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2020 NABTEB CHEMISTRY PRACTICAL QUESTIONS/ANSWERS VERIFIED EXPO

 

(1)
NOTE: dont forget to use your school titre value, indicator and volume of pipette
Volume of pipette used = 25.00cm³
Indicator used = methyl orange

TABLE OF OBSERVATION

TABULATE

Burette Rading ǁ Rough ǁ 1st ǁ 2nd ǁ 3rd

final burette reading(cm³) ǁ 25.20 ǁ 24.60 ǁ 24.40 ǁ 24.50

Initial burette reading(cm³) ǁ 0.00 ǁ 0.00 ǁ 0.00 ǁ 0.00

volume of acid used(cm³) ǁ 25.20 ǁ 24.60 ǁ 24.40 ǁ 24.50

Average volume of Acid used = 1st + 2nd + 3rd / 3
= 24.60 + 24.40 + 24.50 / 3
= 73.50/3
=24.50cm³

(1bi)
Amount of Acid in the Average volume of A
Molar concentration of A, CA= 0.0450mol/dm³
Average volume of A, VA= 24.50cm³
:. Amount of A, nA= CAVA
= 0.450 x 24.50 / 1000
Amount of A = 0.001103mol/dm³

(1bii)
Amount of B, nB = CBVB
Molar conc. of B = Amount/Volume
= 0.00625 x 1000 / 250
CB = 0.025mol/dm³
VB = 25.00cm³

:. Amount of base in 25.00cm³ pipetted in B = CBVB
= 0.025 x 25 /100
Amount of B = 0.000625mol.

(1biii)
Mole ratio of the Acid to Base.
From,
CAVA/CBVB = nA/nB
Let the mole ratio of the acid to the base be N.
Hence,
nA/nB = N
CAVA/CBVB = N
0.001103/0.000625 = 1.76
N = 1.76 : 1
͌ 2 : 1
nA/nB = 1.76/1
͌ 2/1 = 2 : 1

(1ci)
since mole ratio is 2 :1, The equation of the reaction is
2HxY(aq) + NA2CO3(aq) —-> 2NaCL(aq) + H2O(l) + CO2(g)

(1cii)
The value of x from the equation is 1-
The Basicity of the Acid HxY is 1
Hence, it is monobasic
=======================================

(2a)
TABULATE
Test ǁ Observation ǁ Inference

TEST
C + Water + Litmus paper

OBSERVATION
C is soluble in water to give a colourless solution which turns Red litmus paper blue

INFERENCE
C is a soluble salt
It is an Alkaline solution

(2b)
TEST
Solution 2(a) + BaCl2 (aq) + HCL in drop
then in excess

OBSERVATION
White precipitate is prodcued or formed
the precipitate is soluble in Excess HCL(aq)

INFERENCE
CO3² ˉ , SO3³ˉ or SO4² ˉ is present
CO3² ˉ is confirmed

(2ci)
TEST
D + 5cm³ of dilute HCl

OBSERVATION
A colourless, Odourless gas evolved. The gas gives a pop sound with a lighted splint light green solution is produced

INFERENCE
The gas is hydrogen
The residue is a metal above hydrogen in the activated series

Fe²+ suspected

(2cii)
TEST
First portion from (2ci) + NaOH (aq) in drops
Then in excess

OBSERVATION
Dirty green precipitate formed
The precipitate is insoluble in excess NaOH(aq)

INFERENCE
Fe²+ Confirmed

(2ciii)
TEST
Second portion from (2ci) + KSCN(aq)

OBSERVATION
No visible reaction is observed

INFERENCE
Fe²+ confirmed
=======================================

(3a)

First fraction is benzene at 80°C and the second fraction is methylbenzene at 110°C. The mixture is separated by using fractional distillation at different boiling point

(3b)
C1=0.6moldm³
V1=500
C2=4×0.6
=2.4dm³
V2=?
C1V1=C2V2
0.65×500 / 2.4 = 2.4V2 /2.4
V2= 125cm³
=======================================
Completed.

Posted by on 13th September 2020.

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Categories: NABTEB

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