WAEC 2017/2018 Chemistry Practical (Alternative A) Answers – May/June Expo

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

[9:03AM, 4/11/2017] Lijoka CTE: 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=106x0.0475gdm-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 =106x8.565
5.035x18
=10
[9:37AM, 4/11/2017] Lijoka CTE: (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