Home » Education News » 2012 Waec examination Expo: Waec examination CHEMISTRY SPECIMENS- Biology Questions and Answers.

2012 Waec examination Expo: Waec examination CHEMISTRY SPECIMENS- Biology Questions and Answers.

Chemistry 2012 Specimen: Questions And Answers.

2012 WAEC
PRACTICAL
CHEMISTRY


This is the latest and hottest current waec 2012 specimen chemistry examination questions and answers.
After sharing this post with several of my friends, i recieve hundreds of thank you text message.
It is wonderful, for you to recieve private and exclusive 2012 specimen examination.
join our fanpage today, and like this post.
Then, you’re IN.!

(1)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
(2)
A. 150cm of
tetraoxosulp£ (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 REAL AND CONFIRMED WAEC
CHEMISTRY PRACTICAL QUESTION
AND ANSWER
Question 2a (salt analysis)=>
c&d are acqeous solution of two
simple salts. carry out the fll
exercises on c&d, record your
observations identify any gas
evolved. State the conclusion you
draw from the result of each test. A.
Divide solution c into four portions. 1.
1st portion of c= add NaOH solution in
drops and then in excess.
(ii). 2nd portion of solution c = add
acqeous amonia in drops and then in
excess. (iii). 3rd portion of c= add
NaOH solution and warm. (iv). 4th
portion of c= add barium chloride
solution followed by dilute
hydrochloric acid in excess
2b. divide solution d into five
portions.i. to the first portin add NaOH
in drops then in excess ii. to the
second portion add acqeous amonia
in droos then in excessiii. to the third
portion add dilute trioxonitrate (v)
acid followed by silver trioxonitrate
(v) solutioniv. to the fourth portion
add silver trioxonitrate (v) solutionv.
add few drops of solution D into
solution C and warm.

2012 CHEMISTRY ANALYSIS 1
TEST
1a)Small portn of sample C+NaOH
+Heat.
b)Anoda portn of sample C+BaCl
+dil.HCl.
2a)Small portn of spec.D+NaOHaq
drop den in
excess.
b)Small portn of sample D+NH3aq
drop den in
excess.
3a)Small portn of sample D+dil.HNO3
drop
+AgNO3aq.
b)Soln formed above+NH3aq drop
den in excees.
OBSERVATION
1a)A colourless gas with pungent
smell dat turn
moist litmus paper blue.
b)White ppt is formed which is
insoluble in dil.HCL.
2a)Redish brown ppt is formed which
is insoluble in
excees NaOH.
b)Brown ppt which is insoluble in
excees NH3.
3a)White ppt is formed.
b)White ppt dissolve to form a clear
soln.
INFERENCE
1a)NH3g 4rm NH4+ salt.
b)SO4 raise to power 2- is
comfirmed/present.
2a)Fe^3+ is comfirmed.
b)Fe^3+ is comfired.
3a)Cl^- is present.
b)Cl^- is present.

> VOLUMENTRIC ANALYSIS Na2X + 2H2O.
Burrete reading (cm3)
Final
Initial
Volume
Rough | 1st | 2nd | 3rd
28.50 | 26.40 | 25.60 | 26.50
1.00 | 2.00 | 1.00 | 2.00
27.50 |24.40 |24.60 | 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
NOTE:-
1. Even if you don’t Undestand the
Question and Answer, Don’t worry
yourself, Just chill till tommorrow until
you see the Question, Its gonna be
easy with the above Post.
2. Number 3 Question is not Included
because we can’t get it.. Try and Get
it done yourself in the Exam Hall

4 comments

  1. I need question for chemistry weac 2013

  2. I need chemistry question for waec 2013

  3. Mehn tnkz 2 waec board,many studntz made t

Leave a Reply

Your email address will not be published. Required fields are marked *

*