NABTEB GCE

2021 NABTEB GCE Chemistry Practical Answers

Posted by Hedowho on 17 November, 2021

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NABTEB GCE CHEMISTRY PRACTICAL SOLUTIONS (O’LEVEL)

(1)
TABULATE
Final reading: 24.50| 48.80 | 25.60
Initial burette readings: 0.00| 24.40 | 1.00
Volume of acid used: 24.50| 24.40 | 24.60

(1ai)
Volume of acid used = Final burette readings – Initial burette readings

1st titration = 24.50cm³ – 0.00cm³ = 24.50cm³

2nd titration = 48.80cm³ – 24.40cm³ = 24.40cm³

3rd titration = 24.50cm³ – 1.00cm³ = 24.60cm³

(1aii)
Average titre value = 1st + 2nd + 3rd/3
24.50 + 24.40 + 24.60/3
73.50/3 = 24.50cm³

(1bi)
MA = 0.08mol/dm³, VA = 24.5cm³, MB =? , VB = 25.0cm³, NA= 1, nB= 2
MAVA/MBVB = nA/nB
0.08×24.50/25MB = 1/2
25MB = 3.92
MB = 3.92/25 = 0.157
MB = 0.157mol/dm³

(1bii)
Mass concentration = Molar concentration × Molar mass
Molar concentration of acid = 0.08mol/dm³, Molar mass = 2×1+3+32+16×4 = 98g/mol
Mass concentration = 0.08 × 98 = 7.84g/dm³

(1biii)
Molar mass = Mass concentration/Molar concentration
Molar mass = 0.08/0.157 = 56g/mol

(1biv)
YOH = 56
Y + 16 + 1 = 56
Y + 17 = 56
Y = 56 – 17 = 39g

(1bv)
Methyl orange

(1c)
(i) Air bubbles in the burette
(ii) No indicator is use
(iii) Funnels should be removed before titration starts to avoid droplets
(iv) incorrect of reading and scale
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(2a)
=TEST=
(i) C + NaOH (aq) in drops then in excess
(ii) C + NH3(aq) in drops then in excess
(iii) C + BaCl(aq) then followed by dil.HCl

=OBSERVATION=
(i) Blue gelatineous precipitate
(ii) Precipitate dissolves in excess white precipitate
(iii) Precipitate Insoluble

=INFERENCE=
(i) Cu²+ present
(ii) Cu²+ present
(iii) SO4²+ present

(2b)
=TEST=
(i) D + H2O
(ii) Solution of D + NaOH2(aq) in drops then in excess
(iii) Solution of D + NH3(aq) in drops then in excess
(iv) Solution of D + FeSO4(aq) then followed by H2SO4(aq)

=OBSERVATION=
(i) D dissolves in H2O
(ii) White gelatineous precipitate dissolves in excess
(iii) White gelatineous precipitate dissolves in excess
(iv) A brown ring is observed in the rest tube

=INFERENCE=
(i) D is a soluble salt
(ii) Zn²+ present
(iii) Zn²+ present
(iv) NO3+ present
NO3- Confirmed

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(3ai)
P – sulphur
Q – Sulphur (iv) oxide or SO2
S – Sulphur trioxide or SO3
T – Barium tetraoxosulphate (iv) or BaSO4

(3aii)
R could be vanadium (v) oxide or platinum
2SO2(g) + O2(g) ⇌2SO3(g)

(3aiii)
No, this is because hydrochloric acid cannot dissolve tetraoxosulphate (iv) acids from its salt

(3bi)
A black coloration is observed due to the formation of lead (ii) sulphide
Pb (CH3COO)2 + H2S ⇌ Pbs + 2CH3COOH

(3bii)
A yellow deposit of sulphur is observed due to a redox reaction
2H2S + SO2 ——-> 3S + 2H2O

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