Here is the waec gce chemistry practical answers 2021. PLS share and allow notifications so you can get notified when we update this post.
1)
TABULATE
BURETTE READING(cm): 1| 2| 3|
FINAL READING: 21.50| 22.40| 32.20|
INITIAL READING: 2.50| 4.00| 13.60|
VOLUME OF A USED: 19.00| 18.40| 18.60|
Average of A used = 18.40+18.60/2 = 18.50cm³
(1bi)
Using CAVA/CBVB = nA/nB
0.086*18.50/CB*20.00 = 2/1
CB = 0.086*18.50/2*20.00 = 0.039775mol/dm³
(bii)
No. Of moles of I2 = Volume * molarity
= 1*0.06
= 0.06moles
(biii)
Gram Conc. of hydrocarbon = 4.8/1 = 4.8g/dm³
Molarity = Gram Conc./Molar mass = 4.8/80 = 0.06mol/dm³
(3a)
(i) Fehling’s solution
(ii) Ferric chloride
(iii) Million’s reagent
(iv) Ammoniacal solution of copper(i)chloride
(3b
This can be done by pouring hot water on the neck of the bottle which loosens the stopper by weakening the inter-molecular forces in between.
(3ci)
Equivalence point is the point in a titration where the amount of titrants added is enough to completely neutralize the analyst solution.
(3cii)
(i) Spatula; It is used for scraping, transferring, or applying powders and paste-like chemicals or treatments
(ii) Crucible; It is used to burn, melt or mix solid chemical compounds over a burner.
(iii) White tile; it is placed underneath a conical flask to aid with the ease of spotting the end point colour change.
(3ciii)
(i) This is because distilled water change the concentration of the initial solution
(ii) In order to measure the amount of solution added in or drained out
(3di)
(i) Ethene
(ii) Butyne
(3dii)
N₂, CO₂, NH₃
Molar Conc. = 0.06mol/dm
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