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Tuesday, April 3, 2012

CHEMISTRY QUESTION AND ANSWER


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.

ANSWER
> QUANTITATIVE ANALYSIS <
TEST
Solution C + aqueous NaOH + Heat
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Portion of Solution C + BaCl2 + dil. HCl.
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Portion of Solution D + NaOH in drops. Then in excess.
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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.
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white ppt insoluble in dilute HCl
--------
Reddish brown ppt insoluble
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Reddish brown ppt insoluble
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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 | 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.

3 comments:

  1. Tanx alot..GOD bless u, icekid9ja@gmail.com

    ReplyDelete
  2. may the loard continued to bless u ........ May u be hapi in every mod u meet ur self

    ReplyDelete