# 2020 NABTEB GCE CHEMISTRY VERIFIED ANSWERS

(1ai)
To have two or more concurrent result or values

(1aii)
Average Volume titre=20.00 + 20.10 / 2
=20.05

(1aiii)
HNO3 + XOH —–> XNO3 + H2O

(1bi)
When number of acid (na) = 1, and the number of base (nb) = 1
Using CaVa/CbVb=na/nb

Given:
Ca=0.0499
na=1
nb=1
Va=20.05
Vb=20.00cm³
Use your school titre value and volume of base used(i.e 25cm³)
Cb=?

Using the equation
CaVa/CbVb=na/nb
Cb= 0.0499 × 20.05 × 1 ÷ 20 × 1
Cb=0.050 mol/dm³
≈0.1mol/dm³

(1bii)
1.4g –––– 250cm³
x –––– 1000cm³
x=1.4×1000/250
x=5.6g

Mass conc = molarity × Molar mass
5.6=0.1 × Molar mass
Molar mass= 5.6/0.1
=56g

(1biii)
Molar mass = 56g
The Molar mass of x = x + 16 + 1
:. 56 = x + 17
x = 56 – 17
x = 39

Relative atomic mass of x = 39, the element x js potassium and the alkali is potassium hydroxide

(1biv)
Methyl orange

(1ci)
To Ensure High Accurate of the titre Value

(1cii)
Using an acid will alter the volume of the acid used and affect the reading in the final result.

(1ciii)
This is to avoid the movement of air, which may lead to the sucking of the base solution

(2i)
(i) W is Copper (Cu)
(ii) X is Iron (fe)
(iii) Y is Zinc (Zn)

(2ii)
(i) W is Cu^2+
(ii) X is Fe^3+
(iii) Y is Zn^2+
(iv) Z is Pb^2+

(2iii)
TABULATE:

• W

TEST APPLIED
W + NH3 in drop
And in excess

OBSERVATION
Blue precipitate form in a drop of NH3
And Blue precipitate in excess

INFERENCE
Cu^2+ is present

• X

TEST APPLIED
X + NH3 in drop
And in excess

OBSERVATION
Light green precipitate form both in drop and in excess of NH3

INFERENCE
Fe^2+ is present

• Y

TEST APPLIED
Y + NH3 in drop
And in excess

OBSERVATION
White Gelatinous precipitate soluble in excess

INFERENCE
Zn^2+ is present

• Z

TEST APPLIED
Z + NH3 in drop
And in excess

OBSERVATION
White powdery precipitate insoluble in excess

INFERENCE
Pb^2+ is present

(3a)
(i) Pick the solute
(ii) Dissolve the solute
(iii) Decolorize the solution
(iv) Filter any solids from the hot solution
(v) Crystallize the solute
(vi) Collect and wash the crystals
(vii) Dry the Crystals

(3bi)
TEST TUBE A
The color of tube A changes to blue-black because it is in carbohydrate and starch was present

(3bii)
TEST TUBE B
The solution TUBE B turned clear, Meaning that starch was gone

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