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#Post#: 224--------------------------------------------------
FREE 2013 WAEC GCE CHEMISTRY PRACTICAL ANSWERS
By: Ebenezer Date: September 2, 2013, 1:16 am
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Chemistry A. Answers
Titration
Burette reading...1st 2nd
3rd.final burette . 22.20,
22.40,
22.30.
Initial burette 0.00, 0.20,
0.10.
Volume of acid used.
22.20,
22.20,
22.20.
Average volume of acid
use =
22.2+22.2+22.2/3=22.20cm^3
Therefore 25cm^3
volume of
pipette use gives
22.20cm^3 of
acid = 22.20cm^3
Eqn of d rxn= 2HY +
Na2co3-
>2NaY + H2O +CO2. Coc.
Of B in
g/dm3=5*.1
5*1000/250=20.0g/dm3
Conc of mol/dm3 of B=
conc in
g/dm3/molar
mass=20.0g/
dm3/106g/
dm3=0.188mol/dm3
Conc of A in mol/
dm3 .using
CaVA/cbvb=Na/Nb
Ca*22.20/0.188*25=2/1
Ca=9.4/22.20
Ca=0.423mol/dm3
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(2) Test..a. A point of c
+water,stirred and fitered
bi. A
portion of
the fitrate +naoh in drop
and in
excess.
Test..a. A point of c
+water,stirred
and fitered bi. A portion
of the
fitrate +naoh in drop and
in
excess.
Biii. Another portion of
the
filtrate +hcl +bacl
Ci. Residue+ naoh in drop
and in
excess.
----Observation..---
-A clear filtrate and a
black
residue were obtained
-A white gelatinous
precipitate
was formed
-A white precipitate
soluble in
excess was formed
-A white precipitate was
formed
-A blue gelatinous
precipitate
was formed.
----Inference.-----
-C is a mixture of soluble
and
insoluble salt
-Pb^2+,AL^3+,or zn^2+
present
-Zinc ii ion,zn^2+ present
-Tetraoxosulphate vi
ion,so4^2-
present
-Copper ii ion,cu2+
present
(1a).table
rogh-24.00,0.00,24.00
1-23.50,0.00,23.50
2-25.50,2.00,23.50
3-27.50,4.00,23.50.
VA=23.50+23.50/3=70.50/3
VA=23.50
[1a continues]
Equation for the
reactionbr /> 2HY
(aq)+Na2Co3(aq)-----2NaY
(aq)
+H2O(1)+CO2(g)
(1b).CAVA/
CBVB=CA=?,CB=0.05
VA=23.50,VB=25
CA*23.50/0.05*25=2/1
CA*23.50*1=0.05*25*2
=2.5/23.50
CA=2.5/23.5
CA=0.106moldm^3.
(3ai)
S is hydrogen sulphate
(H2S),
T is Iron (III) sulphid
V is sulphur(IV) oxide
(3aii)It decolourizes the
Solution
(3bi)
Na2CO3 is soluble in
water while
NaHCO3 is insoluble in
water
Na2CO3 is a basic salt
which
changes red litmus paper
to blu
while NahCO3 is a acidic
salt that
changes blue litmus
paper to
read.
(3bii) Na2CO3 cannot be
decomposed on heating
while
NaHCO3 can be
decompose on
heating
2NaHCO3(g)=>Na2CO3(aq)
+H20+CO2(g)
(2Ai) tabulate
Observation: partially
dissolve to
give colourless filtrate and
black
residue
Inference: C is a mixture
of
soluble and insoluble salts
(2bi)
Observation: whire
gelationous
ppt
Inference: Zn ^z+ or
Al^3+ likely
present
(2bii)
Observation: white
gelation ppt
and ppt dissolve solube
Inference: Zn ^z+ or
Al^3+ likely
present
(2biii)
Obersation: Insoluble in
dil. Hcl
Inference: Co3^-2, So4^2-
is
present
(2ci)
Observation: Residue
dissolved to
give a pale blue solution.
Inference: Cu^2+ present
(2cii)
Observation: ppt dissolves
to give
deep blue solution.
Inference: Cu^2+
conformation
[br]TITRATION,SALT ANALYSIS&GENERAL CHEMISTRY
This are the 3 questions that waec n0rmaly ask in PRACTICAL
CHEMISTRYTITRATION
==>Formular used in titration
*.conc. Of A in mol/dm3 = mass conc./molar mass
*.conc. Of A in g/dm3 = molar conc x molar mass
*.Number of mol of Acid = vol x mol of acid/1000
*.Number of mol of Base = vol x mol of base/1000
*.Molar conc. Of Acid or Base = [CaVa/CbVb =Na/Nb]
*.Percentage purity = mass of pure/mass of impure x 100
*.Mass of pure = Molar conc x Molar mass
*.Molar mass = Mass conc/Molar conc.
*.Percentage impurity = Mass of impure /Total mass x 100
==>INDICATOR USED
Strong acid + Strong base = any suitable indicator
Strong acid + Weak base = methyl orange
Weak acid + Strong base = phen0phtalen
Weak acid + Weak base = No indicator
==>EXERCISE / SOLUTION
If you are to s0lve this questions
[a.] In an experiment 20.0cm3 of 0.065mol/dm3, Na0H were
titrated against dilute Hcl. The table sh0w the result of the
titration
*.BURETE READING: 1st, 2nd, 3rd
*.FINAL READING: 23.50, 46.60, 47.40
*.INITIAL READING: 0.00, 23.50, 24.00
*.VOL OF A USED: 23.50, 23.10, 23.00
(1)Name the suitable indicator and give reas0n
(2)Give the color of the indicator
(3)What type of reaction is in the experiment
(4)Write the balancing equation
(5)Determin the Average titre value
(6)Cal. Conc of A in mol/dm3
SOLUTION
(1.) Any suitable indicator i.e Methyl orange or Phen0pthaline.
REAS0N:becos, the reaction given to use indicateSTRONG ACID AND
STRONG BASE
==>(2.)
*.INDICATOR: Starting color, Ending color
*.METHYL ORANGE: yellow/orange, pink @endcolor
*.PHEN0LPHTALINE: red/purple, colourles@the end
==>
(3.) The type of rxn is
NEUTRALIZATION REACTION
(4.) HCL + NA0H ---> NACL + H20
(5.) Average titre value is
Va = (23.50 + 23.10 + 23.00 divided by 3) = 23.45cm3
(6.) conc of A is { Ca=? , Cb=0.065, Va=23.45, Vb=20.0cm , Na=1,
Nb=1 }
FormularCa Va/Cb Vb x Na/Nb
i.e Ca x 23.45/0.065x 20.0 x 1/1=0.06mol/dm3
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#Post#: 225--------------------------------------------------
Re: FREE 2013 WAEC GCE CHEMISTRY PRACTICAL ANSWERS
By: Ebenezer Date: September 2, 2013, 9:33 am
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COMPLETE ANSWERS
(2Ai) tabulate
Observation: partially dissolve to
give colourless filtrate and black
residue
Inference: C is a mixture of
soluble and insoluble salts
(2bi) Observation: whire gelationous
ppt
Inference: Zn ^z or Al^3 likely
present
(2bii) Observation: white gelation ppt
and ppt dissolve solube
Inference: Zn ^z or Al^3 likely
present
(2biii) Obersation: Insoluble in dil. Hcl
Inference: Co3^-2, So4^2- is
present
(2ci) Observation: Residue dissolvedto
give a pale blue solution.
Inference: Cu^2 present
(2cii)
Observation: ppt dissolves to give
deep blue solution.
Inference: Cu^2
[3-] electrochemical or votaic cell.
3i-elctrolyte e.g cuso4.
3ii- glucose solution.
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3B-evaporating dish
stirrer bubsen burner
thermometer.
3c- white presoluble in NH3
ii. White pewdetry precipotate
iii.white precipitate soluble in eccessive
gaseouse NH2
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