2019 waec chemistry practical answer
(1a)
DRAW A TABLE WITH THE FOLLOWING
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
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
=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. 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
=0.1mol/dm³
(1bii)
Using CaVa/CbVb = na/nb
0.1*12.50/Cb * 25.00 = 2/1
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
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
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%
= 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
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
====================
(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
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
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
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
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.
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
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
Q is reducing agent like SO2, H2S, CO
(3bi)
calcium oxide(Cao)
calcium oxide(Cao)
(3bii)
concentrated H2SO4
concentrated H2SO4
(3c)
NaOH pellet is delinquescent because it absorb moisture from the atmosphere to form solution
NaOH pellet is delinquescent because it absorb moisture from the atmosphere to form solution
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