Welcome to Mercy Gist's Blog: Waec 2019 Chemistry 3 (Practical) (Alternative A) Answer Now

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COMPLETED!!!!

MORE EXPLANATION:
You should use ur school average titre value in number 1a
bcos each school have their general titre value for their
students which should differ a little with their fellow
classmates due to errors encountered during titration in the
lab. So when you see 12.50cm3, you should check the range
ur school got during the titration in the lab, ur chemistry
teacher must have taught how to input ur values and he/she
must have given you an average titre value. So, use it to edit
our solution. Some school got 13.00cm3, 12.90cm3,
15.50cm3, etc. You should edit and recalculate the Number
1bii by editing our 12.50 there with ur school value. But if ur
school, got ranges like 12.30cm3-12.70cm3, then u can use
my 12.50cm3. Its difference is always +- 0.2... Just take note.
This is applicable in number 1a and 1bii.

Thanks

CHEMISTRY PRATICAL ANSWER ALTERNATIVE A

(1a)
DRAW A TABLE WITH THE FOLLOWING
Burette reading|ROUGH|1st |2nd |3rd
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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Tabulate

(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.

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(3ai)
On adding BaCl2 solutions to a portion of saturated Na2CO3
precipitate is formed,precipitate dissolves on adding excess
dilute HCl (3aii)
Q is reducing agent like SO2, H2S, CO
(3bi)
calcium oxide(Cao)
(3bii)
concentrated H2SO4
(3c)
NaOH pellet is delinquescent because it absorb moisture from

the atmosphere to form solution