NABTEB GCE CHEMISTRY PRACTICAL

(1ai)
Concentration of K₂CO₃ in gdm⁻³ =13.8/1 = 13.8gdm⁻³
Molar mas of K₂CO₃ = (39 x 2) + 12 (16×3) = 138gmol⁻¹
Therefore Concentration of K₂CO₃ in mold⁻³ = concentration of K₂CO₃ in gdm⁻³/molar mass of K₂CO₃
= 13. 8/18.8
= 0.100moldm⁻³

(1aii)
HCL(aq) + K₂CO₃(aq) –> KCL(aq) + H₂O(l) + CO₂(g)

(1aiii)
Using CAVA/CBVB = nA/nB
CA = 0.250mold⁻³
CB = 0.100mold⁻³
VA = ?
VB = 25.0cm³
nA = 2 nB = 1
0.250 x VA/0.100 x 25 = 2/1
VA = 0.100 x 25 x 2/0.250 x 1
= 5/0.25
= 20.0cm³

(1aiv)
1 mole of K₂CO₃ —> 1 mole of C0₂
Mole of K₂CO₃ = 0.100 x 25/1000 = 0.0025mold
Therefore 0.0025 moles of K₂CO₃ —> 0.0025moles of C0₂
Volume of C0₂ at S.T.P = 0.0025 x 22.4dm³
= 0.056dm³
= 56cm³ s.t.p

(1b)
(i) it is rinsed with distilled water to make the burrette clean and it is rinsed with the acid in order not to decrease the concentration
(ii) This is to prevent air aspiration

Examination

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(2ai)
An orange colour of K₂Cr₂ 0₄ is observed

(2aii)
PLOT THE GRAPH [https://i.imgur.com/JVKGcvM.jpg]

-Graph of height of the product against Volume of K₂Cr₂ 0₄

-Scale 1cm rep 1 unit on both axis

(2aiii)
(I) volume of HCL = 8.0cm³; Amount = 0.500 x 8/1000 = 0.004 mole

(II) volume of K₂Cr₂ 0₄ = 8.0cm³ ; Amount = 0.500 x 8/1000 = 0.004mole
Hence, mole ratio of HCL to K₂Cr₂ 0₄ = 0.004/0.004 = 1/1 = 1:1

(2aiv)
BaCL₂ + K₂Cr₂ 0₄ —> BaCr0₄(g) + 2KCL

(2bi)
(i) Add aqueous potassium iodine (KI) to the solution
(ii) A bright yellow precipitate indicates he presence of pb²

(2bii)
DRAW THE DIAGRAM [https://i.imgur.com/kABz3tN.jpg]

===================================

(3ai)
Dessicator

(3aii)
Grease should be applied on the lid of the desiccator to make it air tight

(3aiii)
(i) Silica gel
(ii) Fuse calcium chloride
(iii) Calcium oxide

(3b)
DELIQUESCENCE absorb moisture from the atmosphere and then turns into solution while HYGROSCOPY absorb moisture from the atmosphere but does not form solution

(3c)
P –> Sulphur
Q –> Sulphur(iv) oxide
R –> Vanadium oxide
S –> Sulphur(vi) oxide

Completed!!!!!!!!!

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