NECO Chemistry Practical 2020. Many NECO candidates find it difficult to answer the NECO Practical Chemistry Alternative A and B correctly due to not understanding the questions. I will show you the best way to answer the 2020 NECO Chemistry Practical questions and make a good grade in your Chemistry examination.
It is pertinent you understand the NECO Chemistry Practical making scheme before attempting any questions so as to present your answers in such a way that it will look attractive to the examiner to earn you big marks.
Bear it in mind that every single procedure is awarded marks. So, ensure you do not skip any step while reporting and presenting your practical results.
The Chemistry Practical question paper will consist of two sections and general Chemistry questions namely:
- Quantitative Analysis
- Qualitative Analysis
These involve titration and test of ions. You will be expected to carry out an experiment and report your inference and observation. What you will see will not be far from the NECO Chemistry Specimen given to your school.
Note: Your titration endpoint will be given to you by your Chemistry teacher. It is not the same for every school but within a range.
Do not use the endpoint here. It may vary from your center endpoint but you can follow the calculation procedures.
NECO Practical Chemistry Questions and Answers (Expo)
Note: The 2020 NECO Chemistry Practical answers will be posted here today Wednesday, 21st October during the NECO Practical Chemistry exam. Keep checking and reloading this page to know when the answers are posted. Do not forget to reload this page in order to see the answers.
NECO GCE Chemistry Practical Answers
Note: The answers below are not today’s answers.
2020 NECO Practical Chemistry answers loading…
Indicator used is methyl orange.
Volume of pipette used is 25cm³
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
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
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
Molar mass of B in glmol
Concentration of B(moldm-³)= conc gldm³/molar mass
0.092 * molar mass =10.60gldm³
Molar mass =10.60gldm³/0.092
Relative atomic mass of X in X²CO³
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
TEST: To the first portion add NaOH solution in drys
OBSERVATION: Dark green gelatinous was formed
INFERENCE: Fe²+ present
OBSERVATION; Dark green gelatinous was formed
INFERENCE; Fe²^+ present
TEST; then in excess
OBSERVATION; precipitate is insoluble
INFERENCE; Fe²^+ present
TEST; To the second portion add K3Fe(CN)6 solution
OBSERVATION; A dark blue precipitate formed
INFERENCE; Fe²^+ confirmed
TEST: then in excess
OBSERVATION: Precipitate is insoluble
INFERENCE: Fe²+ present
TEST: To the second portion add K3Fe(CN)6 solution
OBSERVATION: A dark blue precipitate formed
INFERENCE: Fe²+ confirmed
TEST: To the third portion add AgNO³ solution
OBSERVATION: White Precipitate formed
INFERENCE: SO4²- , CL-, CO²- present
TEST: To the results obtained in add dilute HNO3 in drops , then excess
OBSERVATION: White Precipitate is insoluble
INFERENCE: CL- present
TEST: To the results obtained in add NH3 solution
OBSERVATION: White Precipitate dissolves
INFERENCE: CL- Confirmed
TEST; To the third portion add AgNO³ solution
OBSERVATION; White Precipitate formed
INFERENCE; SO4²^- , CL^- CO²^- present
TEST; To the results obtained in add dilute HNO3 in drops , then excess
OBSERVATION; White Precipitate is insoluble
INFERENCE; CL^- present
TEST; To the results obtained in add NH3 solution
OBSERVATION; White Precipitate dissolves
INFERENCE; CL^- Confirmed
I. Methyl Orange
II. Methyl orange
A white Precipitate is observed. Effervescence occurs a gas is evolved which change lime water milky.
More Answers Loading… Keep checking and reloading this page
NECO Chemistry Specimen 2020
CHEMISTRY PRACTICAL SPECIMEN:
(a) One burette (50cm³)
(b) One pippete (20cm³/25cm³). However, all
candidates in a center must use pipettes of the
(c) The usual apparatus and reagents for
qualitative work including:
(i) Dilute sodium hydroxide solution
(ii) Dilute ammonia solution
(iii) Dilute hydrochloric acid
(iv) Barium chloride solution
(v) Distilled water
(vi) Red and Blue litmus paper
(d) Methyl orange
(e) One boiling tube
(f) Five test tubes
(g) Filtration apparatus
(h) Source of heat
Each candidate should be supplied with the
following: Labelled An, Bn, Cn, where ‘n’ is the
candidate’s serial number.
(a) 150cm³ of chloride acid solution in a bottle
labelled “An”. The acid solution which should
be the same for all candidates will contain
3.4cm³ of the concentrated hydrochloric acid
per dm³ solution.
(b) 150cm³ of sodium hydroxide solution in a
bottle labelled “Bn”. The solution which should
be the same for all candidates will contain
4.0g of sodium hydroxide per dm³ of solution.
(c) One spatulaful of copper (ii)
tetraoxosulphate (vi) salt in a specimen bottle
The questions below are strictly for practice not the 2020 Chemistry expo.
All your burette readings (initials and final) as well as the size of your pipette must be recorded but no account of experimented procedure is required. All calculations must be done in your booklet.
- A is O.200 moldm3 of HCI. C is a solution containing 14/3g of Na2CO3. x H2O in 500 cm3 of solution. (a) Put A into the burette and titrate it against 20.0 cm3 or 25.0 cm3 portions of C using methyl orange as indicator. Repeat the titration to obtain Consistent titre values. Tabulate your results and calculate the average volume of A used. The equation for the reaction is: Na2CO3 x H2O + 2HCL(aq) → 2NaCI(aq) + CO2(g) + (x+1) H2O(I).
(b) From your results and the information provided. Calculate the: (i) concentration of C in moldm-3; (ii) concentration of C in gdm-3; (iii) molar mass of Na2CO3, xH2O; (iv) the value of x in Na2CO3 xH2O. [H = 1.0; C=12.0; O = 16.0; Na =23.0]
Credit will be given for strict adherence to the instruction, for observations precisely recorded and for accurate inferences. All tests, observations and influences must be clearly entered in the booklet in ink at the same time they are made.
2. F is a mixture to two inorganic salts. Carry out the following exercise on F. record your observation and identify any gas(s) evolved. State the conclusions you draw from the result of each test. (a) Put all of F in a beaker and add about 10cm3 of distilled water. Stir well and filter. Keep the filtrate and the residue. (b)(i) To about 2cm3 of the filtrate, add NaOH(aq) in drops and then in excess (ii) To another 2cm3 portion of the solution, add a few drops of NHO3(aq)followed by few drops of AgNO3(aq). (d)(i) Put all the residue into a clean test-tube and add NHO3(aq) followed by few solution from 2(d)(i) add NaOH(aq) in drops and then in excess.
3. State what would be observed if the following reactions are carried out in the laboratory: (i) methyl orange is dropped into a solution of lime juice (ii) hydrogen sulphide gas is bubbled through iron (III) chloride solution; (iii) sulphur (IV) oxide gas is bubbled into acidified solution of KMnO4; (iv) ethanoic acid is added to a solution of K2CO3.
CHEMISTRY PRACTICALS ANSWERS
- Indicator = Methyl Orange
Volume of the base used = 25.00cm3
|Titration||Rough Titre||1st Titre||2nd Titre||3rd Titre|
|Final Burette readings cm3||24.70||24.80||24.70||24.90|
|Initial Burette reading cm3||0.00||0.00||0.00||0.00|
|Volume of acid used cm3||24.70||24.80||24.90||24.90|
Average Titre = 1st + 2nd + 3rd/3
= 24.80 + 24.70 + 24.90/3
Alternatively 2 concordant titres can be used to calculate average titre.
Equation of the reaction: Na2CO3 XH2O + 2HCI(aq) → 2NacI(aq) + CO2(aq) + (x+1)H2O(I)
CAVA/CBVB = nA/nB
CA = Molar concentration of HCI(aq) in moldm3
VA = Volume of acid used in cm3 = 24.80cm3
CB = Molar concentration of Na2Cu3 xH2O in moldm3
nA = 2
nB = 1
(b)(i) concentration of C in moldm-3 = ? From the equation of reaction:
CAVA/CBVB = nA/nB
CA = 0.200 moldm3 VA = 24.80cm3
CB = ? VB = 25.00cm3
Substitution of known values
0.200 x 24.80/ CB x 25.00 = 2/1
CB = 1×0.200×24.80/2×25.00
CB = 0.0992 moldm-3
(ii) Concentration of C in g dm-3 = ?
500cm2 → 14.3 g
1000cm3 →14.3/ 500 x 10002 g
= 28.6 g dm-3
(iii) Molar mass of Na2CO3 xH2O
Molar conc in moldm-3 = conc in g dm-3 / molar mass
Therefore: Molar mass g mol-1 = conc in g dm-3 / molar conc in moldm-3
= 28.6 g dm-3/0.0992 moldm-3
Approximately 288 g mol-1
(iv) Value of x in Na2CO3 xH2O?[H = 1.0, C = 12.0, O = 16.0, Na = 23.0]
Na2CO3 xH2O = 288
2(23) + 12 + 3 (16) + x (2(I) + 16) = 288
46 + 12 + 48 + 18x = 288
106 + 18x = 288
18x = 288
18x = 182
x = 182/18
x = 10.11
x approximately 10
|(a)||F + distil water + stir + filter||Effervescence (bubbling) occurs, colourless and odourless gas evolved after filtration pale blue/colourless filtrate and green residue|
|(b)(i)||Filtrate + NaOH(aq) in drops then in excess||White precipitate formed Precipitate insoluble in excess NaOH(aq)||Ca2+/Pb2+ Ca2+ present|
|(b)(ii)||Filtrate + NH3(aq) in drops then in excess||No visible reaction No visible reaction OR||Ca2+ present Ca2+ confirmed|
|(b)(i)||Filtrate + NaOH(aq) in drops then in excess||Blue gelatinous precipitate formed Precipitate insoluble in excess||Cu2+ is present|
|(b)(ii)||Filtrate + NH3(aq) in drops then in excess||Pale (light) blue gelatinous precipitate formed Precipitate dissolves to form a deep blue solution||Cu2+ Cu2+ is confirmed|
|(c)||Filtrate + NHO3(aq) + AgNO3(aq)||No visible reaction No gas involved White precipitate formed||CuI- is present|
|(d)(i)||Residue + HNO3(aq)||Efferveness occurs, colourless and odourless gas evolves that turn blue litmus paper to red and turns lime water milky||CO2(g) CO32- present|
|(d)(ii)||Solution from 2(d)(i) + NaOH(aq) in drops then in excess||Blue precipitate formed Precipitate insoluble in excess NaOH(aq)||Cu2+ Cu2+ present|
3 (a)(i) Solution of lime juice turns to pink or red. (ii) The FeCI3 solution changes colour from brown to greenn and a yellow deposit. (iii) The purple colour of KmnO4 solution turns colourless or decolourised. (iv) The solution of K2CO3 reacts with the addition of ethanoic acid to evolve a colourless, odourless gas with bubbling of effervescence.
If you have any question about the 2020 NECO Chemistry Practical questions and answers, kindly let us know in the comment box.
Last Updated on February 10, 2021 by Admin