Year : 
1993
Title : 
Chemistry
Exam : 
JAMB Exam

Paper 1 | Objectives

11 - 20 of 49 Questions

# Question Ans
11.

The air around a petroleum refinery is most likely to contain?

A. CO2, SO3 and N2O

B. CO2, CO and N2O

C. SO2, CO and NO2

D. PH3 H2O and CO2

Detailed Solution

Topic is Impurities of Crude-oil
12.

Water can be identified by the use of?

A. anhydrous copper (II) tetraoxosulphate (VI)

B. anhydrous sodium trioxocarbonate (IV)

C. potassium heptaoxochromate (VII)

D. copper (II) trioxocaronate (IV)

Detailed Solution

Anhydrous CuSO4 will turn blue with water
13.

The phenomenon whereby sodium trioxocarbonate (IV) decahydrate loses some of its water of crystallization on exposure to the atmosphere is known as?

A. deliquescene

B. hygroscopy

C. effervescence

D. efflorescence

Detailed Solution

Topic is Salts With Water of Crystallization.
14.

A student prepares 0.5 M solution each of hydrochloric and ethanoic acids and then measured pH. The result would show that the?

A. pH values are equal

B. HCI solution has a higher pH

C. sum of the values is 14

D. ethanoic acid solution has a higher pH

D
15.

NH3 + H3O+
→ NH4+ + H2O. It may be deduced from the reaction above that?

A. a redox reaction has occurred

B. H3O+ acts as an oxidizing agent

C. H3O+ acts as an acid

D. water acts as ab acid

C
16.

4.0 g of sodium hydroxide in 250 cm3 of solution contains?

A. 0.40 moles per dm3

B. 0.10 moles per dm3

C. 0.04 moles per dm3

D. 0.02 moles per dm3

Detailed Solution

Unit of concentration = (moles)/(dm3)
If 250 cm3 solution NaOH contains 4.0 g NaOH
i.e -1cm3 solution NaOH (4.0)/(250) g
∴ 1000 cm3 (1 dm3) will contain (40 x 1000) / (250 x 1) = 16 g/dm3
16 g/dm3 = 16g/dm3 + molar mass NaOH
because no of moles = (mass of substance given) / (molar mass)
∴ (16 g) / (40 g) dm3 = (16 g)/(dm3 x (1 mole)/ (40g) = 0.4 (moles)/(dm3)
17.

During the electrolysis of a salt of a metal M, a current of 0.5 A flows from for 32 minutes 10 seconds and deposits 0.325 g of M, What is the charge of the metal ion?

(M = 65, If = 96,500 C per mole of electron)

A. 1

B. 2

C. 3

D. 4

Detailed Solution

Ampere = (Coulomb) x (Sec) x 1930 secs = 965 coulombs (unit of current)
But quantity of electricity = current x time = (Coulomb)/(Sec) x time
If 0.5 A for 32 mins 10 secs; Quantity of electricity = (0.5 coulomb)/(Sec x 1930) secs
= 965 coulombs
Since 965 coulombs give 0.325 g M
i.e1 coulombs = (0.325)/(965) g
∴96500 coulombs = (0.30)/(965) x 96500 =32.5 g
∴(65 g)/(32.5) = 2

inference: it takes 2 Faraday of electricity to deposit 1 mole of M. Therefore, the change on M ion = 2.
18.

Which of the reactions occurs at the anode during the electrolysis of a very dilute aqueous solution of sodium chloride?

A. OH- - e → OH

B. CIsup>- - e → CI

C. OHsup>- + CIsup>-HCI + O2

D. Na+ - esup>- Hg (Na)/(Hg) amalgam

A
19.

The oxidation states of chlorine in HOCL, HCLO3 and HCLO4 are respectively?

A. -1, +5, and +7

B. -1, -5 and 7

C. +1, +3 and +4

D. +1, +5 and +7

D
20.

A reaction takes places spontaneously if?

A. ∆G = O

B. ∆S < O and ∆H > O

C. ∆H < T ∆S

D. ∆G > O

C
11.

The air around a petroleum refinery is most likely to contain?

A. CO2, SO3 and N2O

B. CO2, CO and N2O

C. SO2, CO and NO2

D. PH3 H2O and CO2

Detailed Solution

Topic is Impurities of Crude-oil
12.

Water can be identified by the use of?

A. anhydrous copper (II) tetraoxosulphate (VI)

B. anhydrous sodium trioxocarbonate (IV)

C. potassium heptaoxochromate (VII)

D. copper (II) trioxocaronate (IV)

Detailed Solution

Anhydrous CuSO4 will turn blue with water
13.

The phenomenon whereby sodium trioxocarbonate (IV) decahydrate loses some of its water of crystallization on exposure to the atmosphere is known as?

A. deliquescene

B. hygroscopy

C. effervescence

D. efflorescence

Detailed Solution

Topic is Salts With Water of Crystallization.
14.

A student prepares 0.5 M solution each of hydrochloric and ethanoic acids and then measured pH. The result would show that the?

A. pH values are equal

B. HCI solution has a higher pH

C. sum of the values is 14

D. ethanoic acid solution has a higher pH

D
15.

NH3 + H3O+
→ NH4+ + H2O. It may be deduced from the reaction above that?

A. a redox reaction has occurred

B. H3O+ acts as an oxidizing agent

C. H3O+ acts as an acid

D. water acts as ab acid

C
16.

4.0 g of sodium hydroxide in 250 cm3 of solution contains?

A. 0.40 moles per dm3

B. 0.10 moles per dm3

C. 0.04 moles per dm3

D. 0.02 moles per dm3

Detailed Solution

Unit of concentration = (moles)/(dm3)
If 250 cm3 solution NaOH contains 4.0 g NaOH
i.e -1cm3 solution NaOH (4.0)/(250) g
∴ 1000 cm3 (1 dm3) will contain (40 x 1000) / (250 x 1) = 16 g/dm3
16 g/dm3 = 16g/dm3 + molar mass NaOH
because no of moles = (mass of substance given) / (molar mass)
∴ (16 g) / (40 g) dm3 = (16 g)/(dm3 x (1 mole)/ (40g) = 0.4 (moles)/(dm3)
17.

During the electrolysis of a salt of a metal M, a current of 0.5 A flows from for 32 minutes 10 seconds and deposits 0.325 g of M, What is the charge of the metal ion?

(M = 65, If = 96,500 C per mole of electron)

A. 1

B. 2

C. 3

D. 4

Detailed Solution

Ampere = (Coulomb) x (Sec) x 1930 secs = 965 coulombs (unit of current)
But quantity of electricity = current x time = (Coulomb)/(Sec) x time
If 0.5 A for 32 mins 10 secs; Quantity of electricity = (0.5 coulomb)/(Sec x 1930) secs
= 965 coulombs
Since 965 coulombs give 0.325 g M
i.e1 coulombs = (0.325)/(965) g
∴96500 coulombs = (0.30)/(965) x 96500 =32.5 g
∴(65 g)/(32.5) = 2

inference: it takes 2 Faraday of electricity to deposit 1 mole of M. Therefore, the change on M ion = 2.
18.

Which of the reactions occurs at the anode during the electrolysis of a very dilute aqueous solution of sodium chloride?

A. OH- - e → OH

B. CIsup>- - e → CI

C. OHsup>- + CIsup>-HCI + O2

D. Na+ - esup>- Hg (Na)/(Hg) amalgam

A
19.

The oxidation states of chlorine in HOCL, HCLO3 and HCLO4 are respectively?

A. -1, +5, and +7

B. -1, -5 and 7

C. +1, +3 and +4

D. +1, +5 and +7

D
20.

A reaction takes places spontaneously if?

A. ∆G = O

B. ∆S < O and ∆H > O

C. ∆H < T ∆S

D. ∆G > O

C