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Waec 2024 Chemistry Practical Questions And Answers
By on May 16th, 2024. Waec
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(2)


(2a)
When all of C is added to a test tube with about 10cm³ of distilled water and shaken, the mixture is filtered. The residue and the filtrate are obtained.

Observations:
The residue: This will contain the insoluble salt(s) present in C.

The filtrate: This will contain the soluble salt(s) present in C.

(2bi)
Heating about 2cm³ of the filtrate in a boiling tube strongly.

Observations:
If there is effervescence (bubbling), it indicates the presence of a carbonate salt in the filtrate.

The gas evolved is likely carbon dioxide (CO₂).

(2bii)
Adding dilute HCl to half of the residue in a test tube.

Observations:
If there is effervescence (bubbling), it indicates the presence of a carbonate salt in the residue.
The gas evolved is likely carbon dioxide (CO₂).

(2biii)
Adding aqueous NaOH in drops, then in excess, to about 2cm³

Observations:
If a white precipitate forms upon adding NaOH drops, and it dissolves upon excess addition of NaOH, it indicates the presence of a metal hydroxide in the solution.

(2biv)
Adding aqueous ammonia in drops, then in excess, to another 2cm³ of the clear solution from (ii).

Observations:
If a colored precipitate forms upon adding ammonia drops, and it dissolves upon excess ammonia addition, it indicates the presence of a metal hydroxide in the solution.

(1a)
TABULATE PLEASE

Burette | Rough| 1st | 2nd

Final reading(cm³) | 19.00 | 18.10| 18.10|

Initial reading(cm³) | 0.00| 0.00| 0.00

Volume of A used(cm³) | 19.00| 18.10| 18.10|

Average titre = 18.10+18.10/2
= 18.10cm³

(1bi)
Conc. of B in mol/dm³ = ??
Ca = 0.020mol/dm³, Va = 18.10cm³
Na= 1, Nb = 5, Vb= 25.00cm³ , Cb= ??

CaVa/CbVb= Na/Nb

Cb = CaVaNa/VbNb

Cb= (0.020*18.10*5)/(25.0*1)
Cb= 0.0724mol/dm³

(1bii)
Con. of B in g/dm³ = ??
Con.(g/dm³) = Mass*1000/volume

= 3.8*1000/250
= 15.2g/dm³

(1biii)
FeSO₄<—> Fe²⁺ + SO₄²⁻
1 mol of FeSO₄ 1mol Fe²⁺
0.0724mol/dm³ xmol/dm Fe²⁺

x= 1*0.0724 = 0.0724mol/dm³

Mole= Con.*Vol/1000

Mole = 0.0724*250/1000
Mole= 0.0181mol

Mass= mole * m.m
= 0.0181*56
= 1.0136g

(1biv)
Percentage= 1.0136*100/15.2
= 6.67%

 

(3a)
(i)ammonia (NH₃)
(ii)carbon dioxide (CO₂).

(3b)
(i)A graduated pipette
(ii) A burette can be used.

(3c)
A fume cupboard in a laboratory is used to provide ventilation that limits exposure to toxic fumes, vapors, or dust.

(3d)
(i)safety goggles.
(ii)lab coats.
(iii)gloves.

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