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Post-UTME Chemistry 2005 Objective Past Questions

All 27 questions from the Post-UTME Screening (Post-UTME) Chemistry 2005 Objective paper, with the correct answer and a full explanation for each. Free, no signup needed.

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Chemistry 2005 Objective — Question 1

During the electrolysis of copper (II) sulphate solution, an ammeter showed a steady current reading of 1.0A for 1/2 hr. 6.6x10^-4 kg of copper were liberated. Calculate the % error in ammeter reading

  • A. 12%
  • B. 11%Correct
  • C. 1.2%
  • D. 1.1%

Explanation

Theoretical mass (using Faraday's laws) is about 5.97x10^-4 kg; comparing to the actual 6.6x10^-4 kg gives an error of approximately 11%.

Chemistry 2005 Objective — Question 2

With the Daniel cell the depolarizer is

  • A. Tetra-oxo-sulphate(IV)acid
  • B. Ammonium chloride solution
  • C. Copper sulphate solutionCorrect
  • D. Manganese (IV) Oxide

Explanation

In a Daniell cell, copper(II) sulphate solution acts as the depolarizer at the copper electrode.

Chemistry 2005 Objective — Question 3

Which of the following is a physical change?

  • A. The bubbling of chlorine into water
  • B. The passing of steam over heated iron
  • C. The burning of substances in air
  • D. The dissolution of sodium chloride in waterCorrect

Explanation

Dissolving sodium chloride in water is a physical (reversible) change, unlike the chemical reactions described in the other options.

Chemistry 2005 Objective — Question 4

A mixture of oil and water can be easily separated by

  • A. paper chromatography
  • B. evaporation to dryness
  • C. using a separating funnelCorrect
  • D. fractional distillation

Explanation

Immiscible liquids like oil and water are easily separated using a separating funnel.

Chemistry 2005 Objective — Question 6

To separate a mixture of benzene, ethanol and water use

  • A. filtration
  • B. decantation
  • C. simple distillation
  • D. fractional distillationCorrect

Explanation

These three liquids have different boiling points, so fractional distillation is used to separate them.

Chemistry 2005 Objective — Question 7

Hydrogen forms two oxides. In the first oxide, 2g of hydrogen combine with 16g of oxygen. In the second oxide, 2g of hydrogen combine with 32g of oxygen. What law is illustrated?

  • A. The law of constant proportion
  • B. The law of conservation of matter
  • C. The law of multiple proportionCorrect
  • D. The law of reciprocal proportion

Explanation

Fixed masses of one element (hydrogen) combining with different masses of another (oxygen) in a simple whole-number ratio (16:32=1:2) illustrates the law of multiple proportions.

Chemistry 2005 Objective — Question 8

Lead forms two chlorides. The ratio of chloride atoms in the two chlorides is 1:2. If the mass of chlorine in the second chloride is 42.0g, find the mass of chlorine in the first compound

  • A. 71g
  • B. 72g
  • C. 21gCorrect
  • D. 73g

Explanation

Since the ratio is 1:2, the first chloride contains half the chlorine mass of the second: 42/2 = 21g.

Chemistry 2005 Objective — Question 9

According to the balanced equation below, which gas is in excess and by how much? H2(g) + F2(g) -> 2HF(g); 50dm^3, 60dm^3

  • A. F is in excess by 6dm^3
  • B. F is in excess by 10dm^3Correct
  • C. H is in excess by 10dm^3
  • D. HF is in excess by 100dm^3

Explanation

The reaction is 1:1; with 50dm^3 H2 and 60dm^3 F2, F2 is in excess by 60-50=10dm^3.

Chemistry 2005 Objective — Question 10

Which is heavier: 0.2mol SO2 or 0.2mol CO2? [S=32.0, O=16, C=12]

  • A. SO2 is heavier by 4gCorrect
  • B. CO2 is heavier by 44g
  • C. SO2 is heavier by 0.4g
  • D. CO2 is heavier by 40g

Explanation

Mass SO2 = 0.2 x 64 = 12.8g. Mass CO2 = 0.2 x 44 = 8.8g. Difference = 4g, with SO2 heavier.

Chemistry 2005 Objective — Question 11

A gas jar of air was placed on top of a gas jar of nitrogen IV oxide. After sometime, a uniform colour was seen throughout both gas jars. What process has occurred?

  • A. evaporation
  • B. oxidation
  • C. decomposition
  • D. diffusionCorrect

Explanation

The gradual mixing/spreading of gas molecules to achieve a uniform colour/concentration is diffusion.

Chemistry 2005 Objective — Question 12

At 17 degrees C, a sample of hydrogen occupies 125cm^3. What will the volume be at 100 degrees C. If the pressure remains constant?

  • A. 161cm^3Correct
  • B. 1.61cm^3
  • C. 131cm^3
  • D. 16.1cm^3

Explanation

V2 = 125 x (373/290) = 160.8cm^3, approximately 161cm^3.

Chemistry 2005 Objective — Question 13

Which of the following atoms A,B,C and D would readily form an ion with a charge of 2? Mass No.: (a)12 (b)16 (c)24 (d)31; Atomic No.: (a)6 (b)8 (c)12 (d)15

  • A. 12, 6
  • B. 16, 8
  • C. 24, 12Correct
  • D. 31, 15

Explanation

An atomic number of 12 (magnesium, a group 2 metal) readily loses 2 electrons to form a 2+ ion.

Chemistry 2005 Objective — Question 14

The ion X contains 23 particles in the nucleus and 10 electrons outside the nucleus. What does the nucleus of the ion X contain?

  • A. Protons 10, Neutrons 13
  • B. Protons 11, Neutrons 12Correct
  • C. Protons 12, Neutrons 11
  • D. Protons 13, Neutrons 10

Explanation

A sodium (Na) atom, mass number 23, has 11 protons and 12 neutrons; losing one electron to form Na+ gives 10 electrons, matching the description.

Chemistry 2005 Objective — Question 15

Which of the following are properties of carbon in the form of diamond?

  • A. Conductivity: nil, Melting point: -101 degrees C
  • B. Conductivity: nil, Melting point: 119 degrees C
  • C. Conductivity: nil, Melting point: 3500 degrees CCorrect
  • D. Conductivity: good, Melting point: 2600 degrees C

Explanation

Diamond does not conduct electricity (nil conductivity) and has an extremely high melting point, around 3500 degrees C.

Chemistry 2005 Objective — Question 16

What particles are present in solid bromide and molten lead II bromide?

  • A. Solid PbBr2: ions, Molten: ionsCorrect
  • B. Solid: molecules, Molten: molecules
  • C. Solid: ions, Molten: atoms
  • D. Solid: atoms, Molten: molecules

Explanation

PbBr2 is an ionic compound; it consists of ions in both the solid lattice (immobile) and the molten state (mobile, hence conducting).

Chemistry 2005 Objective — Question 17

The non-luminous Bunsen flame has the following characteristics except

  • A. the flame is hotter
  • B. the flame is cleaner
  • C. the flame enters the tube to consume incoming fuelCorrect
  • D. only three/ones can be seen in the flame

Explanation

The flame does not enter the burner tube to consume incoming fuel; this describes flashback, an abnormal/undesirable condition, not a normal characteristic of the non-luminous flame.

Chemistry 2005 Objective — Question 18

Water of crystallization is responsible for

  • A. shapes of crystalsCorrect
  • B. melting salts
  • C. purity of crystals
  • D. boiling point of salts

Explanation

Water of crystallization, incorporated in a fixed ratio within the crystal lattice, is responsible for the characteristic shape of hydrated crystals.

Chemistry 2005 Objective — Question 19

The substance which loses its water of crystallization on exposure to air is

  • A. deliquescent
  • B. efflorescentCorrect
  • C. hygroscopic
  • D. crystalline

Explanation

An efflorescent substance loses its water of crystallization when exposed to air.

Chemistry 2005 Objective — Question 20

15g of water dissolves 3.5g of sodium chloride at 25 degrees C. What is the solubility of the sodium chloride at that temperature? [Na=23, Cl=35.5]

  • A. 0.4mol/dm^3
  • B. 4.0mol/dm^3Correct
  • C. 40mol/dm^3
  • D. 400mol/dm^3

Explanation

Solubility = 3.5/15 x 100 = 23.3g per 100g water. In 0.1dm^3 (approx 100g water), moles = 23.3/58.5 = 0.399mol, giving a concentration of about 4.0mol/dm^3.

Chemistry 2005 Objective — Question 21

To determine the solubility of a salt in the laboratory, you need a/an

  • A. saturated solution of the saltCorrect
  • B. unsaturated solution of the salt
  • C. supersaturated solution of the salt
  • D. homogenous solution of the salt

Explanation

Solubility determination requires a saturated solution (containing the maximum dissolved solute) at a given temperature.

Chemistry 2005 Objective — Question 22

The major water pollutant is

  • A. spillage of oil
  • B. pesticide
  • C. detergent
  • D. sewageCorrect

Explanation

Sewage (from the sheer volume discharged) is generally regarded as the major water pollutant overall.

Chemistry 2005 Objective — Question 23

Which of the following acids is monobasic?

  • A. H2CO3
  • B. H2SO3
  • C. CH3COOHCorrect
  • D. H2SO4

Explanation

Ethanoic (acetic) acid, CH3COOH, has only one replaceable/ionizable hydrogen, making it monobasic.

Chemistry 2005 Objective — Question 24

Which is the function of silica SiO2 in the equation shown? CaO + SiO2 -> CaSiO3

  • A. a basic oxide
  • B. a reducing agent
  • C. an acidic oxideCorrect
  • D. an oxidizing agent

Explanation

Silica acts as an acidic oxide, reacting with the basic oxide CaO.

Chemistry 2005 Objective — Question 25

Which of the following substances is not a polymer

  • A. Nylon
  • B. Polyethane
  • C. PropaneCorrect
  • D. Starch

Explanation

Propane is a simple small hydrocarbon molecule, not a polymer, unlike nylon, polyethene, and starch.

Chemistry 2005 Objective — Question 26

The oxidizing agent in the reaction H2S(g) + Br2(aq) -> 2HBr(aq) + S(s) is

  • A. H2S
  • B. Br2Correct
  • C. HBr
  • D. S

Explanation

Bromine (Br2) is reduced to Br- (in HBr), meaning it is the oxidizing agent.

Chemistry 2005 Objective — Question 27

The preferential discharge of sodium during the electrolysis of brine using mercury as cathode is due to one of the following

  • A. High concentration of Na
  • B. Na is lower than H in the electrochemical series
  • C. Discharge of sodium as the amalgam requires less energyCorrect
  • D. Concentration of H+ is higher

Explanation

Mercury forms a sodium amalgam, which significantly lowers the energy required to discharge sodium at the cathode compared to hydrogen.

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