Free NECO Chemistry Practical Expo 2020 - ZAZZAFA TV
(1)
Volume of pipette is 25.00cm³
Titrations |Rough| 1 | 2 | 3
Final Reading|23.00|21.50|21.40|21.50
Initial Reading|0.00|0.00|0.00|0.00
Vol. Of acid used|23.00|21.50|21.40|21.50
(1a)
Vol of acid used VA = 21.50+21.40+21.50/3
= 21.47cm³
(1bi)
Conc. Of A in g/dm³ = 3.65÷500/1000 = 7.3g/dm³
Conc. of A in mol/dm³ = gram conc/molar mass
= 7.3/[1+35.5] = 7.3/36.5
=0.2mol/dm³
(1bii)
Using CAVA/CBVB = nA/nB
0.2×21.47/CB×25.00 = 2/1
CB = 0.2×21.47/25.00×2
= 0.08588mol/dm³
(1biii)
Molar mass of B = gram conc/molarity = 10.60/0.08588
=123.43g/mol
(1biv)
X2CO3 = 123
2[X] + 12 + 3[16] = 123
2x + 60 = 123
2x = 123 - 60 = 63
X = 63/2 = 31.5
(1bv)
No of moles of X2CO3 = molarity × volume
= 0.08588 × 25
= 2.147moles
Mole ratio of X2CO3 to CO2 is 1:1
Therefore, No of moles of CO2 released = 2.147moles
Volume of CO2 released = 2.147 × 22.4
=48.1dm³
Volume of pipette is 25.00cm³
Titrations |Rough| 1 | 2 | 3
Final Reading|23.00|21.50|21.40|21.50
Initial Reading|0.00|0.00|0.00|0.00
Vol. Of acid used|23.00|21.50|21.40|21.50
(1a)
Vol of acid used VA = 21.50+21.40+21.50/3
= 21.47cm³
(1bi)
Conc. Of A in g/dm³ = 3.65÷500/1000 = 7.3g/dm³
Conc. of A in mol/dm³ = gram conc/molar mass
= 7.3/[1+35.5] = 7.3/36.5
=0.2mol/dm³
(1bii)
Using CAVA/CBVB = nA/nB
0.2×21.47/CB×25.00 = 2/1
CB = 0.2×21.47/25.00×2
= 0.08588mol/dm³
(1biii)
Molar mass of B = gram conc/molarity = 10.60/0.08588
=123.43g/mol
(1biv)
X2CO3 = 123
2[X] + 12 + 3[16] = 123
2x + 60 = 123
2x = 123 - 60 = 63
X = 63/2 = 31.5
(1bv)
No of moles of X2CO3 = molarity × volume
= 0.08588 × 25
= 2.147moles
Mole ratio of X2CO3 to CO2 is 1:1
Therefore, No of moles of CO2 released = 2.147moles
Volume of CO2 released = 2.147 × 22.4
=48.1dm³
(2ai)
Tabulate:
TEST: C + 5cm³ of distilled water and shake thoroughly. Divide the solutions into three portions.
OBSERVATION: A pale green solution results
INFERENCE: Salt is soluble
(2aii)
TEST: To the first portion add NaOH solution in drys
OBSERVATION: Dark green gelatinous was formed
INFERENCE: Fe²+ present
(2aiii)
TEST: then in excess
OBSERVATION: Precipitate is insoluble
INFERENCE: Fe²+ present
(2aiv)
TEST: To the second portion add K3Fe(CN)6 solution
OBSERVATION: A dark blue precipitate formed
INFERENCE: Fe²+ confirmed
(2bi)
TEST: To the third portion add AgNO³ solution
OBSERVATION: White Precipitate formed
INFERENCE: SO4²- , CL-, CO²- present
(2bii)
TEST: To the results obtained in add dilute HNO3 in drops , then excess
OBSERVATION: White Precipitate is insoluble
INFERENCE: CL- present
(2biii)
TEST: To the results obtained in add NH3 solution
OBSERVATION: White Precipitate dissolves
INFERENCE: CL- Confirmed
TEST: then in excess
OBSERVATION: Precipitate is insoluble
INFERENCE: Fe²+ present
(2aiv)
TEST: To the second portion add K3Fe(CN)6 solution
OBSERVATION: A dark blue precipitate formed
INFERENCE: Fe²+ confirmed
(2bi)
TEST: To the third portion add AgNO³ solution
OBSERVATION: White Precipitate formed
INFERENCE: SO4²- , CL-, CO²- present
(2bii)
TEST: To the results obtained in add dilute HNO3 in drops , then excess
OBSERVATION: White Precipitate is insoluble
INFERENCE: CL- present
(2biii)
TEST: To the results obtained in add NH3 solution
OBSERVATION: White Precipitate dissolves
INFERENCE: CL- Confirmed
Filtration
(3aii)
Diagram
(3bi)
I. Methyl Orange
II. Methyl orange
III. Phenolphthalein
(3bii)
A white precipitate soluble in excess dilute HCl is formed
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