2018 WAEC Chemistry Practical Questions/Answers

         

Free 2018 WAEC Chemistry Practical Questions/Answers Posted Here.

2018 WAEC Chemistry Practical Questions/Answers

2018 WAEC Chemistry Practical Questions/Answers. 08176387240

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WAEC 2018 CHEMISTRY PRACTICAL 3 (Practical) (Alternative A) 

(3a) 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

(3b) Iron II sulphide reacts with hydrochloric acid, releasing the malodorous and very toxic gas, hydrogen sulphide
FeS(aq) + 2HCl -> Fecl2(aq) + H2S(g)

(3c) 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)


 

2a) -Observation-
part of the C dissolves,colourless filtrate salt

-Inference-
C is a mixture of soluble salt

2bi) -observation-
white precipitate formed

-Inference-
Cl is present

2bii) -Observation-
precipitate dissolves

-Inference-
Cl confirmed

2biii) -Observation-
colourless gas evolved with characteristic choking smell, turns litimus paper blue

-Inference-
Gas is NH3

2c) -Observation-
Residue dissolves liberating a colourless gas which turns lime water milky

-Inference-
gas is CO2
CO^2-3 present


 

(1a) Volume of pipette used=25.0cm^3
TABULATE
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

(1bi) 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

(1bii) 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

(1biii) No of moles of Fe²+ in the volume of B pipetted
n= cv/1000 =
0.151×25.0/1000 =
3.775/1000
= 0.003775moles of Fe²+


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