2024 NECO GCE CHEMISTRY PRACTICAL ANSWER

(2ai)
OBSERVATION: Thre is water droplets around the test tube. It is colourless and odourless. It turns white anhydrous copper II tetraoxosulphate VI blue. It is neutral to litmus

(2biii)
INFERENCE: Reducing sugar ie glucose is confirmed

(3a)
-Filtration removes Pb(SO4)2
-Sublimation removes NH4Cl
-Filtration and evaporation is used to recover NaCl

(3bi)
V1 = 50.0cm³
C1 = 0.05moldm
C2 = 0.01
C1V1 = C2V2
Using dilution formula
0.05×50 = 0.01×V2
V2 = 0.05×50/0.01 = 250cm³

Hence volume of water that should be added
V = V2 – V1 = 250 – 50 = 200cm³

(3ci)
Shine a beam of light to each of solutions of glucose and starch. Starch will disperse the ray of light (Tyndall effect) while glucose will not.

(3cii)
Chemical Test: Use Benedict’s solution on starch; No positive result WHILE on glucose, It shows a positive test of brick red or orange colour.
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(1ai)
In a tabular form:

Titration | 1 | 2 | 3
Final burette reading |25.60|30.35|25.30
Initial burette reading|0.85 |5.65 |1.50
Vol. of acid used |24.75|24.70|23.80

(1aii)
Average volume of acid used
VA = (VA1 + VA2 )/2 cm³
VA = (24.75+24.70)/2
VA = (49.45)/2 cm³ = 24.73cm³

(1aiii)
Neutralization reaction

(1aiv)
Draw the diagram

(1av)
The end point will not be affected if 10cm³ of distilled water is added to the reaction mixture while the titration progresses because the specific amount of Na2CO3 in 25.0cm³ of the mixture remains the same.

(1bi)
To find the CA
Conc of Na2CO3 in moldm-³
CB = 0.10moldm-³

VB = 25.0cm³
Using CAVA/CBVB = na/nb
CA × 24.73/0.1×25 = 2/1
CA = 2×0.1×25/24.73
= 0.202moldm-³

(1bii)
Concentration of acid in gdm-³
= molarconc × molarmass
Where the molar mass of Hcl = 1 + 35.5
= 36.5gmol-¹

Therefore mass conc = 0.202moldm-³ × 36.5gmol = 7.37gdm-³

(1biii)
From the equation
2Hcl(aq)+Na2CO3(aq) –> 2NaCl(aq)+H2O(l)+CO2(g)

1 mole of base librates 2 moles of NaCl
No of moles reacted
n = CV/1000 –> 0.1×1000/1000 = 0.1mol Na2CO3

Hence
0.1mole of Na2CO3 = 2×0.1mole NaCl
= 0.2mol NaCl

Hence
Mass conc of NaCl = Amount × molar mass
= 0.2×(23 + 35.5)
= 0.2 × 58.5
= 11.7g/dm³

(1biv)
volume of CO2 liberated
1 mole of Na2CO3 = 1 mole of CO2
0.1 mole = 0.1 mole of CO2 = 22.4×0.1
= 2.24dm³ = 2240cm³

Posted by admin on 25th November 2019.

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