2024 NECO CHEMISTRY: 2024 NECO CHEMISTRY PRACTICAL ANSWERS (4703)
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VERY IMPORTANT INFORMATION FOR NUMBER 1!!!
PLEASE REMEMBER TO USE YOUR'S SCHOOL'S AVERAGE TITRE VALUE(AVERAGE VOLUME OF ACID). WE USED 23.00cm3, TRY TO KNOW YOUR SCHOOL'S AVERAGE TITRE VALUE. ASK YOUR CHEMISTRY TEACHER. THEN, ANYWHERE YOU SEE 23.00cm3 IN MY CALCULATION, PUT YOUR SCHOOL'S OWN AND RE-CALCULATE. THIS IS VERY IMPORTANT!!!
1biv corrected
1bv Added.
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Welcome to official 2024 Chemistry NECO answer page. We provide 2024 Chemistry NECO Questions and Answers on Essay, Theory, OBJ midnight before the exam, this is verified & correct NECO Chems Expo. NECO Chemistry Questions and Answers 2024. NECO Chems Expo for Theory & Objective (OBJ) PDF: verified & correct expo Solved Solutions, 2024 NECO CHEMISTRY PRACTICAL ANSWERS. 2024 NECO EXAM Chemistry Questions and Answers
(1)
Indicator used is methyl orange.
Volume of pipette used is 25cm³
Tabulate.
Titration; Rough|1st|2nd|3rd|
Final burette reading cm³|23.50|23.00|23.00|33.00|
Initial burette reading cm³| 00|0.00|0.00|10.00|
Volume of acid used(cm³) |23.50|23.00|23.00|23.00|
Average volume of A used = 23.00+23.00+23.00/3 =23.00cm³
Equation for the reaction
X²Co³(aq)+2HCl(aq)---->2XCl(aq)+H²O(s)+Cl(s)
(1i)
Concentration of A in moldm³ =concentration in gldm³.
3.6g of A =500cm²
Xg of A will be = 10000³
Xg=3.65*1000/500 =36.5/5 =7.3gldm³
Molar Mass of A HCl=1+35.5=36.5
Concentration of A =7.3gldm³/36.5gldm³
=0.20moldm³
(1ii)
Concentration of B in moldm³
concentration of A in moldm³ CA=0.20
Volume of used VA=23.00
Concentration of B in moldm³ CB=?
Volume of B used VB= 25.00
CAVA/CBVB = n(A)/n(B) =2/1
0.20*23.00/CB*25.00=2/1
2*CB*25.00=0.20*23.00*1
CB=0.20*23.00/50
Cb=46/50=0.092moldm-³
(1iii)
Molar mass of B in glmol
Concentration of B(moldm-³)= conc gldm³/molar mass
0.092=10.60gldm³/molar mass
0.092 * molar mass =10.60gldm³
Molar mass =10.60gldm³/0.092
=115.2glmd
(1biv)
Relative atomic mass of X in X²CO³
X²+12+3(16)=115.2
2x+12+48=115.2
2x+60=115.2
2x=115.2÷60
2x=55.2
x=55.2/2=27.6
(1bv)
No of moles of X2CO3 = molarity × volume
= 0.092 × 25
= 2.3moles
Mole ratio of X2CO3 to CO2 is 1:1
Therefore, No of moles of CO2 released = 2.3moles
Volume of CO2 released = 2.3 × 22.4
=51.52dm³
===============
(2ai)
Tabulate:
TEST: C + 5cm³ of distilled water and shake thoroughly. Divide the solutions into three portions.
OBSERVATION: A pale green solution results
INFERENCE: Salt is soluble
(2aii)
TEST: To the first portion add NaOH solution in drys
OBSERVATION: Dark green gelatinous was formed
INFERENCE: Fe²+ present
(2aiii)
TEST: then in excess
OBSERVATION: Precipitate is insoluble
INFERENCE: Fe²+ present
(2aiv)
TEST: To the second portion add K3Fe(CN)6 solution
OBSERVATION: A dark blue precipitate formed
INFERENCE: Fe²+ confirmed
(2bi)
TEST: To the third portion add AgNO³ solution
OBSERVATION: White Precipitate formed
INFERENCE: SO4²- , CL- CO²- present
(2bii)
TEST: To the results obtained in add dilute HNO3 in drops , then excess
OBSERVATION: White Precipitate is insoluble
INFERENCE: CL- present
(2biii)
TEST: To the results obtained in add NH3 solution
OBSERVATION: White Precipitate dissolves
INFERENCE: CL- Confirmed
=======================
(3ai)
Filtration
(3aii)
Diagram
(3bi)
I. Methyl Orange
II. Methyl orange
III. Phenolphthalein
(3bii)
A white Precipitate is observed. Effervescence occurs a gas is evolved which change lime water milky.
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