2025 WAEC CHEMISTRY: 2025 WAEC MAY/JUNE Chemistry Practical Answers (9305)
2025 WAEC MAY/JUNE Chemistry Practical Answers Password/Pin/Code: 9305.
Tuesday 18th August 2020
Chemistry 3(Practical) (Alternative A)(1st set) – 9:30am – 11:30pm (2hrs)
Chemistry 3 (practical)(Alternative A) (2nd set) – 12:00pm – 2:00pm
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Refresh in every 5 minutes starting from 1 hour to the exam.
For No. 1... PLEASE REMEMBER TO USE YOUR'S SCHOOL'S AVERAGE TITRE VALUE(AVERAGE VOLUME OF ACID). WE USED 15.26cm3, TRY TO KNOW YOUR SCHOOL'S AVERAGE TITRE VALUE. ASK YOUR CHEMISTRY TEACHER. THEN, ANYWHERE YOU SEE 15.26cm3 IN MY CALCULATION, PUT YOUR SCHOOL'S OWN AND RE-CALCULATE. THIS IS VERY IMPORTANT!!!
Welcome to official 2025 Chemistry WAEC answer page. We provide 2025 Chemistry WAEC Questions and Answers on Essay, Theory, OBJ midnight before the exam, this is verified & correct WAEC Chems Expo. WAEC Chemistry Questions and Answers 2025. WAEC Chems Expo for Theory & Objective (OBJ) PDF: verified & correct expo Solved Solutions, 2025 WAEC MAY/JUNE Chemistry Practical Answers. 2025 WAEC EXAM Chemistry Questions and Answers
Pls, use your school endpoints.
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(1a)
In a tabular form
Burette Reading|1st reading|2nd reading|3rd reading
Final |15.25|0.00|45.79|
Initial|0.00||15.25|30.53|
Volume of acid used |15.25|15.28|15.26
Average volume of acid used = 15.25 + 15.26/2
= 15.255cm³
=15.26cm³
(1bi)
Given: Mass of conc of A = 5g/500cm³ = 5g/0.5dm³
Ca = 10g/dm³
A is HNO3
Therefore; Molar mass = 1+14+(16×3)
= 15+48
=63g/mol
Molarity of A = gram conc/molar mass
Ca = 10/63 = 0.1587mol/dm³
(1bii)
Using CaVa/CbVb = nA/nB
With reacting equation:
HNO3 +NaOH-->NaNO3 + H2O
nA = 1, nB = 1
0.1587×15.26/Cb×25.00 = 1/1
25Cb = 0.1587×15.26
CB = 0.1587×15.26/25
CB = 0.09687mol/dm³
(1biii)
B is NaOH
Molar mass = 23+16+1
=40g/mol
Conc of B in g/dm³ = molarity×molar mass
=0.09687×40
=3.8748g/dm³
(1biv)
No of moles present in 250cm³ of NaOH is
= molar conc. × volume
= 0.09687 × 250/1000
= 0.0242 moles
Mole ratio of NaOH and NaNO3 is 1:1
No of moles of NaNO3 which reacted is 0.0242
Mass of NaNO3 formed = molar mass × no of moles
= 85 × 0.0242
=2.057grams
(2a)
TEST: C+burning splint
OBSERVATION: Sample C burst into flame. It burns with non-smoky blue flame without soot. Colourless gas that turns wet blue litmus paper faint red and turns like water milky is present.
INFERENCE: C is volatile and flammable. The gas is CO2 from combustion of a saturated organic compound.
(2bi)
TEST: C + distilled water + shake
OBSERVATION: Clear or colourless solution is observed
INFERENCE: C is miscible with water
(2bii)
TEST: C + Acidified K2Cr207
OBSERVATION: Orange colour of K2Cr207 solution turns pale green and eventually pale blue on cooling
INFERENCE: C is a reducing agent
(2C)
TEST: D + C + 10% NaOH + Shake
OBSERVATION: D dissolves slowly in C and produces a reddish-brown solution. The reddish-brown solution turns yellow precipitate. The precipitate has an antiseptic odour
INFERENCE: D is soluble in organic solvents
(2d)
The compound belongs to the class of secondary alkanol
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(3ai)
Zinc nitrate
(3aii)
2 Zn(NO 3 )2 ----->2 ZnO + 4 NO 2 + O 2
(3aiii)
It turns white when cold from its yellow colour when it was hot.
(3b)
Pipette/measure 50.0cm3 of the stock solution into a 250
cm3 volumetric flask (containing some distilled water). Shake/swirl and add more distilled water until the mark is reached.
(3c)
Al2(SO4)3 - turns blue litmus red
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