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Chemistry practical questions for waec 2012
By: Inyavic Date: April 3, 2012, 3:59 pm
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CHEMISTRY WAEC PRACTICAL Question1) burette of 50cm capacity
pipette, either 20cm or 25 (using d
usual apparatus for titratione) Reagents
for qualitative work (i) red and blue litmus paper;
(ii) aqueous ammonia
(iii) dilute hydrochloric acid
(iv)dilute sodium hydroxide solution
(v)barium chloride solution
(vi)dilute trioxonitrate (V) acid
(vii)silver trioxonitrate
(V) solution [AgNO3]
(viii)lime water methyl orange indicator Question 2) A. 150cm of
tetraoxosulphate
(VI) solution in a corked flask or
bottle,labelled 'A' containing 2.8cm of
concentrated H2SO4 (about 98% w/w)
per dm of solution. B. 150cm of NaOH solution, in a corked
flask or bottle labelled 'B' containing
3.9g of NaOH per dm of solution. C.
10cm of (NH4)2SO4 solution in a bottle
labelled 'C' containing 66g of (NH4)
2SO4 per dm of solution. D. 10cm of FeCl3 solution in a bottle
labelled D containing 40g of FeCl3 per
dm of solution. 2012 WAEC PRACTICAL CHEMISTRY
ANSWERS ,>QUANTITATIVE ANALYSIS<
TEST
Solution C + aqueous NaOH + Heat
-----------------------
Portion of Solution C + BaCl2 + dil. HCl.
------------- Portion of Solution D + NaOH in drops.
Then in excess.
----------
Portion of Solutio D + NH3(aq) in drop
then iq excess.
-------- Portion of Solution D + HNO3 + AgNO3
solution.
---------
Mixture frm above + HNO3 in excess.
------ OBSERVATIONS
Evolution of colourless gas with
characteristics choking smell which
turns moist red litmus paper blue and
form white foam with conc. HCl acid
vapour. -----------------
white ppt insoluble in dilute HCl
--------
Reddish brown ppt insoluble
----------
Reddish brown ppt insoluble --------
White ppt insoluble
--------
White ppt dissolved to give a colourless
solution.
----------------- INFERENCE
NH3 gas from NH4+ is present.
-------
SO42-, SO32-, CO32- are present
--------
Fe3+ is present -------
Fe3+ is present
--------
Cl- ion is present
------
Cl- ion is confirmed. (N.B u write a test scroll to its
observation and then its inference. 'cos
it is to be written in tabular form i.e
make a 3 columns: then write
TEST| OBSERVATION|INFERENCE.) ,
==================== >VOLUMENTRIC ANALYSIS<
A solution acid H2X 2.45g in 500cm3 of
solution. Solution B contain 3.90g of
NaOH in 1dm3 of solution. Put A into
the burrette and titrate with 20cm3 or
25cm3 portion of B using methyl orange as indicator.
Record the value of your pipette.
Tabulate your readings and calculate
the average titre value of acid used.
From ur results and information
provided. Cal. the: i) Concentration of solution B in mol/
dm3
ii) Concentration of A in mol/dm3
iii) Molar Mass of acid H2X and hence
determine the relative mass of X.
H2X + 2NaOH===> Na2X + 2H2O. Burrete reading (cm3)
Final
Initial
Volume Rough|1st |2nd
28.50 |26.40 |25.60
1.00 | 2.00 | 1.00
27.50 |24.40 |24.60 3rd |
26.50|
2.00|
24.50|
(N.B folow d order i.e final reading,
initial reading and volume of A used). Average Volume of Acid
used =
(24.40+24.60+24.50) /3
= (73.50/3) =24.50cm3. A=H2X B=NaOH
2.45g ==> 500cm3
xg ==>1000cm3
:. x= (2.45 *100)/500
=4.90g/dm3. i.) Cb = mas concentration/molar
concentration
=(3.90/40)mol/dm3
Cb=0.0975mol/dm3. ii.) CaVa/CbVb =na/nb ; Va =24.50cm3,
Cb=0.0975mol/dm3, Vb=25cm3, na=1,
nb=2.
(Ca*24.50)/(0.0975*25) Ca=
(0.0975*25)/(24.50*2). Ca
=2.4375/49.00 Ca=0.0497mol/dm3. INFERENCE
NH3 gas from NH4+ is present.
-------
SO42-, SO32-, CO32- are present
--------
Fe3+ is present -------
Fe3+ is present
--------
Cl- ion is present
------
Cl- ion is confirmed. (N.B u write a test scroll to its
observation and then its inference. 'cos
it is to be written in tabular form i.e
make a 3 columns: then write
TEST| OBSERVATION|INFERENCE.) ,
==================== >VOLUMENTRIC ANALYSIS<
A solution acid H2X 2.45g in 500cm3 of
solution. Solution B contain 3.90g of
NaOH in 1dm3 of solution. Put A into
the burrette and titrate with 20cm3 or
25cm3 portion of B using methyl orange as indicator.
Record the value of your pipette.
Tabulate your readings and calculate
the average titre value of acid used.
From ur results and information
provided. Cal. the: i) Concentration of solution B in mol/
dm3
ii) Concentration of A in mol/dm3
iii) Molar Mass of acid H2X and hence
determine the relative mass of X.
H2X + 2NaOH===> Na2X + 2H2O. Burrete reading (cm3)
Final
Initial
Volume Rough|1st |2nd
28.50 |26.40 |25.60
1.00 | 2.00 | 1.00
27.50 |24.40 |24.60 3rd |
26.50|
2.00|
24.50|
(N.B folow d order i.e final reading,
initial reading and volume of A used). Average Volume of Acid
used =
(24.40+24.60+24.50) /3
= (73.50/3) =24.50cm3. A=H2X B=NaOH
2.45g ==> 500cm3
xg ==>1000cm3
:. x= (2.45 *100)/500
=4.90g/dm3. i.) Cb = mas concentration/molar
concentration
=(3.90/40)mol/dm3
Cb=0.0975mol/dm3. ii.) CaVa/CbVb =na/nb ; Va =24.50cm3,
Cb=0.0975mol/dm3, Vb=25cm3, na=1,
nb=2.
(Ca*24.50)/(0.0975*25) Ca=
(0.0975*25)/(24.50*2). Ca
=2.4375/49.00 Ca=0.0497mol/dm3. iii.) Ca = mass concentration /
molar
concentration
0.0497=4.90/molar mass
molar mass =4.90/0.0497
molar mass=98.6g/mol
H2X =98.6 (1*2) + X =98.6
X= 98.6 -2
= 96.6
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