There are practical questions and answers available for chemistry that will be helpful for individuals preparing for the next NABTEB GCE.
Continue reading if you're ready to find out more!
Practical Chemistry Questions and Answers for the NABTEB
Question 1
(1ai) Explain how you calculated the molar mass of K?CO?. If the concentration of K?CO? 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 K?CO?(aq).
(iii) What does CAVA/CBVB = nA/nB represent in the context of the reaction between HCl and K?CO??
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 K?CO?) 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.
(c) Assign the correct chemical name to each of the following:
(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 Na?CO?, 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 HNO?(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 H?SO?(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 NH?(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(NO?)?) and hydrogen sulfide gas (H?S).
(bi) Define what is meant by a strong acid and provide an example.
(bii) Compare the ionization behavior of H?SO? with CH?COOH in terms of completeness.
(c) Describe the process of sublimation and condensation that occurs when heating ammonium chloride (NH?Cl).
CHEMISTRY Practical Answers
Solution for Question One
(1ai) Concentration of K?CO? in gdm?³ =13.8/1 = 13.8gdm?³
Molar mas of K?CO? = (39 x 2) + 12 (16×3) = 138gmol?¹
Therefore Concentration of K?CO? in mole?³ = concentration of K?CO? in gdm?³/molar mass of K?CO?
= 13. 8/18.8
= 0.100moldm?³
(ii) HCL(aq) + K?CO?(aq) –> KCL(aq) + H?O(l) + CO?(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 K?CO? —> 1 mole of C0?
Mole of K?CO? = 0.100 x 25/1000 = 0.0025mold
Therefore, 0.0025 moles of K?CO? —> 0.0025moles of C0?
Volume of C0? 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.
It takes more than simply memorization of formulas to understand NABTEB Chemistry Practical Questions; it takes applying that information in a lab environment.
Develop your analytical and observational skills, become familiar with the style of the questions, and conduct pertinent experiments.
In this manner, you can approach the practical exam with assurance.
Related
- Internal Auditor, Tomato Jos Farming and Processing Ltd.
- Design Engineer at Toptech Engineering Limited
- Marketing Management Trainee at Citygate Global Investment Ltd
- Business Development Analyst at Visa Incorporated
- Demand Manager at PZ Cussons Nigeria Plc
Have a question? Ask now!
Follow us on Instagram!
