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# WAEC Chemistry Practical Answer – May/June Expo 2019

WAEC 2019 Chemistry Practical Solutions and Expo Answer

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THERE WILL NOT BE FREE ANSWER FOR OBJ AND THEORY.

Please Note: Make sure you use your school end point Number 1. In my answer, you will see that i used 12.50cm3. Make sure you use what your school used. So try edit my 12.50cm3 and put yours, then recalculate. Meaning that you should replace my 12.50cm3 with your school’s end point anywhere you see it on my work. Some school got 12.90cm3, 13.00cm3, some even got 18.50cm3, etc

All the best

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(1a)
DRAW A TABLE WITH THE FOLLOWING

Final |13.60|15.50|14.80|12.50
Initial |1.00 |3.10 |2.40 |0.00
Vol. acid used |12.60|12.60|12.40|12.50

Volume of S2O3²- =12.60+12.40+12.50/3
=37.50/3
=12.50cm³

(1bi)
Conc. in g/dm³ = molar mass * conc in mol/dm³
But molar mass of Na2S2O3
=(23*2)+(32*2)+(16*3)
=46 + 64 + 48
=158g/mol
Therefore 15.8g/dm³ = 158g/mol * conc of A in mol/dm³

Conc of A in mol/dm³ = 15.8/158
=0.1mol/dm³

(1bii)
Using CaVa/CbVb = na/nb
0.1*12.50/Cb * 25.00 = 2/1

50Cb = 1.25
Cb = 1.25/50
Cb = 0.025mol/dm³

The conc of I2 in B = 0.025mol/dm³

(1biii)
Gram conc of I2 = molar mass * molar conc
= (127*2)*0.025
=6.35g/dm³

Percentage mass of I2 in sample = 6.35/9.0 *100%
= 0.7056 * 100%
=70.56%

(1c)
There have to be a change in the colour of the mixture before it is added. The end point is known when the blue colour formed as starch is added, changes to colourless

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(2a)
Test: C + Water and filtered

Observation: part of C dissolves, colour less filtrate is formed

Inference: C is a mixture of a soluble salt

(2bi)
Test: 2cm^2 of filterate of salt C + AgNO3(aq) then dilute HNO3

Observation: white precipitate is formed

Inference: Cl^- present

(2bii)
Test: Solution im 2(b)(i) + excess NH3 solution

Observation: precipitate disorder in excess solution of NH3

Inference; Cl^- confirmed

(2ci)
Test: Resident + dilute HCl + shake

Observation; residue dissolved to form a blue solution

Inference; Cu^2+ present

(2cii)
Test; solution in 2(c)(i) + NH3 solution in drops and then in excess

Observation: light pale blue precipitate is formed precipitate dissolved in excess solution of NH3 to give a deep blue solution

Inference; Cu^2+ confirmed.
==========

3(i)On adding BaCl2 solutions to a portion of saturated Na2CO3 precipitate is formed,precipitate dissolves on adding excess dilute HCl

3 (ii) Q is reducing agent like SO2, H2S, CO

3bi calcium oxide(Cao)
(ii) concentrated H2SO4

3c NaOH pellet is delinquescent because it absorb moisture from the atmosphere to form solution

CHEMISTRY PRACTICAL COMPLETED.

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