2020/2021 WAEC/NECO Answers: WAEC GCE 2021. Chemistry 3 answers



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Exam Time: Monday, 17th September, 2018
Literature-In-English 2(Prose) 9.30am 10.45am
Literature-In-English 1 (Objective) 10.45am 11.45am
Literature-In-English 3 (Drama & Poetry ) 2.00pm 4.30pm

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Literature OBJ:
1-10: ABCDBCBCDB
11-20: CACDDCABAD
21-30: ABACCCBCCC
31-40: CBDACADDDB
41-50: CDBCBAACA

This is actually the WAEC GCE 2018. Chemistry 3 answersAnswers as questions sent to our subscribers on WHATSAPP, if you are looking for 2020 WAEC/NECO Answers midnight before the exam, try contacting us.

(2)

A is HCl(aq)
B is ZnSO4(aq)
C is NaOH(aq)
D is Pb(NO3)2(aq)

Reasons:
Row 1: Hcl is used to distinguish pb²+ from Zn²+ ions since both behave alike with NaOH. It gives a white precipitate with pb²+ ions but not with Zn²+. The first row obviously shows that. Also there is usually no visible reaction when Hcl(A) and NaOH are mixed (c) 

Row 2: When A and B mixes, the products Zncl2 and H2SO4 are soluble hence no visible reaction is seen. Between B and D, a white precipitate is seen because PbSO4 is insoluble. Between B and C, Zn²+ ions yield a while gelatinous precipitate with NaOH

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

(3ai)
Sodium is stored under paraffin oil to prevent its oxidation by atmospheric gases

(3aii)
Desiccator lid must be greased with a thin layer of grease, to ensure an airtight seal.

(3aiii)
Because it is hygroscopic

(3bi) 
Add an acid eg HCL and this gives a neutralization reaction and the indicator react with the acids. Heat the solution to obtain the NaCL(s) 

(3bii)
Dip a rod in aqueous Ammonia and dip in a beaker containing the acids, only HCl produces dense white fumes of ammonium chloride

(3ci)
(i)Beaker
(ii)glass rod
(iii)volumetric flask

(3cii)
On adding the dilute HCL to Na3SO3 a gas with a chocking smell evolves



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

(1ai)
Tabulate:
Burette reading |1 |2 |3 |
final reading |24.80|34.50|20.65|
initial reading |0.00 |10.50|6.55 |
vol of acid used|24.80|24.00|24.10|

(1aii) Average volume of A used
Va = 24.00 + 24.10/2
= 24.05cm³


(1bi) 
Concentration of A in moldm-³
A contains 0.79g of KMnO4 per 250cm³ of solution. 

Hence since 250cm³ = 0.79g
100cm³ = Xg
X = 1000×0.79/250 = 3.16g

Hence Conc in g/dm³ of A = 3.16gdm-³
But molar conc = mass conc(gdm-³)/molar mass(g/mol)

Hence molar Conc.Of A= 3.16gdm-³ /molar mass ofA
But molar mass of A = KMnO4 
= 39 + 55 + 4(16) = 39 + 55 + 64
= 158g/mol

Hence Conc in moldm-³ of A = 3.16gdm-³/158g/mol
= 0.020moldm-³


(1bii)
To get concentration of B in moldm-³, we use the relation 
CAVA/CBVB = na /nb
Where,
CA = 0.02moldm-³
VA = 24.05cm-³
na = 1
CB = ?
VB = 25.0cm³
nb = 5
0.02×24.05/CB×25 = 1/5
CB = 5 × 0.02×24.05/25
= 2.405/25
=0.0962
CB = 0.096moldm-³

(1biii)
From the balanced equation 
5 moles of B contains 8 moles of SO²-
4
Hence 5 moles of B = 8moles of H+
0.09moles of B = X
5 × X = 0.09 × 8
X = 0.09 × 8/5 = 0.72/5 = 0.144
=0.144moles/dm-³
= 0.144mol/dm-³


(1biv)
m(SO4²-)=C×M×Vdm³
SO4²-
32+16×14=96gmol
= 0.144×96×250/1000
= 3.456g


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