Waec 2017 Chemistry Practical Answers – May/June Expo

Tuesday, 11th April, 2017
Chemistry 3 (Practical Alternative A) 09.30am – 11.30am

WAEC

ATTENTION:- NOTE THAT SUBSCRIBERS GETS ANSWERS BEFORE THE EXAM STARTS..
ALWAYS SUBSCRIBE IF YOU NEED FAst and early runz

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COMPLETED
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(1a)
Pls Tabulate:
Burette reading|Final burette reading(cm^3)|Initial burette reading (cm^3)|Volume of acid used(cm^3)|
Rough- 24.10,0.00,24.10
First- 25.02, 0.00, 25.05
Second- 25..03, 0.00, 25.03
Third- 25.03, 0.00, 25.03
Average volume of A used = 25.02 + 25.03+ 25.03cm^3/3
=25.03cm^3

(1bi)
Frm the equation, mole ratio of acid to base is 2:1
Using CAVA/CBVB = nA/nB
0.2*25.03/CB*25 = 2/1
CB= 0.2*25.03/2*25
=0.100moldm^-3

(1bii)
C contains 14.3g in 500cm3
.:. Conc of c in gdm^-3 is 28.6dm^-3.

(1biii)
Conc in dm^-3 = conc in mol/dm3 * molar mass
Molar mass = 28.6gdm^-3/0.100moldm^3
=286gmol^-

(1biv)
The hydrated salt is NA2CO3.XHO=286
2(23)+12+3(16)+X(2+16)=286
46+60+18X=286
18X+106=286
18X=286-106
X=180/18
X=10
The number of water molecules in the hydrated salt is 10

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Pls Tabulate:
(2ai)
TEST: F + water
OBSERVATION: Partially soluble salt
Inference: F is a mixture of soluble and insoluble salts

(2bi)
TEST: Filtrate + NaOH(aq)in drops,then in excess
OBSERVATION: A pale blue gelatinous precipitate which is insoluble in excess NaOH(aq).
INFERENCE: Cu2+ present

(2bii)
TEST: Filtrate + NH3(aq) in drops,then in excess.
OBSERVATION: A pale blue precipitate was formed. The precipitate is soluble in excess NH3(aq)to give a deep blue solution.
INFERENCE: Cu2+ confirmed

(2c)
TEST: Filtrate + dilute HNO3(aq) + AgNO3(aq)
OBSERVATION: White precipitate formed which turns grey on exposure to light.
INFERENCE: Cl- present

(2di)
TEST: Residue + HNO3(aq)
OBSERVATION: Effervescence of a colourless, odourless gas which turns moist blue litmus paper faint and turns lime water milky.
INFERENCE: CO2(g) from CO3^2- or HCO3- present

(2dii)
TEST: Residue + NaOH(aq)
OBSERVATION: White powdry ppt insoluble in NaOH(aq)
INFERENCE: Ca2+ present.

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(3i)
The colour of methyl orange will change from orange to yellow

(3ii)
Iron(III)chloride will be reduced from brown iron to green iron
Fe^3+(aq)->Fe^2+(aq)

(3iii)

The color of KM2O4 is decolorized because SO2(g) acts as a reducing agent.

(3iv)
Addition of ethanoic acid to KCO3 results to the liberation of a colorless and odorless gas CO2 which turns lime water milky

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16 Comments

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