WAEC 2018 || Chemistry Practical Answers


Volume of pipette used=25.0cm^3
Reading: Rough|1st|2nd|3rd
Final reading: 16.00|15.40|17.20|15.30
Initial reading: 0.00|0.00 |2.00 |0.00
Vol of acid used: 16.00|15.40|15.20|15.30

Average volume of A used = (VA1 + VA2 + VA3) cm³
VA = (15.40 + 15.20 + 15.30)/3cm³
VA = 45.90/3 = 15.30cm³.

B contains 4.80g/250cm³
Hence x8 = 1000cm³
X = 4.80 × 1000/250 = 4.80 × 4 = 19.2gdm³

Hence molar concentration of B in moldm3 
Mass Conc(8dm³) = molar Conc × molar mass 
Molar Conc of B = mass conc/molar mass = 19.2gdm³/127gmol
Cb = 0.151moldm3

Concentration of A in moldm-³, CA is given by CAVA/CBVB = na/nb
Where CA = ?
VA = 15.30cm³, 
na = 1
CB = 1.51moldm-³, VB = 25.0cm³, nb = 5
CA × 15.30/0.151 × 250= 1/5
Therefore CA = 0.151×25×1/15.30×5 = 3.775/76.50 = 0.049moldm-³
Hence Conc. Of A in moldm-³ = 0.049moldm-³

No of moles of Fe²+ in the volume of B pipetted 
n= cv/1000 =
0.151×25.0/1000 =
= 0.003775moles of Fe²+

More Coming Soon.

Keep Refreshing Steady.
TEST: C + about 5cm3 of distilled water
OBSERVATION: C dissolves to form a clear solution; 
Turns blue litmus paper to pink
INFERENCE: C is a soluble salt;
C is fairly acidic

TEST: 1st portion of C + NaOH(aq) in drops;
then excess
OBSERVATION: pale-blue precipitate forms in drops of NaOH(aq) which persists in excess of NaoH(aq)
INFERENCE: cu²+ present. 

TEST: 2nd portion of C + NH3(aq) in drops and then in excess 
OBSERVATION: pale blue precipitate forms in drops of NH3(aq).
Ppt dissolves to form a deep blue coloration. 
INFERENCE: Cu²+ present. 
Cu²+ confirmed

TEST: 3rd portion of c + AgNO3(aq) + HCL(aq)
OBSERVATION: a creamy white ppt forms. 
INFERENCE: Halogen/Halides present. 

TEST: D + about 5cm³ of distilled H2O
OBSERVATION: D dissolves to form a clear solution. 
The test tube feels warmth 
INFERENCE: D is a soluble salt. 

TEST: 2cm³ of D + HCL(aq)
OBSERVATION: effervescence occurs, an odourless and colourless gas evolved. Precipitate dissolves in dil HCL(aq)
INFERENCE: CO32-, so32-present 
CO32- present

A white precipitate of barium sulphate will be formed by the instant reaction between sulphuric acid and barium chloride. 
H2SO4(aq) + BaCl2(aq) -> BASO4 + 2HCl

Iron II sulphide reacts with hydrochloric acid, releasing the malodorous and very toxic gas, hydrogen sulphide 
FeS(aq) + 2HCl -> Fecl2(aq) + H2S(g)

When solid iron fillings are added to dilute aqueous hydrochloric acid, Iron(II)Chloride OR ferrous chloride is formed, with the liberation of hydrogen gas. 
Fe(s) + 2HCL(aq) -> Fecl2 + H2(g)

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