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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
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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 + 23.75cm^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
2x25
=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=106x0.0475gdm-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 =106x8.565
5.035x18
=10
1b
CaVa Na
_____ =____
CcVc Nc
CaVa 1
______ = ___
CcVc 2
Cc= CaVa
_____
2Vc
= 0.100 × 23.75 mol dm^3
________________
2 × 25.00
= 0.0475 mol dm^3
Conc. Of C in mol dm^-3 = 0.0475
(a)(i)Fn+H2O,then
filter
White residue and blue
filtrate was observed
Fn is a mixture of
soluble and insoluble
salts
(ii)Filtrate+NaOH(aq)in
drops,then in excess
A Blue gelatinous
precipitate which is
insoluble in excess
NaOH(aq)was formed
Cu2+present
(iii)Filtrate+NH3(aq)in
drops,then in excess
A pale blue gelatinous
precipitate was
formed.The precipitate
dissolves or is soluble
in excess NH3(aq)to give
a deep blue solution
Cu2+confirmed
(iv)Filtrate+dil.HNO3
+AgNO3(aq)
+NH3(aq)in excess
No Visible Reaction
White Precipitate
formed
Precipitate dissolved in
excess NH3(aq)
Cl-present
Cl-confirmed
(b)(i)First Portion Of
residue+NaOH(aq)in
drops,then in excess
White Powdery
precipitate which is
insoluble in excess
NaOH(aq)
Ca2+present
(ii)Second Portion Of
residue+dil.HCl
Effervescence/bubbles;
colourless,odourless
gas evolved.Gasturns
lime water milky and
turns damp blue litmus
paper red.
GasisCO2
CO3
2-orHCO3
-present
(a)(i)+H2O,then
filter
White residue and blue
filtrate was observed
Fn is a mixture of
soluble and insoluble
salts
(ii)Filtrate+NaOH(aq)in
drops,then in excess
A Blue gelatinous
precipitate which is
insoluble in excess
NaOH(aq)was formed
Cu2+present
(iii)Filtrate+NH3(aq)in
drops,then in excess
A pale blue gelatinous
precipitate was
formed.The precipitate
dissolves or is soluble
in excess NH3(aq)to give
a deep blue solution
Cu2+confirmed
(iv)Filtrate+dil.HNO3
+AgNO3(aq)
+NH3(aq)in excess
No Visible Reaction
White Precipitate
formed
Precipitate dissolved in
excess NH3(aq)
Cl-present
Cl-confirmed
(b)(i)First Portion Of
residue+NaOH(aq)in
drops,then in excess
White Powdery
precipitate which is
insoluble in excess
NaOH(aq)
Ca2+present
(ii)Second Portion Of
residue+dil.HCl
Effervescence/bubbles;
colourless,odourless
gas evolved.Gasturns
lime water milky and
turns damp blue litmus
paper red.
GasisCO2
CO3
2-orHCO3
-present
================================
(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
2bi) Firstportionof
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 CO 2
CO 3 ^2- or HCO3
- present
3i)Lime Juice is acidic in Nature and the colour of methyl orange in acidic medium is red
3ii)Iron(II)Chloride will be reduced to Iron(II) with the yellow deposit of sulphur
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
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