2017 WAEC Chemistry Practical Answers

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WAEC 2017 Chemistry Practical Answers

Tabulate
2ai)
Test:F+H2O,thenfilter
Observation:Partially soluble salt
Inference:F is a mixture of soluble and insoluble salts
2aii)
Test:Filtrate+NaOH(aq)in drops,then in excess
Observation:A blue gelatinous precipitate which is insoluble in drops, in excess NaOH(aq)was formed
Inference:Cu^2+present
2aiii)
Test: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
2aiv)
Test:Filtrate+dil.HNO3+AgNO3(aq)
Observation:No visible reaction White precipitate formed
Inference:Cl^-present
2av)
Test:+NH3(aq)in excess
Observation:Precipitatedissolved in excessNH3(aq)
Inference:Cl^- confirmed

3ai)
The colour of methyl orange will change from orange to yellow
3aii)
Iron(III)chloride will be reduced from brown iron to green iron
Fe^3+(aq)->Fe^2+(aq)

NO 3
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NOTE THAT ^ MEANS Raise to power
Pls Draw Your Table As Usual And
Input The Following:-
Volume of pipette=25.00cm^3
indicator used- Methyl orange
colour change at end point-yellow to orange/purple
Note Use Your School End Point.

TABULATE

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
1bii)
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 =0.0475mol/dm^3
1biii)
Molar mass of Bing mol-1:
Molar mass of Na2CO3.yH2O=mass concentration of B in g/dm^3
molar concentration of B in mol/dm3=13.6gdm-3
0.0475mol/dm3=286gmol^-1
1biv)
The hydrated salt is NA2CO3.XHO=286
2(23)+12+3(16)+X(2+16)=286
46+60+18X=286
18X+106=286
18X=286-106
X=180/18
X=10
The number of water molecules in the hydrated salt is 10

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