NABTEB Chemistry Practical Questions and Answers for Students

Filed in Exam by The Elites on February 28, 2024

The NABTEB Chemistry Practical Questions play a vital role in the exam. This allows individuals to display their hands-on skills, experimental techniques, and understanding of chemical principles.

For those gearing up for the upcoming NABTEB GCE, valuable insights can be gained from the chemistry practical questions and answers that will be shared.

If you’re eager to learn more, keep reading!

NABTEB Chemistry Practical Questions

Question 1

(1ai) Explain how you calculated the molar mass of K₂CO₃. If the concentration of K₂CO₃ in gdm⁻³ is given as 13.8, why did you divide it by 18.8 in the calculation for the concentration in mole⁻³?

(ii) Write a balanced chemical equation for the reaction between HCl(aq) and K₂CO₃(aq).

(iii) What does CAVA/CBVB = nA/nB represent in the context of the reaction between HCl and K₂CO₃?

Why is it necessary to use the given formula for stoichiometry calculations in this context?

(iv) Explain the reasoning behind multiplying 0.100 (concentration of K₂CO₃) by 25 (volume of HCl) and then dividing by 0.250 (concentration of HCl) in the calculation for VA.

(b) Discuss the importance of rinsing the burette with distilled water before titration.

Question 2

(2ai) What’s the laboratory apparatus used for drying or preserving moisture-sensitive substances?

(ii) Explain the purpose of applying grease to the lid of the desiccator.

(iii) Differentiate between the use of silica gel, fused calcium chloride, and calcium oxide as drying agents.

(b) Differentiate between deliquescence and hygroscopy.

(i) P

(ii) Q

(iii) R

(iv) S

Question 3

(3a) Calculate the average volume of acid used (VE) and explain its importance in the titration results.

(bi) Calculate the concentration of substance E (CE) in mole dm⁻³. Show each step of your calculation.

(ii) In the expression CEVE/CFVF = Ne/Nf, explain the meanings of CE, VE, CF, and VF.

(iii) Calculate the concentration of substance F (CF) in moldm⁻³. Show each step of your calculation.

(iv) Determine the molar mass of Na₃CO₃, and explain how it is calculated.

Question 4

(4) Calculate;

Under Test

Why was 5cm³ of distilled water added in the first step?

Explain the inference drawn from the observation that Sample G dissolves partially in water, resulting in a colorless filtrate and a black residue.

Describe the gas liberated during the reaction with HNO₃(aq) and its effect on moist red litmus paper.

What is the significance of the residue dissolving completely on warming?

Under Observation

Explain the observation of a pale blue precipitate and its insolubility in excess NaOH(aq).

What inference can be made from the formation of a pale blue precipitate?

Under Inference

Describe the reaction that occurs when H₂SO₄(aq) is added to the residue.

What inference can be drawn from the fact that the residue is soluble in acid?

What inference can be made from the observation of a deep blue solution upon the addition of NH₃(aq) in drops and excess in the second portion?

Question 5

(5a) Write the balanced chemical equation for the reaction between lead(II) nitrate (Pb(NO₃)₂) and hydrogen sulfide gas (H₂S).

(bi) Define what is meant by a strong acid and provide an example.

(bii) Compare the ionization behavior of H₂SO₄ with CH₃COOH in terms of completeness.

CHEMISTRY Practical Answers

Solution for Question One

(1ai) 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

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

(iii) 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

(iv) 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

(1b) (i) It is rinsed with distilled water to make the burette clean and it is rinsed with acid in order not to decrease the concentration
(ii) This is to prevent air aspiration

Solution for Question Two

(2ai) Desiccator

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

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

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

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

Solution for Question Three

(3a) Volume of Pipette = 25.0cm

TABULATE
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

(b) Concentration of E (CE) in mole
= mole x 1000/volume
4.90g ==> 500cm³
xg ==> 1000cm³
x = 4.90x1000gdm³/500
=9.80gdm³
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
=106gmol³
Conc (gdm³) = 106gmol x 0.0980
=10.39gdm³

Solution for Question Four

(4) TABULATE
Test | Observation| Inference

UNDER TEST
– 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

UNDER OBSERVATION

– 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 dissolves 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

UNDER INFERENCE

– 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

Solution for Question Five

(6a) 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)

(bi) – H2SO4 ==> strong acid
– CH3COOH ==> weak acid

(ii) – H2SO4 is strong because it ionizes completely in water
– While CH3COOH is a weak acid because it ionizes partially in water

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

Understanding NABTEB Chemistry Practical Questions isn’t just about memorizing formulas; it’s about applying that knowledge in a lab setting.

Get comfortable with the question format, do relevant experiments, and sharpen your observation and analytical skills.

This way, you can confidently handle the practical exam.

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