Waec 2017 Chemistry 3 (Practical Alternative B) Questions & Answers

Expo 4 Comments 86

CHEMISTRY PRAT SOLVED BY BOMB
AS AT 12:am last night!!
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
================================
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
1bi)
CAVA /CcVc = 2 /1
Cc= CAVA /2 VC
=0 . 100* 23 . 75 Moldm ^ – 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
1000cm 3 ofCcontained 0 . 002375x
1000mol
2x 25
=0 . 0475mol
concentration of C in moldm – 3 =
0 . 0475 moldm – 3
1bii )
Molar mass of Bing mol- 1 :
Molar mass of Na2 CO 3. yH 2 O
=mass concentration of
Bingdm- 3
molar concentration of Binmoldm –
3
=13 . 6 gdm – 3
0. 0475 moldm- 3
=286gmol – 1
1biii )
Molar mass of Na2 CO 3 =[( 2 ×
23 ) + 12 + ( 16 × 3)]= 106gmol – 1
Mass of anhydrous Na 2 CO3 =
106x 0 . 0475 gdm – 3
=5 . 035gdm – 3
Mass of water= 13 . 6 – 5 . 035gdm –
3
=8 . 565gdm – 3
Mass of Na 2 CO3 = Molar mass of
Na2 CO 3
Mass of water y × Molar mass of
water
5. 035= 106
8. 565 18 y
y =106x 8 . 565
5. 035x 18
=10
2a )
Tabulate
Test
(i ) Fn + H 2 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 . HNO 3
+AgNO 3( 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
2bi)
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
2bii )
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 CO 2
CO3 ^ 2- or HCO3
– present
3i)
lime juice is acidic in nature and
the colour of methyl orange in
acidic medium is red
3ii)
Iron(iii)chloride will be reduced to
iron(ii) with yellow deposit of
sulphur
3iii)
The color of KM 2 O 4 is
decolorized because SO 2 (g ) acts
as
a reducing agent.
3iv )
Addition of ethanoic acid to KCO 3
results to the liberation
of a colorless and odorless gas CO
2 which turns lime
water milky
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4 Comments

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