2015 WAEC Expo: Chemistry Practical 3 Questions & Answers – Now Available

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Wednesday, 15th April, 2015:
Chemistry 3(Practical) (Alt A) ————– 8.30am-10.30am(1st Set)
Chemistry 3(Practical) (Alt A) ————– 11.00am-1.00pm(2nd Set)
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PRACTICAL CHEMISTRY

(2)
Test | Observation | Inference
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1. Sample C + water 5cm^3 of C + NaoH in drop and in excess | soluble white geletenous precipitate in drop. White soluble or dissolve (in excess) | soluble salt. Zn^2+ or Al^3+ present.
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2. 5cm^3 of C + NH4OH in drop. In excess | white geletenous precipitate in drop. White geletenous ppt dissolve (in excess) | Zn^2+ or Al^3+ present. Zn^2+ confirmed (in excess).
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3. Solution of sample C + dilute HN3 + Bacl2(AQ) + HCL(AQ) | No visible reaction. White ppt form. White ppt insoluble in dilute Hcl | SO3^2- or SO4^2- may be present. SO4^2- confirmed.
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4. Sample D + dilute Hcl(aq) | A colourless and odourless gas evolve with a effervescence. The gas turn like water milky. The gas turn blue litmus to red | The gas in CO2 from CO3^2- or Hco3^-

PLS IN NUMBER 1, TAKE NOTE OF UR SCHOOL END POINT OR TITRE VALUE
EACH SCHOOL HAS A DIFFERENT TITRE VALUE FROM UR CHEMISTRY LAB.
OUR END POINT HERE IS 21.70cm3.
ANYWHR U SEE IT IN OUR WORK, EDIT IT WITH UR SCHOOL’S AVERAGE TITRE VALUE.
THEN RE-CALCULATE ONCE U EDIT IT IN ANY SOLUTION.

(1)
TABULATE:
Burrette Readings(cm3),1st Reading | 2nd Reading | 3rd reading|

Final burette Reading|23.70|45.00|26.00|

Initial burette Reading|2.00|23.50|4.30|

Volume of acid used|21.50|20.70|20.90|

Average titre value = (21.70 21.50 21.70)cm^3 /3
=(64.90)/3
= 21.70cm3

(1B)
A IS 0.100 mol
2.50g of Na2CO3 and Na2SO4 in 250cm^3
Na2CO3+2HNO3->2NaNO3+H2O+CO2
Conc of B in mol/dm^3
Conc of Na2CO3 in the mixture
Na2CO3 =106
VA=21.7cm^3
CA=0.100
VB=25cm^3
CB-=?
NA=2
NB=1
CAVA/CBVB=NA/NB
0.1*21.7/CB*25=2/1
CB=(0.1*21.7*1)/(25*2)
=0.0434M

ii)
molar concentration =mass conc/molar mass
mass conc=molar conc*molar mass
=0.0434*106
=4.69g/dm^3

iii)
2.50g of Na2CO3+Na2SO4->250cm^3
Xg of Na2CO3+Na2SO4->1000cm^3
Xg=2.50*1000/250
Xg=10g
% of Na2CO3=4.6/10*100
=46%

(3ai)
fountain experiment

(3aii)
to show that HCL is extremely soluble in water

(3aiii)
Ammonia gas (NH3)

(3aiv)
The water turn red immediately

(3av) No

(3avi)
because of its solubility in water

(3b)
u=>(heated) v + CO2
CaCo3(s) =>(heated) Cao(s) + Co2(g)
v + H20 => w
Cao(s) + H20(l) => Ca (OH)2(aq)
U = CaCa(3), V = Cao, W = Ca (OH)2

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