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Waec Chemistry Practical Answers 2023 for Thursday 18th May 2023

Waec Chemistry Practical Answers 2023 for 18th May 2023 : Access free live 2023 WAEC May/June Chemistry Practical Questions and Answers for school candidates, without any charges! Join the WAEC May/June Free Chemistry Practical Questions and Answers EXPO on the 18th of May, 2023. Join the WAEC May/June 2023 FREE Chemistry Practical ANSWER ROOM for school candidates now!




TEST; C + distilled water + shake

OBSERVATIONS; It dissolves to give a colourless or clear solution

INFERENCE; Soluble salt suspected


TEST; Solution C + litmus paper

OBSERVATIONS; It has no effect on both blue and red litmus paper

INFERENCE; Neutral solution


TEST; solution C + Fehling solution A and B + heat

OBSERVATIONS; It gives a black red colour

INFERENCE; Reading agent confirmed


TEST; D + heat

OBSERVATIONS; It turns yellow on heating and white on cooling

INFERENCE; ZnO is present


TEST; D + dilute Hcl + heat

OBSERVATIONS; It dissolves completely to give a clear or colourless solution

INFERENCE; Solution chlorides of Zn³+, Al³+ is present


TEST; Solution D + NaOH in drops and then in excess

OBSERVATIONS; It gives white gelatinous precipitate. The precipitate is soluble in excess.

INFERENCE; Al³+, Zn²+ present.

Al³+, Zn²+ present .


TEST; Solution D + NH3 in drops and then in excess

OBSERVATIONS; It gives white gelatinous precipitate. The precipitate is soluble in excess NH3

INFERENCE; Al³+, Zn²+ is present.

Zn²+ confirmed.

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Waec Chemistry Practical Answers 2023

Thursday, 18th May, 2023

  • Chemistry 3 (Practical) (Alternative A) – 09:30am – 11:30am (1st Set)
  • Chemistry 3 (Practical) (Alternative A) –12:00pm – 2:00pm (2nd Set)

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WAEC Chemistry Practical Specimen 2023

Welcome to our guide on the WAEC Chemistry Practical Specimen for the 2023/2024 exams. If you’re a secondary school student preparing for the Chemistry practical exam, you’re in the right place! In this article, we’ll provide you with easy-to-understand information and useful tips to help you do well in this important part of your Chemistry examination. Let’s get started and explore the specimen together.

  1. Great care should be taken to ensure that the information given in the items 2 and 3 below does not reach the candidate either directly or indirectly before the examination.
  2. In addition to the fittings and reagents normally found in a chemistry laboratory, the following apparatus and materials will be required by each candidate:
    • (a) One burette of 50 cm3 capacity;
    • (b) One pipette, either 20 cm3 or 25 cm3. (All candidates at one centre must use pipettes of the same volume. These should be clean and free from grease)
    • (c) The usual apparatus for titration;
    • (d) The usual apparatus and reagents for qualitative work including the following with all reagents appropriately labelled;
      • (i) Dilute sodium hydroxide solution;
      • (ii) Dilute hydrochloric acid;
      • (iii) Dilute trioxonitrate(V) acid;
      • (iv) Silver trioxonitrate(V) solution;
      • (v) Acidified potassium dichromate solution;
      • (vi) Aqueous ammonia;
      • (vii) Lime water;
      • (viii) Red and blue litmus paper
      • (ix) Dilute tetraoxosulphate (VI) acid;
      • (x) Fehlings solution A & B.
    • (e) Spatula;
    • (f) Filtration apparatus;
    • (g) One beaker
    • (h) One boiling tube;
    • (i) Four test tubes;
    • (j) Methyl orange as indicator;
    • (k) Glass rod;
    • (l) Wash bottle containing distilled/deionized water;
    • (m) Burning splint;
    • (n) Watch glass;
    • (o) Bunsen burner/source of heat;
    • (p) Droppers;
    • (q) Mathematical table/calculator.
  1. Each candidate should be supplied with the following, where ‘n’ is the candidiate’s serial number
    • (a) 150 cm3 of a solution of HCl in a corked flask or bottle labelled ‘An’. These should all be the same containing 8.5 cm3 of concentrated HCl per dm3 of solution.
    • (b) 150 cm3 of Na2CO3.10 H2O in a corked flask or bottle labelled ‘Bn’. These should all be the same containing 5.0 g of the hydrated salt per dm3 of solution.
    • (c) One spatulaful of glucose in a specimen bottle labelled ‘Cn’. This must be the same for all candidates.
    • (d) One spatulaful of zinc oxide powder in a specimen bottle labelled ‘Dn’. This must be same for all candidate.

Past Chemistry Practical Answers

Titration; | rough | 1 | 2 | 3 |

Final; | 20.50 | 14.00 |14.00 | 14.00 |

Initial; | 0.00 | 0.00 | 0.00 | 0.00 |

Volume of A used; | 20.50 | 14.00 |14.00 | 14.00 |

Volume of pipette used

Average volume of A used = 14.00+14.00+14.00/3
=14.00cm³ (Va)

2HNO³ + X²CO³ —> 2XNO³ + H²O + CO²

Conc – of B in molldm³ = ?
Ca= 0.15molldm³
Vb= 25.00cm³
Va= 14.00cm³
Cb= ??
Na= 2
Nb= 1

Cb= CaVaNb/VbNa
Cb= 0.15*14.00*1/25.00*2
Cb= 2.1/50
Cb= 0.042molldm³.

Molar mass of B =?
Conc(molldm³)= Conc(gldm³)/molar mass
0.042/1 * 4.20/m.m

Value of x in x²(i)³=?
(2x)+12 + (16*3)= 100

(i)I ensure that the readings are taken at the lower meniscus
(ii)I ensure that my saliva did not enter the pipette while taking the base



Under Observation;
An effervescence occurs in which a colorless and odourless gas which burns blue litmus red is evolved.

Under Inference;
CO² gas from CO3^²is suspected

Under Observation;
The gas turns lime water milky.

Under Inference;
CO² gas is confirmed.

Under Observation;
Y dissolves completely in water

Under Inference;
Soluble salt

Under Observation;
White gelatinous PPT is formed.

Under Inference;
Al³+, Zn²+, Pb²+, present

Under Observation;
The white gelatinous PPT dissolve in excess NaOH

Under Inference;
Al³+, Zn²+, Pb²+, present

Under Observation;
White gelatinous PPT formed.

Under Inference;
Al³+, Zn²+, Pb²+, present

Under Observation;
The white gelatinous PPT remain insoluble in excess NH³

Under Inference;
Al³+, Pb²+, present.

It increases the boiling point of water

It requires heat to occur.

This is because it is deliquescent.

I. Calcium oxide
II. Anhydrous copper(ii) tetraoxosulphate(vi)

Hydrogen Chloride

I. To separate a mixture of immiscible liquids
II. To collect gases.
Teacher’s Report filling
1: A=0.15, B= 4.20
2 (i=14.00, ii=14.00, iii=14.00)


For question 1

Use your school’s endpoint and use it wherever you see our endpoint (13.20cm³) …and calculate with it.

Use the answer you got in 1b(i) to solve in 1b(ii) i.e we got 0.0396 in 1bi after calculating with our average titre (endpoint) value 0.0396, as u can see, we then used the 0.0396 in 1bii which we got from our calculation in 1bi.

NB: If you use our endpoint and your school teacher submits a different one with a wide margin to what u used, you will either fail or lose some mark in this particular question. Your school teacher should have in one way given you the average titre value(endpoint) or you might have gotten it in the lab from your past experiment.

If your school wants to use mine, then the chemistry teacher must submit endpoint similar to ours without much difference…eg 13.10, 13.20, 13.30, etc

According to Lewis, an acid can be defined as a substance that can accept a pair of non-bonding electrons i.e an electron pair acceptor.


Salting out is a process through which soap is precipitated as solid from the suspension by adding common salt to the suspension. It is a purification process.

Reagent – Silver trioxonitrate(v)
Condition – The reaction takes place at high temperatures and pressure.

This is the proportion at which isotopes of an element are to each other in its composition.

(i) It has a stable configuration
(ii) This is because they have a covalent bond.

They can be differentiated using barium chloride. When barium chloride is added to concentrated H₂SO₄ a white precipitate is formed with concentrated HNO₃


(i) There are two electrolytes
(ii) Salt bridge is present.

(i) There is only one electrolyte
(ii) There is no salt bridge present.

The lower the ionisation energy, the higher the reactivity of metals. Since it decreases down the group, the reactivity increases down the group I.

This can be defined as a formula that shows the actual number of atoms in a compound.

(i) NH₃
(ii) This is because its interaction is perfectly inelastic.


Ionization energy can be defined as the energy required to remove a loosely bound electron from the outermost shell of a gaseous covalently bonded atom.

This is because B has more shells than Be which makes the valence electrons to be far from the nucleus making it require less energy due to lesser nuclear attraction.

I = 0.12A, t = 500seconds
m = 0.015g, F =96500C
M = 48.0, Charge = ?
m = MIt/CF
C = MIt/mF
C = (48×0.12×500)/(0.015×96500)
C = 2


(i) Electricity supply
(ii) Nearness to source of material

This can be defined as a property of metal which are weakly attracted to an applied magnetic field

(i) I → ₂₄Cr²⁺ → 2
₂₄Cr⁶⁺ → 0

(ii) ₂₄Cr²⁺

(iii) It has unpaired electrons.
This can be defined as the number of atoms or molecules in one mole of a substance which is equal to 6.02 ×10²³

Mass/m.m = no. of molecules/Avogadro’s number
(2.30/m.m) × (3.01×10²²/6.02×10²³)
m.m = (2.30×6.02×10²³)/3.01×10²²
m.m = 46g/mol

Since the molar mass = 46g/mol
The formula is NO₂
:. 14+(16×2)
14+32 = 46g/mol
:. NO₂

(i) There is no reaction but rather forms a layer underneath the water
(ii) It reacts violently with water forming white silicon dioxide and hydrogen chloride gas.

The reaction is different because CCl₄ is an organic substance while SiCl₄ is not.

Extraction of copper

(i) CuSO₄
(ii) CuCl₂

Cu²⁺ + 2e⁻ —> Cu

Using; m = ZIt
Z = m/It
Z = 3.2/(50x(3×60)+13)
Z = 3.2/(50×193)
Z = 3.2/9650
Z = 0.000332g/Asec


Oxygen can be prepared by heating KClO₃ in the presence of manganese(iv)oxide which acts as a catalyst to produce KCl and oxygen gas. The reaction takes place at a lower temperature and much faster rate.
KClO₃ (MnO₂)heat—–> KCl + O₂
(i) It melts to clear mobile liquid
(ii) A brownish gas is evolved.

(i) Carbon(ii)oxide
(ii) Lead(ii)oxide


(i) It leads to erosion
(ii) It causes pollution

This is because it has more surface area than it occupies leading to a faster rate of reaction.

Ammoniacal liquor

When water is added to white anhydrous CuSO₄ it turns blue.

(i) It removes the hardness by precipitating the trioxocarbonate(iv)

(ii) It removes the hardness by removing the calcium or magnesium ion and precipitating trioxocarbonate(iv).

Ca(HCO₃)₂ → CaCO₃ + H₂O + CO₂

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Waec Chemistry Practical Answers 2023 for 18th May 2023

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