NECO Chemistry Practical Questions and Answers 2022/2023 Examination (Complete Solution)

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NECO Chemistry practical

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NECO Chemistry Practical Specimen 2022

TITRATION

(i) Dilute sodium hydroxide solution

(ii) Dilute ammonia solution

(iii) Dilute hydrochloric acid

(iv) Dilute trioxonitrate(V)acid

(v) Distilled water

(vi) Red and blue litmus paper

(vii) Barium chloride solution

(viii) Phenolphthalein solution

(ix) Methyl orange

(x) One boiling tube

(xi) Five test tubes

(xii) Source of heat

(xiii) Wash bottle containing distilled water

(xiv) Filtration apparatu

NECO Practical Chemistry Questions and Correct Answers – Expo

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NECO GCE Practical Answers

Note: The answers below are not today’s answers but 2021 Answers.

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(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

(1iv)
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=276

(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

OBSERVATION; Dark green gelatinous was formed

INFERENCE; Fe²^+ present

(2iii)
TEST; then in excess

OBSERVATION; precipitate is insoluble

INFERENCE; Fe²^+ present

(2iii)
TEST; To the second portion add K3Fe(CN)6 solution

OBSERVATION; A dark blue precipitate formed

INFERENCE; Fe²^+ confirmed

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

(2iv)
TEST; To the third portion add AgNO³ solution

OBSERVATION; White Precipitate formed

INFERENCE; SO4²^- , CL^- CO²^- present

(2v)
TEST; To the results obtained in add dilute HNO3 in drops , then excess

OBSERVATION; White Precipitate is insoluble

INFERENCE; CL^- present

(2vi)
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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NECO Chemistry Specimen 2022

CHEMISTRY PRACTICAL SPECIMEN:

Requirements:
(a) One burette (50cm³)
(b) One pippete (20cm³/25cm³). However, all
candidates in a center must use pipettes of the
same volume.
(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
(vii) Phenolphtalein
(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
labelled “Cn”.

The questions below are strictly for practice not the 2021 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

TitrationRough Titre1st Titre2nd Titre3rd TitreFinal Burette readings cm324.7024.8024.7024.90Initial Burette reading cm30.000.000.000.00Volume of acid used cm324.7024.8024.9024.90

Average Titre              =          1st + 2nd + 3rd/3

=          24.80 + 24.70 + 24.90/3

                                    =          24.80cm3

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

C = 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

                                                 = 288.3065

                                    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

3 (a)(i) Solution of lime juice turns to pink or red. (ii) The FeCI3 solution changes colour from brown to green and a yellow deposit. (iii) The purple colour of the 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.

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