(1)
Tabulate;
Burette readings;
|Rough |1st |2nd|3rd|
Final readings in cm³;
|22.00|20.10 |20.20|20.30|
Initial reading in cm³;
|0.00|0.00|0.00 |0.00|
Volume of A (acid) used cm³;
|22.00|20.10|20.20| 20.30|
The average titre
= 20.10+20.20+20.30/3
=60.60/3
= 20.20cm³
;Va 20.20cm³
(1bi)
Concentration of A in mol/dm³, Ca = ?
Concentration of A (g/dm³) = 4.40g/dm³.
.m.m of A, H²SO⁴ =(12)+(321)+(16*4)
=2+32+64
=98g/mol
Using concentration (mol/dm³) = Concentration (g/dm³)/m.m
Concentration (mol/dm³)
=4.40/98
=0.045mol/dm³
;: Concentration of A(mol/dm³) = 0.045mol/dm³
(1bii)
Concentration of B(mol/dm³) =???
Ca=0.045, Va=20.20, Vb=25.00, na=1, nb=1, cb=?
Using; CaVa/CbVb = na/nb
Cb=CaVanb/Vbna =0.04520.201/25.00*1
Cb=0.036mol/dm³
(1biii)
Conc. of B(g/dm³) that reacted =??
molar mass of B, Na₂CO₃ = (232)+(121)+(16*3) = 106g/mol
Conc. of B(mol/dm³) = 0.036mol/dm3
Conc. (mol/dm³) = Conc. (g/dm³)/molar mass
(0.036/1) = Conc. (g/dm³)/106
Conc. (g/dm³) = 0.036*106 = 3.82g/dm³
Conc. of B(g/dm³) = 3.82/dm³
(1biv)
mass of the impurity in 1.0dm³ of B=??
Conc. of impurity (g/dm³) = 7.50 – 3.82 = 3.68g/dm³
:. Mass of impurity= Conc. (g/dm³) x vol. (dm³)
= 3.68×1 = 3.68g
(1bv)
Vol. of CO₂ produced in 1.0dm³ of B = ??
Mole of B= Conc. x Vol. (dm³)
Mole of B = 0.036*1
Mole of B =0.036mol
By proportion,
1mol of Na₂.10₃ = 1mol of CO₂
0.036mol of Na₂.10₃ = xmol of CO₂
mole of CO₂ produced = 1*0.036 = 0.036mol
:. Volume = mole * 22.4 = 0.81dm³
(2)
[FILL THE TABLE WITH THE INFORMATION BELOW]
(2ai)
-TEST: C+5 cm³ of distilled water and shake thoroughly. Divide the solution into two portions
-OBSERVATION: It dissolves completely to give a colourless solution
-INFERENCE: C is a soluble salt
(2aii)
-TEST: To the first portion, add NaOH solution in drops.
-OBSERVATION: A white gelatinous precipitate is formed
-INFERENCE: Al³⁺ and Zn²⁺ present
(2aiii)
-TEST: Then add NaOH in excess
-OBSERVATION: The precipitate dissolve to give a colourless solution
-INFERENCE: Zn²⁺ present
(2aiv)
-TEST: To the second portion, add NH₃ solution in drops
-OBSERVATION: A white precipitate is formed
-INFERENCE: Al³⁺ and Zn²⁺ present.
(2av)
-TEST: then in excess
-OBSERVATION: The white precipitate dissolves in excess NH₃
-INFERENCE: Zn²⁺ is confirmed
(2bi)
-TEST: 2cm³ of solution D + soda lime, heat, test with litmus paper.
-OBSERVATION: Ammonia gas (NH₃) is evolved, turning red litmus paper blue.
-INFERENCE: Presence of NH₄⁺ ions.
(2bii)
-TEST: Solution D + NaOH solution + CuSO₄ solution
-OBSERVATION: A blue precipitate formed
-INFERENCE: Presence of peptide bonds (proteins or peptides).
(3ai)
I- to avoid the drop of the acid into the burette which will alter the reading
II- to avoid change in concentration and contamination
III- to avoid bubbles in the pipette
(3aii)
This can be done through crystallization. The solution is dissolved, heated to a high temperature and then allowed to cool. During cooling,the sodium chloride crystallized out.
(3bi)
(i)evaporating dish
(ii)Bunsen burner
(3bii)
Solubility
TO SUBSCRIBE FOR 2023 WAEC, NECO & NABTEB EXAM ANSWERS VIA WHATSAPP JOIN OUR GROUP
LIVE NECO QUESTIONS AND ANSWERS SUBSCRIBE NOW
We are the Top Examination Expo Website for WAEC, NECO and NABTEB Exams Runs and Verified Exam Answers provider in Nigeria, Ghana, Gambia, Sierra Leone, Liberia, Cameron, etc.
Do you know that if you are seeking admission into University, Polytechnic or College Of Education, you need to have A or B in your WAEC, NECO or NABTEB Result?
Because it gives you a better chance than those WITH C.
If you really want to have A or B in your results, make sure you subscribe to our exam runs, that is the only way to secure your admission.
Note: Our WAEC/NECO/NABTEB Exam Expo and RUNZ Answers come 3hours before the exam time (Verified Answers).