# 2018 NECO GCE Chemistry Practical Answers

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# 👑 2018 NECO GCE CHEMISTRY PRACTICAL ANSWERS👑

(1ai)
Titrations| 1 | 2 | 3
Final | 21.80 | 21.50 | 24.00
Initial | 01.30| 01.00 | 03.50
Vol of Acid used |20.50|20.50|20.50

(1aii)
Average volume of A used = (VA1 + VA2 + VA3)cm³
VA = (20.50+20.50+20.50)/3
= 61.50/3 = 20

(1aiii)

(1aiv)
Methyl orange is used because we are titrating a strong acid and a strong base

(1av)
(i) Do not blow air into the pipette when releasing its constant
(ii) Use the lower meniscus when checking your base in the pipette

(1avi)
Sodium hydroxide

(1bi)
No of moles of NaOH in the volume of the pipette
Recall that mass conc = molar conc × molar mass NaOH
Molar conc = mass/molar mass = 4.0/40
= 0.1dm-³
Hence 0.1mol = 1000cm³
X mol = 25.0cm³
X = 25×0.1/1000
= 0.0025mol of NaOH

(1bii)
Concentration of Acid CA = 0.06moldm-³
Mass conc. of NaOH = 4.0gdm-³

(1biii)
mole =0.1mol1dm^3*25cm^3/1000
=2.5/1000=0.0025mole

(1biv)
Mole ratio of acid to base in the reaction
Using CAVA/CBVB = 0.06×20.50/0.1×25
= 1.23/2.5 = 0.412

The mole ratio na/nb = 0.5 = 1/2

na/nb = 1/2
Where na = no of moles of acid
nb = no of moles of NaOH

(1ci) It is a dibasic acid because it produces a valency of +2, which multiples the no of molecules of the base.

(1cii)
H2X + NaOH –> Na2X + H2O
=====================================

C is soluble salt

(2ai)
Zn^2+ or Al^3+ present

-Zn^2+ or Al^3+ present

(2aii)
Zn^2+ or Al^3+ present

-Zn^2+ is confrim

(2bi)
SO3^2- or SO4^2-present

(2bii)
SO3^2- present

(2biii)
SO3^2- confrim

(2c)
ZnSO3
====================================

(3ai)
(I) CO2 can be removed by adding caustic soda ie sodium hydroxide.

(ii) Purity of water can be determined through its boiling point as it is noted that pure water boils at 100°c

(3aii)
AgNO3 produces brown fumes as NO2 is produced on heating as
2AgNO3(s) –> 2Ag(s) + O2(s) + 2NO2(g)

(3aiii)
It changes blue litmus paper red as the hydrogen ion in the wet litmus forms as acid on exposure to the litmus paper.

(3bi)

(3bii)
It cannot be collected by displacement of air because Nitrogen constitutes to large percentage of the air constituents(78%).

(3biii)
HCl gas or ammonia gas

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