Waec Chemistry Practical

NOTE:

Tita means θ

^ means Raise to power

Pie means π

/ (means) division or divide

2 whole no 3/4 means 2¾
* means‎
multiplication (×)

Sqr root means √

Proportional means ∝

==================

1) 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- 23.80, 0.00, 23.80
Second- 23.75, 0.00, 23.70
Third- 23.75, 0.00, 23.75
Average volume of A used = 23.80 +
23.70
+ 23.75cm^3/3
=23.75cm^3

1bi)
CAVA/CcVc=2/1
Cc=CAVA/2VC
=0.100*23.75Moldm^-3/2*25.00
=0.0475moldm^-3
amount of A used = 0.100x
VA/1000=0.100*23.75/1000 =0.00237
2moles Of A = 1mole of C
0.002375mol of A = 0.002375mol/2
100cm^3 of C contain
0.00237*100mol/2*25 =0

1bii )
Molar mass of Bing mol- 1:
Molar mass of Na 2CO 3. yH2O = mass
concentration of Bingdm- 3
molar concentration of Binmoldm – 3
=13 .6gdm – 3
0. 0475 moldm – 3
=286 gmol -1

1biii )
Molar mass of Na 2CO 3 =
[( 2×23) + 12+( 16 ×3)]=106 gmol- 1
Mass of anhydrous
Na 2CO 3= 106x 0 .0475 gdm -3
=5 .035 gdm -3
Mass of water =13. 6- 5.035 gdm -3
=8 .565 gdm -3
Mass of Na2CO 3 =Molar mass of Na 2CO 3
Mass of water y × Molar mass of water
5. 035= 106
8. 565 18y
y =106 x 8.565
5. 035x 18
=10

WAEC CHEMISTRY PRACTICAL ANSWERS BY WWW.SOLUTIONWAP.NET

===============

NOTE THAT ^ MEANS Raise to

power.

Pls Draw Your Table As Usual And Input The Following:

Volume of pipette=25.00cm^3

indicator used- Methyl orange

colour change at end point-yellow to orange/purple

Note Use Your School End Point.

Tabulate

==================

1) 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- 23.80, 0.00, 23.80

Second- 23.75, 0.00, 23.70

Third- 23.75, 0.00, 23.75

Average volume of A used = 23.80 +

23.70

+ 23.75cm^3/3

=23.75cm^3

1bi)

CAVA/CcVc=2/1

Cc=CAVA/2VC

=0.100*23.75Moldm^-3/2*25.00

=0.0475moldm^-3

amount of A used = 0.100x

VA/1000=0.100*23.75/1000 =0.00237

2moles Of A = 1mole of C

0.002375mol of A = 0.002375mol/2

100cm^3 of C contain

0.00237*100mol/2*25 =0

1bii )

Molar mass of Bing mol- 1:

Molar mass of Na 2CO 3. yH2O = mass

concentration of Bingdm- 3

molar concentration of Binmoldm – 3

=13 .6gdm – 3

0. 0475 moldm – 3

=286 gmol -1

1biii )

Molar mass of Na 2CO 3 =

[( 2×23) + 12+( 16 ×3)]=106 gmol- 1

Mass of anhydrous

Na 2CO 3= 106x 0 .0475 gdm -3

=5 .035 gdm -3

Mass of water =13. 6- 5.035 gdm -3

=8 .565 gdm -3

Mass of Na2CO 3 =Molar mass of Na 2CO 3

Mass of water y × Molar mass of water

5. 035= 106

8. 565 18y

y =106 x 8.565

5. 035x 18

=10

============+++++=======

2)

Tabulate

Test

(a ) ( i ) Fn + H 2O , then

filter

Observation

White residue and blue

filtrate was observed

Inference

Fn is a mixture of

soluble and insoluble

salts

Test

(ii ) Filtrate + NaOH(aq ) in

drops , then in excess

Observation

A blue gelatinous

precipitate which is

insoluble in excess

NaOH(aq ) was formed

Inference

Cu 2 + present

Test

(iii) Filtrate +NH 3 (aq ) in

drops , then in excess

Observation

A pale blue gelatinous

precipitate was

formed . The precipitate

dissolves or is soluble

in excess NH 3 (aq ) to give

a deep blue solution

Inference

Cu 2 + confirmed

Test

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=======================

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