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WAEC: OFFICIAL WASSCE 2021 TIMETABLE


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Free WAEC Chemistry Practical Answer – May/June 2017 Expo

Free Verified and Correct Waec 2017 Chemistry Alternative A Practical Expo Answer Solutions – 2017 May/June

NOTE THAT ^ MEANS Raise to
power.
Pls Draw Your Table As Usual And
Input The Following : –
Volume of pipette= 25 .00 cm ^ 3
indicator used- Methyl orange
colour change at end point- yellow to orange / purple
Note Use Your School End Point .

Tabulate

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Tabulate
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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 . 75 cm ^ 3 / 3
= 23 .75 cm ^ 3
1 bi)
CAVA / CcVc=2 / 1
Cc =CAVA / 2 VC
= 0 .100 * 23 .75 Moldm ^ – 3 / 2 * 25 .00
= 0 .0475 moldm ^ – 3
amount of A used = 0 .100 x VA / 1000 = 0 .100 * 23 .75 / 1000 =0 .00237
2 moles Of A = 1 mole of C
0 .002375mol of A = 0 .002375 mol/ 2
100 cm ^ 3 of C contain 0 .00237 * 100 mol/ 2 * 25 =0
1000 cm 3 ofCcontained 0 . 002375x 1000 mol
2 x25
= 0 .0475 mol
concentration of C in moldm – 3 = 0 .0475 moldm – 3
1 bii )
Molar mass of Bing mol – 1 :
Molar mass of Na2 CO3 . yH 2 O =mass concentration of Bingdm – 3
molar concentration of Binmoldm – 3
= 13 .6 gdm- 3
0 .0475 moldm – 3
= 286 gmol – 1
1 biii)
Molar mass of Na2 CO3 =[( 2 × 23 )+12 + (16 × 3 )]=106 gmol – 1
Mass of anhydrous Na 2 CO3 = 106 x 0 .0475 gdm – 3
= 5 .035 gdm – 3
Mass of water =13 .6 – 5 .035 gdm – 3
= 8 .565 gdm – 3
Mass of Na 2 CO3 =Molar mass of Na 2 CO3
Mass of water y × Molar mass of water
5 .035 =106
8 .565 18 y
y = 106 x 8 .565
5 .035 x 18
= 10
More Typing .. ..
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2 a )
Tabulate
Test
(i )Fn+ H2 O, 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
Cu 2 +present
Test
(iii )Filtrate + NH 3 (aq)in
drops , then in excess
Observation
A pale blue gelatinous
precipitate was
formed . The precipitate
dissolves or is soluble
in excess NH 3 (aq)to give
a deep blue solution
Inference
Cu 2 +confirmed
Test
(iv )Filtrate + dil .HNO3
+ AgNO3 (aq)
Observation
No visible reaction
White precipitate
formed
Inference
Cl – present
Test
+ NH 3 (aq)in excess
Observation
Precipitatedissolvedin
excessNH 3 (aq)
Inference
Cl – confirmed
Test
2 bi)Firstportionof
residue +NaOH(aq)in
drops , then in excess
Observation
White powdery
precipitate which is
insoluble in excess
NaOH(aq)
Inference
Ca 2 + present
Test
2 bii )Second portion of
residue +dil . HCl
Observation
Effervescence/ bubbles;
colourless , odourless
gas evolved. Gas turns
lime water milky and
turns damp blue litmus
paper red.
Inference
Gas is CO2
CO3 ^ 2 – or HCO 3
– present
3i) Lime juice is cacitic in nature and the color of methyl orange in active medium is Red
3ii) Iron(iii)chloride will be reduced to iron(ii) with the yellow deposit of Sulphur
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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