NOTE THAT ^ MEANS Raise to power.
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.
==============
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 iro(ii) with
yellow deposit of sulphur
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
===============
2a)
Tabulate
Test
(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
Precipitate dissolved in excess NH3(aq)
Inference
Cl- confirmed
2bi)First portion of residue +NaOH(aq) in
drops, then in excess
Observation
White powdery precipitate which is insoluble in
excess NaOH(aq)
Inference
Ca2+ 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 CO2
CO3^2- or HCO3- present
===============
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
1000cm3ofCcontained0.002375x1000mol
2×25
=0.0475mol
concentration of C in moldm-3
=0.0475moldm-3
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=106×0.0475gdm-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 =106×8.565
5.035×18
=10
=====COMPLETED=====
GOOD LUCK
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