2021 NECO CHEMISTRY PRACTICAL Questions & Answers: 2020 NECO CHEMISTRY PRACTICAL ANSWERS Night before Exam

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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.

Refresh always... 

(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

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