NECO 2024 CHEMISTRY(PRACTICAL) ANSWERS – EXAMKING.NET

Final readings in cm³;
|22.00|20.10 |20.20|20.30|

Initial reading in cm³;
|0.00|0.00|0.00 |0.00|

Volume of A (acid) used cm³;
|22.00|20.10|20.20| 20.30|

The average titre
= 20.10+20.20+20.30/3
=60.60/3
= 20.20cm³

;Va 20.20cm³

(1bi)
Concentration of A in mol/dm³, Ca = ?

Concentration of A (g/dm³) = 4.40g/dm³.

.m.m of A, H²SO⁴ =(1*2)+(32*1)+(16*4)
=2+32+64
=98g/mol

Using concentration (mol/dm³) = Concentration (g/dm³)/m.m

Concentration (mol/dm³)
=4.40/98
=0.045mol/dm³

;: Concentration of A(mol/dm³) = 0.045mol/dm³

(1bii)
Concentration of B(mol/dm³) =???
Ca=0.045, Va=20.20, Vb=25.00, na=1, nb=1, cb=?

Using; CaVa/CbVb = na/nb
Cb=CaVanb/Vbna =0.045*20.20*1/25.00*1
Cb=0.036mol/dm³

(1biii)
Conc. of B(g/dm³) that reacted =??
molar mass of B, Na₂CO₃ = (23*2)+(12*1)+(16*3) = 106g/mol

Conc. of B(mol/dm³) = 0.036mol/dm3
Conc. (mol/dm³) = Conc. (g/dm³)/molar mass

(0.036/1) = Conc. (g/dm³)/106

Conc. (g/dm³) = 0.036*106 = 3.82g/dm³
Conc. of B(g/dm³) = 3.82/dm³

(1biv)
mass of the impurity in 1.0dm³ of B=??
Conc. of impurity (g/dm³) = 7.50 – 3.82 = 3.68g/dm³

:. Mass of impurity= Conc. (g/dm³) x vol. (dm³)
= 3.68×1 = 3.68g

(1bv)
Vol. of CO₂ produced in 1.0dm³ of B = ??

Mole of B= Conc. x Vol. (dm³)
Mole of B = 0.036*1
Mole of B =0.036mol

By proportion,
1mol of Na₂.10₃ = 1mol of CO₂

0.036mol of Na₂.10₃ = xmol of CO₂

mole of CO₂ produced = 1*0.036 = 0.036mol

:. Volume = mole * 22.4 = 0.81dm³
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(2)
[FILL THE TABLE WITH THE INFORMATION BELOW]

(2ai)
-TEST: C+5 cm³ of distilled water and shake thoroughly. Divide the solution into two portions
-OBSERVATION: It dissolves completely to give a colourless solution
-INFERENCE: C is a soluble salt

(2aii)
-TEST: To the first portion, add NaOH solution in drops.
-OBSERVATION: A white gelatinous precipitate is formed
-INFERENCE: Al³⁺ and Zn²⁺ present

(2aiii)
-TEST: Then add NaOH in excess
-OBSERVATION: The precipitate dissolve to give a colourless solution
-INFERENCE: Zn²⁺ present

(2aiv)
-TEST: To the second portion, add NH₃ solution in drops
-OBSERVATION: A white precipitate is formed
-INFERENCE: Al³⁺ and Zn²⁺ present

(2av)
-TEST: then in excess
-OBSERVATION: The white precipitate dissolves in excess NH₃
-INFERENCE: Zn²⁺ is confirmed

(2bi)
-TEST: 2cm³ of solution D + soda lime, heat, test with litmus paper.
-OBSERVATION: Ammonia gas (NH₃) is evolved, turning red litmus paper blue.
-INFERENCE: Presence of NH₄⁺ ions.

(2bii)
-TEST: Solution D + NaOH solution + CuSO₄ solution
-OBSERVATION: A blue precipitate formed
-INFERENCE: Presence of peptide bonds (proteins or peptides).
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(3ai)
I- to avoid the drop of the acid into the burette which will alter the reading
II- to avoid change in concentration and contamination
III- to avoid bubbles in the pipette

(3aii)
This can be done through crystallization. The solution is dissolved, heated to a high temperature and then allowed to cool. During cooling,the sodium chloride crystallized out.

(3bi)
(i)evaporating dish
(ii)Bunsen burner