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Author Topic: Waec practical chemistry 2021/2022  (Read 438 times)

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Offline NORTOLA

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Waec practical chemistry 2021/2022
« on: September 14, 2021, 10:21:51 am »
•••••••••••••••••••••••••••••••••••••••
CHEMISTRY PRACTICAL ANSWERS
(1a)
Volume of burrete = 50.00cm³
Volume of pipette = 25.00cm³
Indicator = Methyl orange

TABULATE

Final burette readings (cm³); 37.50 | 22.50 | 37.50 | 44.50

Initial burette readings(cm³); 0.00 | 0.00 | 0.00 | 0.00

Volume f Acid used (cm³); 37.50 | 22.50 | 22.50 | 22.50

Volume of A used = 1st + 2nd + 3rd/3
= 22.50+22.50+22.50/3
Average volume of A used = 22.50cm³

(1bi)
Concentration of A ( CA )?
2.03g in 500cm³ of solution

Volume = 500cm³/1000 = 0.500dm³
In g/dm³ = 2.03/0.5 = 4.06g/dm³

To find concentration in mol/dm³
Conc. of A in mol/dm³ = Conc. of A in g/dm³/Molar mass

Conc. of A in mol/dm³ = 4.06/36.5
= 0.1112mol/dm³
Molar mass of A (HCL) = 1+35.5 = 36.5g/mol

(1bii)
Number of mole of the Acid in average titre
Na = CaVa
= 0.1112 × 22.50
= 2.52moles

(1biii)
Nb= ? Na= 3, Ca= 0.1112mol/dm³, Cb = 0.12mol/dm³ , Va = 22.50cm³, Vb = 50.00cm³

CaVa/CbVb= nA/nB

0.1112×22.50/0.12×50 = 3/nB

nB × 2.502 = 18
nB = 18/2.502 =
:. nB = 7.194moles
= 7moles(approx.)

(1biv)
Mole ratio of Acid to base
nA:nB = 7:3

Offline NORTOLA

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Re: Waec practical chemistry 2021/2022
« Reply #1 on: September 14, 2021, 10:47:06 am »
••••••••••••••••••••••••••••••••••••••••
CHEMISTRY PRACTICAL ANSWERS
(1a)
Volume of burrete = 50.00cm³
Volume of pipette = 25.00cm³
Indicator = Methyl orange

TABULATE

Final burette readings (cm³); 37.50 | 22.50 | 37.50 | 44.50

Initial burette readings(cm³); 0.00 | 0.00 | 0.00 | 0.00

Volume f Acid used (cm³); 37.50 | 22.50 | 22.50 | 22.50

Volume of A used = 1st + 2nd + 3rd/3
= 22.50+22.50+22.50/3
Average volume of A used = 22.50cm³

(1bi)
Concentration of A ( CA )?
2.03g in 500cm³ of solution

Volume = 500cm³/1000 = 0.500dm³
In g/dm³ = 2.03/0.5 = 4.06g/dm³

To find concentration in mol/dm³
Conc. of A in mol/dm³ = Conc. of A in g/dm³/Molar mass

Conc. of A in mol/dm³ = 4.06/36.5
= 0.1112mol/dm³
Molar mass of A (HCL) = 1+35.5 = 36.5g/mol

(1bii)
Number of mole of the Acid in average titre
Na = CaVa
= 0.1112 × 22.50
= 2.52moles

(1biii)
Nb= ? Na= 3, Ca= 0.1112mol/dm³, Cb = 0.12mol/dm³ , Va = 22.50cm³, Vb = 50.00cm³

CaVa/CbVb= nA/nB

0.1112×22.50/0.12×50 = 3/nB

nB × 2.502 = 18
nB = 18/2.502 =
:. nB = 7.194moles
= 7moles(approx.)

(1biv)
Mole ratio of Acid to base
nA:nB = 7:3
••••••••••••••••••••••••••••••••••••••••
••••••••••••••••••••••••••••••••••••••••
(2a)
Test; C + 10cm³ of distilled water + filter

Observation; C dissolves partially to give a light green solution. Blue filtrate and green residue

Inference; C is a mixture of soluble and insoluble salt

(2bi)
Test; 2cm³ of filtrate + NaOH in drops in excess

Observation; Blue precipitate remains

Inference; CU²+ is present

(2bii)
Test; 2bi + warm

Observation; a colourless gas with a chocking smell which turns moist red litmus blue and forms dense white fumes with hydrogen chloride gas

Inference; NH³ from NH⁴+

(2biii)
Test; 2cm³ of filtrate + BaCl² + excess HCl

Observation; white precipitate, precipitate Remains

Inference; SO²- CO3²- or SO4²- is present, SO4²- confirmed

(2ci)
Test; Residue + HNOg

Observations; effervescence of a colourless and odourless gas which turns like water milky and turns moist blue to litmus red

Inference; CO2 from CO3²-

(2cii)
Test; 2ci + NH3 in drops in excess

Observations; blue precipitate. Precipitate dissolved to give a deep blue solution

Inference; CU2+ present
CU2+ confirmed
••••••••••••••••••••••••••••••••••••••••
••••••••••••••••••••••••••••••••••••••••
(3ai)
The value will increase

(3aii)
The occur as a result of the decrease in the concentration of base due to the added volume of water

(3bi)
There will be no visible reaction because copper is less than Zinc in the electrochemical series

(3bii)
It absorb water and become sticky because it is hygroscopic

(3biii)
The solution turns pink

(3c)

When NaOH is added to the solution of zn³+, a white precipitate is formed which later dissolve in excess NaOH due to the formation of zinc hydroxide

COMPLETED...

 

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