w Tuesday, 11th April, 2017
Chemistry 3 (Practical) (Alternative A) 09.30am - 11.30am (1st Set)
Chemistry 3 (Practical) (Alternative A) 12.00pm - 2.00pm (2nd Set)
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SORRY FOR THE DELAY WE RECEIVED THE QUESTION PAPER A LITTLE BIT LATE
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 ∝


(1a)
PLZ TABULATE
Burette reading|Final burette reading
(cm³)|Initial burette reading (cm³)|
Volume of acid used(cm³)|
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 + 23.75cm³/3
=23.75cm³
(1bii)
CAVA/CcVc=2/1
Cc=CAVA/2VC
=0.100x23.75Moldm-³/2 x 25.00
=0.0475moldm-³
amount of A used = 0.100x
VA/1000=0.100 x 23.75/1000 =0.00237
2moles Of A = 1mole of C
0.002375mol of A = 0.002375mol/2
100cm³ of C contain
0.00237 x 100mol/2 x 25 =0
1000cm3ofCcontained0.002375x1000mol
2x25=0.0475mol
concentration of C in moldm-3
=0.0475mol/dm³
(1biii)
Molar mass of Bing mol-1
Molar mass of Na2CO3.yH2O=mass
concentration of B in g/dm³
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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(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)
The colour of methyl orange will
change from orange to yellow
(3ii)
Iron(III)chloride will be reduced from
brown iron to green iron
Fe^3+(aq)->Fe^2+(aq)
(3iii)
The color of KM2O4 is decolorized
because SO2(g) acts as a reducing
agent.
(3iv)
Addition of ethanoic acid to KCO3
results to the liberation of a colorless
and odorless gas CO2 which turns lime
water milky
3c. Sulfur dioxide gas (SO2) is bubbled
through a solution of acidified
potassium permanganate (KMnO4).
An oxidation-reduction reaction
occurs.
Potassium permanganate has a purple
colour. Sulphur dioxide is a reducing
agent. Potassium permanganate is an
oxidizing agent. When sulphur
dioxide reacts with potassium
permanganate the solution
decolourizes. Color changes from
Purple to colorless
3iv)Addition of ethanoic acid to k2lo3
results to the liberation of a colourless
or odourless gas co2 Which turns lime
water milky

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