Waec 2018 Chemistry Practical Answers

Chemistry 3 (Practical) (Alternative A) 09.30am – 11.30am (1st Set)
Chemistry 3 (Practical) (Alternative A) 12.00pm – 2.00pm (2nd Set)

NOTE
Tita means θ

^ means Raise to power

Pie means π

/ (means) division or divide

2 whole no 3/4 means 2¾
* means
multiplication (×)

Sqr root means √

Proportional means






1)
Tabulate
Burette reading|Final burette reading
(cm^3)|Initial burette reading (cm^3)|
Volume of acid used(cm^3)|
Rough- 24.10,0.00,24.10
First- 23.80, 0.00, 23.80
Second- 23.75, 0.00, 23.70
Third- 23.75, 0.00, 23.75
Average volume of A used = 23.80 + 23.70 + 23.75cm^3/3
=23.75cm^3

1bi)
CAVA/CcVc=2/1
Cc=CAVA/2VC
=0.100*23.75Moldm^-3/2*25.00
=0.0475moldm^-3
amount of A used = 0.100x VA/1000=0.100*23.75/1000 =0.00237
2moles Of A = 1mole of C
0.002375mol of A = 0.002375mol/2
100cm^3 of C contain 0.00237*100mol/2*25 =0
1000cm3ofCcontained0.002375x1000mol
2×25
=0.0475mol
concentration of C in moldm-3 =0.0475moldm-3

1bii)
Molar mass of Bing mol-1
Molar mass of Na2CO3.yH2O=mass concentration of Bingdm-3
molar concentration of Binmoldm-3
=13.6gdm-3
0.0475moldm-3
=286gmol-1

1biii)
Molar mass of Na2CO3 =[(2×23)+12+(16×3)]=106gmol-1
Mass of anhydrous Na2CO3=106×0.0475gdm-3
=5.035gdm-3
Mass of water=13.6-5.035gdm-3
=8.565gdm-3
Mass of Na2CO3 =Molar mass of Na2CO3
Mass of water y×Molar mass of water
5.035=106
8.565 18y
y =106×8.565
5.035×18
=10


(2)Tabulate
Test
(a)(i)Fn+H2O,then

filter

Observation
White residue and blue
filtrate was observed

Inference
Fn is a mixture of

soluble and insoluble

salts

Test
(ii)Filtrate+NaOH(aq)in

drops,then in excess
Observation
A blue gelatinous
precipitate which is
insoluble in excess
NaOH(aq)was formed

Inference
Cu2+present

Test
(iii)Filtrate+NH3(aq)in

drops,then in excess

Observation
A pale blue gelatinous
precipitate was
formed. The precipitate

dissolves or is soluble

in excess NH3(aq)to give
a deep blue solution

Inference
Cu2+confirmed

Test
(iv)Filtrate+dil.HNO3

+AgNO3(aq)

Observation
No visible reaction
White precipitate

formed

Inference
Cl-present

Test
+NH3(aq)in excess

Observation
Precipitatedissolvedin

excessNH3(aq)

Inference
Cl-confirmed

3i)A white precipitate of barium sulphate will be formed by the instant reaction between barium chloride and sulphuric acid.

BaCl2 + H2SO4 = BaSO4 + 2HCl



3b)Iron sulfide reacts with hydrochloric acid, releasing the malodorous (rotten egg smell) and very toxic gas,hydrogen sulphide. FeS + 2 HCl → FeCl2+ H2S



1)
Tabulate
Burette reading|Final burette reading
(cm^3)|Initial burette reading (cm^3)|
Volume of acid used(cm^3)|
Rough- 24.10,0.00,24.10
First- 23.80, 0.00, 23.80
Second- 23.75, 0.00, 23.70
Third- 23.75, 0.00, 23.75
Average volume of A used = 23.80 + 23.70 + 23.75cm^3/3
=23.75cm^3

2a)
-Observation-
part of the C dissolves,colourless filtrate salt

-Inference-
C is a mixture of soluble salt

2bi)
-observation-
white precipitate formed

-Inference-
Cl is present

2bii)
-Observation-
precipitate dissolves

-Inference-
Cl confirmed

2biii)
-Observation-
colourless gas evolved with characteristic choking smell, turns litimus paper blue

-Inference-
Gas is NH3

2c)
-Observation-
Residue dissolves liberating a colourless gas which turns lime water milky

-Inference-
gas is CO2
CO^2-3 present

3a)A white precipitate of barium sulphate will be formed by the instant reaction between barium chloride and sulphuric acid.

BaCl2 + H2SO4 = BaSO4 + 2HCl



3b)Iron sulfide reacts with hydrochloric acid, releasing the malodorous (rotten egg smell) and very toxic gas,hydrogen sulphide. FeS + 2 HCl → FeCl2+ H2S

3c)When solid iron filings are added to dilute aqueous Hydrochloric acid, Iron (II)chloride or ferrous chloride is formed , with the liberation of Hydrogen gas.

Fe (s) + 2HCl (aq) ……..> FeCl2(aq) + H2 (g)