution
C. a buffer solution
D. an alkaline solution

Correct Answer: Option D

Explanation

This is a solution formed from the hydrolyses of an alkali in water.

19. The ideal gas laws and equations are true for all gases at

A. low pressures and lower temperatures
B. low temperatures and high pressures
C. high pressures and high temperatures
D. low pressure and high temperatures

Correct Answer: Option D

Explanation

At high temperature and low pressure, the vanderwaal equation is reduced to ideal gas equation.

i.e, [P +

\frac{a}{v^{2}}

] [v - b] = RT is reduced to PV = nRT

Generally, a gas behaves more like an ideal gas at higher temperature and lower pressure as the potential energy due to intermolecular forces, it becomes less significant compared with the particles kinetic energy and the size of the molecules, then it becomes less significant compared to the empty space between them.

20.When ΔH is negative, a reaction is said to be

A. endothermic
B. exothermic
C. reversible
D. ionic

Correct Answer: Option B

Explanation

When ΔH is negative, heat is liberated to the surrounding and it connotes an exothermic reaction.
ΔH = −ve

_{88}^{226} R a

→

_{86}^{x} R n

+ alpha particle

A. 226
B. 220
C. 227
D. 222

Correct Answer: Option D

Explanation

_{88}^{226} R a

→

_{86}^{x} R n

_{2}^{4} H e

_{2}^{4} H e

= alpha particle

considering the summation of the mass number

226 = x + 4

x = 226 - 4

x = 222

22. The shape of ammonia molecules is

A. trigonal planar
B. octahedral
C. square planar
D. tetrahedral

Correct Answer: Option A

Explanation

Ammonia = NH_{3

It has three bonds of hydrogen to the Nitrogen.}

23. The mass of silver deposited when a current of 10A is passed through a solution of silver salt for 4830s is – (Ag = 108 F = 96500(mol-1)

A. 54.0g
B. 27.0g
C. 13.5g
D. 108.0g

Correct Answer: Option A

Explanation

Recall that

mass deposited =

\frac{M m I t}{96500 n}

Mm =108, t = 4830s

I = 10A, n = 1

m = 108 × 10 × (

\frac{4830}{96500}

) × 1

m = 54.0g

24.Tin is unaffected by air at ordinary temperature due to its

A. Low melting point
B. Weak electropositive character
C. High boiling point
D. White lustrous appearance

Correct Answer: Option A

Explanation

Tin has a melting point of 232° which enables it to be unaffected by air at ordinary temperature coupled with the fact that it also helps in making a good metal for alloying.

Tin is relatively unaffected by both water and oxygen at room temperature due to its low melting point. It does not rust, corrode, or react in any other way. This explains one of its major uses: as a coating to protect other metals.

25.Calculate the amount in moles of silver deposited when 9650C of electricty is passed through a solution of silver salt [= 96500 Cmol^-1]

A. 0.05
B. 10.80
C. 10.00
D. 0.10

Correct Answer: Option D

Explanation

m =

\frac{M m \times Q}{96500 n}

where Q = IT

M = Mm ×

\frac{Q}{96500 n}

where m = mass

Mm = Molar mass

Q = Quantity of electricity

n = number of change= +1

\frac{M}{M m}

= mole =

\frac{m a s s}{M o l a r m a s s}

\frac{M}{M m}

\frac{Q}{96500 n}

\frac{9650}{96500 n}

× 1

=

\frac{1}{10}

= 0.1mol

26.The reaction of halogens with alkanes in the presence of sunlight is an example of

A. oxidation reaction
B. addition reaction
C. hydrogenation reaction
D. substitution reaction

Correct Answer: Option D

Explanation

Alkanes undergoes substitution reaction and it is an example of halogenation substitution reaction.

Reaction of halogen with alkane in the presence of sunlight (ultraviolet) is termed halogenation. Halogenation is an example of substitution reaction. In addition, alkanes only undergo substitution reaction but not addition reaction

27.The enzyme used in the hydrolysis of starch to dextrin and maltose is

A. amylase
B. diastatse
C. invertase
D. zymase

Correct Answer: Option B

Explanation

This is any amylase or a mixture of amylases that converts starch to dextrin and maltose.

28. The alkyl group is represented by the general formula

A. C_nH₂n
B. C_nH₂n − 2
C. C_nH₂n + 1
D. C_nH₂n + 2

Correct Answer: Option C

Explanation

Alkyl group has the general formula C_nH₂n + 1.

This formed when one hydrogen is removed from the alkane family. It has the general formula C_nH₂n + 1

29.Cu2S(g) + O2(g) → 2Cu + SO2(g)

What is the change in the oxidation number of copper in the reaction?

A. 0 to + 2
B. 0 to + 1
C. C + 1 to 0
D. + 2 to + 1

Correct Answer: Option C

Explanation

In the reactant;

Cu2S

2 Cu - 2(1) = 0

2 Cu = 2

Cu =

\frac{2}{2}

Cu = +1

In the product, Cu

Cu = O

The oxidation number of Cu in Cu2S and Cu respectively is +1 and 0 respectively

In the diagram above. X is

A. SO₃
B. SO₂
C. S
D. H₂S

Correct Answer: Option B

Explanation

The setup represents the production of sulfur dioxide. And the cylinder marked X is SO₂

The diagram above. Y is

A. fused CaO
B. H₂O
C. NaOH
D. Concentrated H₂SO₄

Correct Answer: Option D

Explanation

In the preparation of sulphur dioxide by the action of dilute acids on sulphates and bisulphites. conc H₂SO₄ helps to release SO₂ from the mixture.
The setup represents the production of sulphur dioxide

32.Calculate the mass of copper deposited when a current of 0.5 ampere was passed through a solution of copper(II) chloride for 45 minutes in an electrolytic cell. [Cu = 64, F = 96500Cmol^-1]

A. 0.300g
B. 0.250g
C. 0.2242g
D. 0.448g

Correct Answer: Option D

Explanation

M =

\frac{Molar mass × Quantity of Electricity}{96500 × no of charge}

\frac{M m I T}{96500 n}

Copper II Chloride = CuCl₂

CuCl₂ → Cu²⁺ + 2Cl²

Mass of compound deposited =

\frac{Molar mass × Quantity of Electricity}{96500 × no of charge}

Q = IT

I = 0.5A

T = 45 × 60

T = 2700s

Q = 0.5 × 2700

= 1350c

Molarmass = 64gmol^-1

no of charge = + 2

Mass =

\frac{64 \times 1350}{96500 \times 2}

Mass = 0.448g

33.C₃H₅ - COH₃ =CH₂

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

Correct Answer: Option B

Explanation

CH₂CH₂ - CCH₃CH₂
Start the numbering from the terminal carbon.

34.The reddish–brown rust on ion roofing sheets consists of

A. Fe³ + (H₂O)₆
B. FeO.H₂O
C. Fe₂O₃.3H₂O
D. Fe₃O₄.2H2₂O

Correct Answer: Option C

Explanation

Iron [Fe] reacts with H₂ in the presence of oxygen to form a rust.

4Fe + 3O₂ → 2 Fe₂O₂

Fe₂O₃ + H₂O → Fe₂O₃.H₂O

35. The densities of two gases, X and Y are 0.5gdm^-3 and 2.0gdm^-3 respectively. What is the rate of diffusion of X relative to Y?

A. 0.1
B. 0.5
C. 2.0
D. 4.0

Correct Answer: Option C

Explanation

The rate of dimension of a gas inversely proportional to the square root of its molecular mass or its density, which is Graham's Law of diffusion of gas.

R ∝

\frac{1}{\sqrt{M m}}

or R ∝

\frac{1}{\sqrt{D}}

Dx = 0.5gdm^-3, Dy = 2gdm^-3

R=

\frac{K}{\sqrt{D}}

\sqrt{D}

= k

R₁

\sqrt{D_{1}}

= R₁

\sqrt{D_{2}}

R_x

\sqrt{D_{x}}

= R_y

\sqrt{D_{y}}

\frac{R_{x}}{R_{y}}

\frac{\sqrt{D_{y}}}{\sqrt{D_{x}}}

\frac{\sqrt{2}}{\sqrt{0.5}}

= 2.0

36. The carbon atoms on ethane are

A. sp² hybridized
B. sp³ hybridized
C. sp⁴ hybridized
D. sp hybridized

Correct Answer: Option B

Explanation

Alkane's family are sp³ hybridized

37. According to Charle's law, the volume of a gas becomes zero at

A. −100°c
B. −273°c
C. −373°c
D. 0°c

Correct Answer: Option B

Explanation

Where −273°c = OK

i.e −273°c + 273 = OK

At zero kelvin, the volume of a gas becomes zero

38. An oxide XO₂ has a vapour density of 32. What is the atomic mass of X?

A. 20
B. 32
C. 14
D. 12

Correct Answer: Option B

Explanation

Molecular mass = vapour density X₂

Mm of XO₂ = x + 16(2) = x + 32

vapour density = 32

∴ x + 32 = 32 × 2

x + 32 = 64

x = 64 - 32

x = 32

∴ the relative molecular mass of X is 32
Relative molecular mass = vapour density × 2

39.The gas that can be collected by downward displacement of air is

A. chlorine
B. sulphur (IV) oxide
C. carbon (IV) oxide
D. ammonia

Correct Answer: Option D

Explanation

Upward delivery works well for hydrogen and ammonia, which are both less densed than air. Sometimes, they are collected over water.

40. In the laboratory preparation of trioxonitrate (V) acid the nitrogen(iv) oxide formed as a by-product is removed by

A. further heating
B. adding concentrated H₂SO₄
C. cooling the acid solution with cold water
D. bubbling air through the acid solution

Correct Answer: Option D

Explanation

Bubbling of air through the acid solution removes deposited oxides of nitrogen.

Nitric acid is prepared in the laboratory by heating a nitrate salt with the concentrated acid.
NaNO₃ + H₂SO₄ → NaHSO₄ + HNO₃

Vapours of nitric acid are condensed to a brown liquid in a receiver cooled under cold water. "Dissolved oxides of nitrogen" e.g NO₂ are removed by redistillation or blowing a current of carbondioxide or dry air through the warm acid

41.A particle that contains 9 protons, 10 neutrons and 10 electrons is

A. positive ion
B. neutral atom of a metal
C. neutral atom of a non-metal
D. negative ion

Correct Answer: Option D

Explanation

protons = 9

neutrons = 10

electrons = 10

Electronic configuration = 2, 8

Ground state Electronic configuration=2, 7

It means that the atom has gained an electron thereby making it have a negative ion.

When an atom donates an electron, it becomes positively charged.

When an atom accepts an electron, it becomes negatively charged

42. Ethene is prepared industrially by

A. Reforming
B. Polymerization
C. Distillation
D. Cracking

Correct Answer: Option D

Explanation

Ethene is produced from cracking which involves breaking up large hydrocarbon molecules into smaller and more useful bits. This is achieved by using high temperatures and pressures without a catalyst.

C₁₅H₃₂ → 2C₂H₄ + C₃H₆ + C₈H₁₄
......Δheat...ethene........propene.........octane

43. Calculate the amount in moles of a gas which occupies 10.5 dm³ at 6 atm and 30^oC [R = 082 atm dm³ K^-1 mol^-1]

A. 2.536
B. 1.623
C. 4.736
D. 0.394

Correct Answer: Option A

Explanation

For an ideal gas PV = nRT

Amount in moles = n

Volume v = 10.5dm³

Pressure P = 6atm

Temperature T = 30°C + 273 = 303k

R, Gas constant = 0.082 atmdm³k^-1 mol

Recall from ideal gas equation

pv = nRT

n =

\frac{R V}{R T}

n =

\frac{6 \times 105}{0.082 \times 303}

n= 2.536mol

44. If 100cm³ of oxygen pass through a porous plug is 50 seconds, the time taken for the same volume of hydrogen to pass through the same porous plug is? [O = 16, H = 1]

A. 10.0s
B. 12.5s
C. 17.7s
D. 32.0s

Correct Answer: Option B

Explanation

Rate of diffusion

\frac{α_{1}}{D e n s i t y}

\frac{1}{\sqrt{M m}}

Rate =

\frac{1}{t i m e}

\frac{1}{t i m e}

\frac{α_{1}}{D e n s i t y}

\frac{1}{\sqrt{M o l a r m a s s}}

Time

α \sqrt{D e n s i t y}

\sqrt{M o l a r m a s s}

At constant volume of 100cm³

\frac{t_{o 2}}{\sqrt{M n_{o 2}}}

\frac{t_{n 2}}{\sqrt{1 \times 2}}

\frac{50}{4 \sqrt{2}}

\frac{t_{n 2}}{\sqrt{2}}

t_n2= 12.5s

45. Due to the high reactivity of sodium, it is usually stored under

A. water
B. mercury
C. paraffin
D. phenol

Correct Answer: Option C

Explanation

Na is kept under kerosene (paraffin) to avoid reactivity with air.
Paraffin is also known as Kerosene.
Na(sodium) is kept in kerosene to prevent it from coming in contact with oxygen and moisture. If this happens, it will react with the moisture present in air and form sodium hydroxide.

46. The furring of kettles is caused by the presence in water of

A. calcium hydrogentrioxocarbonate (IV)
B. calcium trioxocarbonate (IV)
C. calcium tetraoxosulphate (VI)
D. calcium hydroxide

Correct Answer: Option B

Explanation

Furring of kettles is caused by the temporary hardness in water. Temporary hardness in water is caused by calcium and magnesium trioxocarbonate (IV)
CaCO₃ causes the furring of kettles

47. Water for town supply is chlorinate to make it free from

A. bad colour
B. bacteria
C. temporary hardness
D. permanent hardness

Correct Answer: Option B

Explanation

Chlorine helps to keep water from germs and bacteria.

48.Ca(OH)_2(s) + 2NH₄Cl_(g) → CaCl_2(s) + 2H₂O_2(l) + X. In the reaction above X is

A. NO₂
B. NH₃
C. N₂O
D. NO₂

Correct Answer: Option B

Explanation

Balancing the chemical equation.

Ca[OH]₂ + 2NH₄Cl → CaCl₂ + 2H₂O + 2NH₃

X = NH₃

49. X_(g) + 3Y_(g) ---- 2z_(g) H = +ve. if the reaction above takes place at room temperature, the G will be

A. negative
B. zero
C. positive
D. indeterminate

Correct Answer: Option D

Explanation

\begin{array}{cc} Enthalpy Change [ΔH] & Entropy Change [ΔS] & Gibbs free Energy[ΔG] \\ Positive & Positive & depends on T, may be + or - \\ Negative & Positive & always negative \\ Negative & Negative & depends on T, may be + or - \\ Positive & Negative & always positive \end{array}

ΔG= ΔH − TΔS.
To determine whether ΔG will be positive or negative, the value of ΔH(change in enthalpy) and ΔS (change in entropy) must be given. Likewise the temperature.