# Waec Gce 2018 CHEMISTRY Practical Answers Now Available Here

(1ai)
Tabulate:
Burette reading |1 |2 |3 |
vol of acid used|24.80|24.00|24.10|

(1aii) Average volume of A used
Va = 24.00 + 24.10/2
= 24.05cm^3

(1bi)
Concentration of A in moldm^-3
A contains 0.79g of KMnO4 per 250cmΒ³ of solution.

Hence since 250cmΒ³ = 0.79g
100cmΒ³ = Xg
X = 1000Γ0.79/250 = 3.16g

Hence Conc in g/dmΒ³ of A = 3.16gdm^-Β³
But molar conc = mass conc(gdm^-Β³)/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^-Β³ of A = 3.16gdm^-3/158g/mol
= 0.020moldm^-3

(1bii)
To get concentration of B in moldm^-Β³, we use the relation
CAVA/CBVB = na /nb
Where,
CA = 0.02moldm^-Β³
VA = 24.05cm^-Β³
na = 1
CB = ?
VB = 25.0cmΒ³
nb = 5
0.02Γ24.05/CBΓ25 = 1/5
CB = 5 Γ 0.02Γ24.05/25
= 2.405/25
=0.0962
CB = 0.096moldm^-Β³

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

Reasons

ROW 1
it is used to distinguish Pb from Zn since both behave alike with NAPth
It gives a white precipitate with Pb^2+ ions but not with Zn^2+. The first
row obviously shows that. Also there actually no visible reaction whne Itcl (A) and NaOlt are mixed(C)

Row 2
when A and B mirces, the products Zncl2 and
H2SO4 one soluble hence no visible reaction
is scene Between B and D, a white precipitate is seen because PbaS04 insoluble
Between B and C, Zn^2+ On yield a white gelatinous precipate with NaOlta

(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