NABTEB Chemistry Practical Questions 2022 Latest Update : Current School News

NABTEB Chemistry Practical Questions 2022 Latest Update

Filed in Exam by on February 10, 2022



– NABTEB Chemistry Practical Questions –

NABTEB Chemistry Practical Questions will guide you before taking part in the NABTEB exams. Chemistry Practical NABTEB Expo Questions are out now on our website.

NABTEB Chemistry Practical Questions 2022 Latest Update


The National Business and Technical Examinations Board was founded in 1992 to domesticate craft level examinations.

Further, it was previously conducted by City & Guilds, Pittman’s, and the Royal Society of Arts throughout the United Kingdom in conformity with the principles of the National Policy on Education.

Its formation was the culmination of a 15-year evolutionary process in which FOUR Government Panels were established at various periods to assess the place and structure of public examinations in our educational system.


Develop as a globally recognised Assessment Body for Craftsmen and Technicians.


NABTEB is a vision-led, mission-driven public institution with the following vision: To be a Globally Recognised Assessment and Certification Body Preparing Candidates for the Workplace and Academic/Professional Excellence.

The management of the National Business and Technical Examinations Board (NABTEB) is yet to release the 2021 November/December NBC/NTC and ANBC/ANTC examinations results.

Candidates who intend to take part in the 2021 NABTEB GCE can learn how to check their results online.

NABTEB is one of the examination bodies set up by the Federal Government in 1992 to reduce the burden of conducting examinations, which involves a lot of technical and trade-related practicals.

The board conducts the National Technical Examination (NTC), National Business Certificate (NBC) and their respective advance level examinations (ANBC and ANTC).

READ ALSO!!!        

NABTEB GCE Guide Nov/Dec 2021

NABTEB GCE Registration Form      


NABTEB Chemistry Practical Questions and Answers 2022



Concentration of K₂CO₃ in gdm⁻³ =13.8/1 = 13.8gdm⁻³
Molar mas of K₂CO₃ = (39 x 2) + 12 (16×3) = 138gmol⁻¹
Therefore Concentration of K₂CO₃ in mole⁻³ = concentration of K₂CO₃ in gdm⁻³/molar mass of K₂CO₃
= 13. 8/18.8
= 0.100moldm⁻³

HCL(aq) + K₂CO₃(aq) –> KCL(aq) + H₂O(l) + CO₂(g)


Using CAVA/CBVB = nA/nB
CA = 0.250mold⁻³
CB = 0.100mold⁻³
VA = ?
VB = 25.0cm³
nA = 2 nB = 1
0.250 x VA/0.100 x 25 = 2/1
VA = 0.100 x 25 x 2/0.250 x 1
= 5/0.25
= 20.0cm³


1 mole of K₂CO₃ —> 1 mole of C0₂
Mole of K₂CO₃ = 0.100 x 25/1000 = 0.0025mold
Therefore, 0.0025 moles of K₂CO₃ —> 0.0025moles of C0₂
Volume of C0₂ at S.T.P = 0.0025 x 22.4dm³
= 0.056dm³
= 56cm³ s.t.p


(i) it is rinsed with distilled water to make the burette clean and it is rinsed with the acid in order not to decrease the concentration
(ii) This is to prevent air aspiration

The orange colour of K₂Cr₂ 0₄ is observed


-Graph of the height of the product against Volume of K₂Cr₂ 0₄

-Scale 1cm rep 1 unit on both axis

(I) volume of HCL = 8.0cm³; Amount = 0.500 x 8/1000 = 0.004 mole.

(II) volume of K₂Cr₂ 0₄ = 8.0cm³ ; Amount = 0.500 x 8/1000 = 0.004mole.
Hence, mole ratio of HCL to K₂Cr₂ 0₄ = 0.004/0.004 = 1/1 = 1:1

BaCL₂ + K₂Cr₂ 0₄ —> BaCr0₄(g) + 2KCL

(i) Add aqueous potassium iodide (KI) to the solution
(ii) A bright yellow precipitate indicates the presence of pb²



Grease should be applied to the lid of the desiccator to make it airtight

(i) Silica gel
(ii) Fuse calcium chloride
(iii) Calcium oxide

DELIQUESCENCE absorbs moisture from the atmosphere and then turns into solution while HYGROSCOPY absorbs moisture from the atmosphere but does not form a solution.

P –> Sulphur
Q –> Sulphur(iv) oxide
R –> Vanadium oxide
S –> Sulphur(vi) oxide


Volume of Pipette = 25.0cm

Burette reading |rough |1st |2nd|3rd|
Final burette reading (cm³) |24.90| 39.50|34.80|24.50
Initial Burette reading (cm³) |0.00|15.00|10.00|0.­00
Volume of E used (cm³) |24.90|24.50|24.80|2­4.50

Average volume of acid used (VE)
= (24.50+24.50)cm/2
= 24.50cm³
VE = 24.50cm³


Concentration of E (CE) in mole
= mole x 1000/volume
4.90g ==> 500cm³
xg ==> 1000cm³
x = 4.90x1000gdm³/500
Molar mass (H2S04) = 1×2 + 32 + 16×4
= 2+ 32 + 64
= 98gmol
Concentration (mole dm³) = 9.80gdm³/98
= 0.100moldm³

(ii) CEVE/CFVF = Ne/Nf
= 0.100×24.50/Cf x 25.0 =1/1
Cf = 0.100×24.50/25×1
Cf = 2.45/25 moldm³
= 0.0980moldm³

(iii) Concentration of F in g/dm
Conc (gdm³) = molar mass x conc
Molar mass (Na3CO3) = 23×2 + 12 + 16 x3
Conc (gdm³) = 106gmol x 0.0980

Test | Observation| Inference

– G + 5cm of distilled water

1st portion + HNO3(aq) + AgNO3(aq)

2nd portion + NaOH + heat

Residue + H2SO4(aq)

1st portion + NaOH(aq) in drops than excess

2nd portion + NH3(aq) in drops than excess


– Sample G dissolves partially in water colourless filtrate black residue – A white precipitate is formed

– A pungent irritating, choking gas is liberated which turns moist red litmus paper blue – Residue dissolve completely on warming

– A pale blue precipitate is formed, which is insoluble in excess

– We formed a pale blue precipitate which dissolved in excess to give a deep blue solution


– G is a mixture of a soluble and insoluble salt

– Cl present

– Gas is NH3 from NH4

– Residue is soluble in acid

– Cu2+ is present

– Cu2+ present

The effervescence of brownish gas turns a moist litmus paper red and also it gives a black residue of lead sulphide. Pb (NO3)(aq) + H2S(g) ==> Pbs(s)black + H2O(i) + NO(g)

– H2SO4 ==> strong acid
– CH3COOH ==> weak acid

– H2SO4 is strong because it ionises completely in water
– While CH3COOH is a weak acid because it ionises partially in water

Heat the mixture, ammonium chloride sublime and it then condenses to solid NHCL.

DISCLAIMER! These are not real NABTEB Chemistry Practical Questions, but likely repeated questions over the years to help candidates understand the nature of their examinations. Ensure to note every question provided on this page.

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