JAMB Chemistry Past Questions 2012

10 questions from the JAMB 2012 Chemistry paper, with answers and explanations. Tap "Show answer" under each one when you are ready.

JAMB 2012Question 1 · Acids, Bases, and Salts

Which of the following will produce a solution with pH less than 7 at equivalent point?

A. $HN O_{3}$ +NaOH
B. $H_{2} S O_{4}$ + KOH
C.HCL +Mg (OH) 2
D. $HN O_{3}$ + KOH

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Answer: C. HCL +Mg (OH) 2

HCl (strong acid) reacting with $Mg (OH)_{2}$ (weak base) gives a salt that hydrolyses to produce a solution with pH less than 7 at the equivalence point.

JAMB 2012Question 2 · Atomic Structure and Chemical Bonding

A hydrogen atom which has lost an electron contains

A.one proton only
B.one neutron only
C.one proton and one neutron
D.one proton, one electron and one neutron

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Answer: A. one proton only

A hydrogen atom (protium) that loses its single electron is left with just its one proton (no neutron, no electron).

JAMB 2012Question 3 · Chemistry of Metals

A common characteristic between zinc and the other transition elements is the ability to

A.have variable oxidation states
B.from complex ions
C.act as a catalyst
D.from coloured ions

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Answer: B. from complex ions

Like other transition elements, zinc can form complex ions (e.g. with ammonia or hydroxide ligands).

JAMB 2012Question 4 · Chemistry of Non-metals and Gases

Which of the following gases has a characteristic pungent smell, turns red litmus paper blue and forms dense white fumes with hydrogen chloride gas?

A. $N_{2}$
B. $N_{2} O$
C. $C l_{2}$
D. $N H_{3}$

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Answer: D. $N H_{3}$

Ammonia gas has a pungent smell, turns red litmus blue (it is alkaline), and forms dense white fumes of ammonium chloride with HCl gas.

JAMB 2012Question 5 · Electrochemistry and Redox Reactions

What number of moles of

C u^{2 +}

will be deposited by 360 coulombs of electricity? [f =

96500 C mo l^{- 1}

]

A. $5.36 \times 1 0^{- 4}$ mole
B.1. $87 \times 1 0^{10}$ -3 mole
C.9. $35 \times 1 0^{- 4}$ mole
D.3. $73 \times 1 0^{- 3}$

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Answer: B. 1. $87 \times 1 0^{10}$ -3 mole

Moles of electrons = Q/F = $\frac{360}{96500}$ = $3.73 \times 1 0^{- 3}$ mol. Since $C u^{2 +}$ + 2e- → Cu, moles $C u^{2 +}$ deposited = $3.73 \times 1 0^{- 3}$ /2 = $1.87 \times 1 0^{- 3}$ mol.

JAMB 2012Question 6 · Environmental and Industrial Chemistry

A piece of filter paper moistened with lead (ll) ethanoate solution turns black when the paper is dropped into a gas likely to be

A.sulphur (VI) oxide
B.hydrogen chloride
C.sulphur (VI) oxide
D.hydrogen sulphide

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Answer: D. hydrogen sulphide

Hydrogen sulphide gas reacts with lead(II) ethanoate on filter paper to form black lead(II) sulphide, a classic test for $H_{2} S$ .

JAMB 2012Question 7 · Organic Chemistry

Carbohydrates are large molecules with the molecular formula C

_{x}

_{2}

_{y}

. In which of the following pairs is x not equal to y?

A.glucose and starch
B.maltose and starch
C.sucrose and fructose
D.galactose and starch

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Answer: B. maltose and starch

Maltose ( $C_{12} H_{22} O_{11}$ : x=12,y=11) and starch (( $C_{6} H_{10} O_{5}$ )n: x=6,y=5) both have x not equal to y in their formula, unlike glucose or fructose $(C_{6} H_{12} O_{6})$ where x=y=6.

JAMB 2012Question 8 · Organic Chemistry

An example of oxidation - reduction enzyme is

A.amylase
B.protease
C.lipase
D.dehydrogenase

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Answer: D. dehydrogenase

Dehydrogenase enzymes catalyze oxidation-reduction reactions by removing hydrogen from substrates.

JAMB 2012Question 9 · Physical Chemistry

CO (g)

+ H

_{2}

_{(g)}

→ CO

_{2 (g)}

+ H

_{2 (g)}

Calculate the standard heat change of the reaction above, if the standard enthalpies of formation of CO

_{2 (g)}

, H

_{2}

_{(g)}

and CO

_{(g)}

in kJmol

^{- 1}

are - 394, -242 and - 110 respectively.

A.+ $262 K J mo l^{- 1}$
B.- $262 K J mo l^{- 1}$
C.+ $42 K J mo l^{- 1}$
D.- $42 K J mo l^{- 1}$

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Answer: D. - $42 K J mo l^{- 1}$

ΔH = [Δ $Hf (C O_{2})^{+}$ Δ $Hf (H_{2})$ ] - [Δ $Hf (CO)^{+}$ Δ $Hf (H_{2} O)$ ] = (-394+0) - (-110-242) = -394-(-352) = $- 42 kJ mo l^{- 1}$ .

JAMB 2012Question 10 · Quantitative Chemistry

How many atoms are present in

6.0 g

of magnesium? [Mg = 24, N A =

6.02 \times 1 0^{23} mo l^{- 1}

]

A. $1.20 \times 1 0^{22}$
B. $2.41 \times 1 0^{22}$
C. $1.51 \times 1 0^{23}$
D. $3.02 \times 1 0^{23}$

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Answer: C. $1.51 \times 1 0^{23}$

Moles Mg = $\frac{6.0}{24}$ = $0.25 mol$ . Atoms = 0.25 x $6.02 \times 1 0^{23}$ = $1.505 \times 1 0^{23}$ , approx $1.51 \times 1 0^{23}$ .

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