Waec May/June 2022 Chemistry Practical Answers    Waec May/June 2022 Chemistry Practical Answers
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""CHEMISTRY PRACTICAL EXAM""
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*TAKE NOTE*

For question 1

Use your school's endpoint and use it wherever you see 24.10cm³....and calculator with it

Use the answer in 1b(i) to solve in 1b(Ii)

(1a)
Volume of pipette used, VA = 25.0cm³

[TABULATE]

Burette Reading | Rough titre | 1st titre | 2nd titre |

Final Burette reading (cm³) | 25.60 | 32.80 | 26.40

Initial Burette reading (cm³) | 1.20 | 8.70 | 2.30

Volume of Na₂S₂O₃ used | 24.40 | 24.10 | 24.10

Average burette reading = (24.10 + 24.10)/2
Average volume of Na₂S₂O₃ used = 24.10cm³
VB = 24.10cm³

(1bi)
Given: CB = 0.1mol/dm³
VB = 24.10cm³
VA = 25.0cm³
CA =?
From equation nA/nB = 1/2
.: Using (CABA)/CBVB = (nA)/nB

(CA x 25)/(0.1x24.10) = 1/2
CA = (0.1x24.10)/(25x2)
CA = 0.0482mol/dm³
.: Concentration of iodine in A = 0.0482mol/dm³

(1bii)
Mass in gramms of iodine in 1dm³ of A = Molarity × Molar mass of iodine
= 0.0482×127(2)
= 12.2428g/dm³
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(2a)
[TABULATE]

=TEST=
C + Distilled water

=OBSERVATION=
It dissolves completely to give a light green solution.

=INFERENCE=
Soluble salt.

(2ai)
=TEST=
Solution C + NaOH in drops and in excess + Heat gently

=OBSERVATION=
A dirty green precipitate is formed which remains insoluble in excess.
Effervescence occurs in which a colourless gas with a pungent smell which turns red litmus blue is given off.

=INFERENCE=
Fe²⁺ is present.
NH₃ gas form.
NH₄⁺ is present.

(2aii)
=TEST=
Solution + BaCl₂ + dilute HCL in excess

=OBSERVATION=
A white precipitate is formed.
The white precipitate remains insoluble and gives a white dense forms

=INFERENCE=
SO₄²⁻, CO₃²⁻, SO₃²⁻, is present.
SO₄²⁻ confirmed.

(2b)
Cations → Fe²⁺ and NH₄⁺
Anions → SO₄²⁻
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(3a)
Carbon (iv) oxide turns like water milky white while sulphur (iv) oxide does not

(3b)
(I) H₂: Downward displacement of air
(II) NH₃: Downward displacement of air
(III) HCL: Upward displacement of air

(3bii)
I - It is less dense than air
II - It is less dense than air
III - It is denser than air

(3c)
(i) Distillation
(ii) Filtration followed by evaporation to dryness

(3d)
This is because KCl react with NaHCO₃ to form two salts
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NUMBER 1  COMPLETE NUMBER 2 NUMBER 3   =======================
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