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Jamb Chemistry 2017

Home » Chemistry jamb past questions » Jamb Chemistry 2017
1
ANS 1
2
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3
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4
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5
ANS 5
1

The general formula of alkanones is 

A. RCHO

B. R2CO

C. RCOOH

D. RCOOR

ANS 1

Answer: B

 

Alkanones ( ketones) have the general formula R2CO. 

Where R can be an alky gp

2

The constituent common to duralumin and alnico is

A. Co

B. Mn

C. Al

D. Mg

ANS 2

Answer: C

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

3

The shape of the S-orbital is

A. elliptical

B. spiral

C. circular

D. spherical

ANS 3

Answer: D

 

4

Aluminium hydroxide is used in the dyeing industry as a

A. dye

B. dispersant

C. salt

D. mordant

ANS 4

Answer: D

 

5

The tincture of iodine means iodine dissolved in

A. ethanol

B. bromine chloride

C. chlorine water

D. water

ANS 5

Answer: A

6
ANS 6
7
ANS 7
8
ANS 8
9
ANS 9
10
ANS 10
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

ANS 6

Answer: A

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

ANS 7

Answer: D


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

CO2 + H2O → H2CO3

H2CO3 is an example of a weak acid

8

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

A. Isotopy

B. Resonance

C. Isomerism

D. Reforming

ANS 8

Answer: B

 

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 H4SO4

ANS 9

Answer: D

This is used in the preparation of ethyl ethanoate to remove the water present since it serves as a dehydrated agent.

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

ANS 10

Answer: D

Note water is importance but going by the options,  D is the best

11
ANS 11
12
ANS 12
13
ANS 13
14
ANS 14
15
ANS 15
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.

C.

D.

ANS 11

Answer: C

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

ANS 12

Answer: D

At constant pressure connotes Charle’s law

V1T1 = V2T2

2dm3/300k = 2 x 2dm3/T2

T2= 600k

13

The oxidation number of iodine in KIO3 is

A. +7

B. +3

C. +5

D. +6

ANS 13

Answer: C

14

An isomer of C5H12 is

A. 2–ethyl butane

B. butane

C. 2–methyl butane

D. 2–methyl propane

 

ANS 14

Answer: C

C5H12 has 5 carbons acid 12 hydrogens.

2–methyl butane have  carbon=5 and  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

ANS 15

Answer: A

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

16
ANS 16
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ANS 17
18
ANS 18
19
ANS 19
20
ANS 20
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 × 10-4

B. 6.3 × 10-11

C. 6.3 × 10-4

D. 1.6 × 10-11

ANS 16

Answer: B

PH = – Log[ H+ ]

PH + POH =14

POH = 14 – 3.8

POH = 10.2

POH = – Log[ OH– ]

10.2 = – Log[ OH– ]

10-10.2 = [ OH– ]

[ OH– ] = 6.3 x 10-11
 
17
17

Incomplete oxidation of ethanol yields

A. CH3COOH

B. CH3COCH3

C. CH3CH2OCH2CH3

D. CH2CHO

ANS 17

Answer: D

 
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 alkaline 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

ANS 18

Answer: D

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

ANS 19

Answern: D

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

i.e, [P + av2] [v – b] = RT is reduced to PV = nRT

Generally, a gas behaves more like an ideal gas at higher temperature and lower pressure

20

When ΔH is negative, a reaction is said to be

A. endothermic

B. exothermic

C. reversible

D. ionic

ANS 20

Answer: B

When ΔH is negative, heat is liberated to the surrounding and it connotes an exothermic reaction.
ΔH = −ve
21
ANS 21
22
ANS 22
23
ANS 23
24
ANS 24
25
ANS 25
21

A. 226

B. 220

C. 227

D. 222

ANS 21

Answer : D

42He = 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

ANS 22

Answer:  A

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

ANS 23

Answer: A

Recall that

m = MmIt / 96500n

Mm =108, t = 4830s

I = 10A, n = 1

m = 108 × 10 × (483096500) × 1

m = 54.0g

Mass deposited = Mm × It/96500n
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
ANS 24

Answer: A

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

ANS 25

Answer: D

Solution
 
m = Mm×Q /96500n

where Q = IT

M = Mm × Q /96500n

where m = mass

Mm = Molar mass

Q = Quantity of electricity

n = number of change= +1

M / Mm = mole = mass / Molarmass

M /Mm = Q / 96500n

= 965096500n × 1

= 110 = 0.1mol
26
ANS 26
27
ANS 27
28
ANS 28
29
ANS 29
30
ANS 30
26

The reaction of halogens in the presence of sunlight is an example of

A. oxidation reaction

B. addition reaction

C. hydrogenation reaction

D. substitution reaction

ANS 26

Answer: D

27

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

A. amylase

B. diastatse

C. invertase

D. zymase

ANS 27

Answer: B

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. CnH2n

B. CnH2n − 2

C. CnH2n + 1

D. CnH2n + 2

ANS 28

Answer:  C

29

CH2S(g) + O2(g) → 2Cn + 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

ANS 29

Answer: C

In the reactant;

Cu2S

2 Cu – 2(1) = 0

2 Cu = 2

Cu = 22

Cu = +1

In the product, Cu

Cu = O

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

30

 

In the diagram above. X is

A. SO3

B. SO2

C. S

D. H2S

ANS 30

Answer: B

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

31
ANS 31
32
ANS 32
33
ANS 33
34
ANS 34
35
ANS 35
31

 

The diagram above. Y is

A. fused CaO

B. H2O

C. NaOH

D. Concentrated H2SO4

 

ANS 31

Answer: D

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

ANS 32

Answer: D

 
M = Molar mass × Quantity of Electricity / 96500 × no of charge

= MmIT /96500n

Copper II Chloride = CuCl2

CuCl2 → Cu2+ + 2Cl2

Mass of compound deposited = 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 = 64×135096500×2

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

ANS 33

Answer : B

 

34
The reddish–brown rust on ion roofing sheets consists of
A. Fe3 + (H2O)6
B. FeO.H2O
C. Fe2O3.3H2O
D. Fe3O4.2H22O
ANS 34

Answer: C

 
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
ANS 35

Answerv: C

Graham’s Law of diffusion of gas. The rate of dimension of a gas inversely proportional to the square root of its molecular mass or its density, which is R ∝ 1 / √Mm or R ∝ 1 /√D Dx = 0.5gdm-3, Dy = 2gdm-3 R= K /√D R √D = k R1D1 = R1D2 RxDx = RyDy RxRy = DyDx = 20.5 = 2.0
36
ANS 36
37
ANS 37
38
ANS 38
39
ANS 39
40
ANS 40
36

The carbon atoms on ethane are

A. sp2 hybridized

B. sp3 hybridized

C. sp4 hybridized 

D. sp hybridized

ANS 36

Answer : B

Alkane’s family are sp3 hybridized. It is tetrahedral.

While: 

Alkenes are sp2 hybridization

Alkyne are sp 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
ANS 37

Answer : B

Solution

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
 
ANS 38

Answer: B

 
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

ANS 39

Answer: D

 

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
ANS 40

Answer: D

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

41
ANS 41
42
ANS 42
43
ANS 43
44
ANS 44
45
ANS 45
41
A particle that contains a 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
ANS 41

Answer: D

42

Ethene is prepared industrially by

A. Reforming

B. Polymerization

C. Distillation

D. Cracking

ANS 42

Answer: D

 
Ethene is produced from cracking which involves breaking up large hydrocarbon molecules into smaller useful ones. This is achieved by using high temperatures and pressures .

C15H32 → 2C2H4 + C3H6 + C8H14
……Δheat…ethene….
43
Calculate the amount in moles of a gas which occupies 10.5 dm3 at 6 atm and 30oC [R = 082 atm dm3 K-1 mol-1]
 
A. 2.536
B. 1.623
C. 4.736
D. 0.394
ANS 43

Correct Option: A

Solution
 
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 = RV /RT

n = 6×105 /0.082×303

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
ANS 44
Answer: B
Rate of diffusion α1/Density or 1/ √Mm Rate = 1 /time 1 /time  α1 /Density or 1 / √Molarmass Time α √Density or √Molarmass At constant volume of 100cm3 to2 /√Mno2 = tn2 / √×2 50/ 4√2 = tn /√2 tn2= 12.5s
45
Due to the high reactivity of sodium, it is usually stored under
A. water
B. mercury
C. paraffin
D. phenol
 
ANS 45

Answer: C

Na is kept under kerosene (paraffin) to avoid reactivity with air.
.

46
ANS 46
47
ANS 47
48
ANS 48
49
ANS 49
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

ANS 46

Answer: B

Furring of kettles is caused by the temporary hardness in water as a result of
CaCO3 
47
Water for town supply is chlorinate to make it free from
A. bad colour
B. bacteria
C. temporary hardness
D. permanent hardness
 
ANS 47

Answer: B

 

48
Ca(OH)2(s) + 2NH4Cl(g) → CaCl2(s) + 2H2O2(l) + X. In the reaction above X is
 
A. NO2
B. NH3
C. N2O
D. NO2
ANS 48

Answer: B

 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
 
ANS 49

Answer: D


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.
 
 
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