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2022/2023 NECO Chemistry Practical Specimen Questions For all Candidates

2020 Neco Chemistry practical specimen questions
2020 Neco Chemistry practical specimen questions
Written by Nkedugist

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  • 2020/2021 NECO Chemistry practical specimen questions
  • NECO Chemistry practical specimen questions
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  •  Chemistry practical specimen questions for GCE
  • 2022 NECO Chemistry practical specimen questions

2021 Neco Chemistry Practical Specimen Questions

TITRATION
(i) Dilute sodium hydroxide solution
(ii) Dilute ammonia solution
(iii) Dilute hydrochloric acid
(iv) Dilute trinitrate(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

Answers To Neco Practical Chemistry 2021

(1a)

Tabulate

Volume of pipette used=25cm3.

Indicator used= Methyl orange

Under Titration, Rough, 1st, 2nd, 3rd,

Final burette reading; |25.00| |23.40| |23.50| 23.60|

Initial burette reading; |0.00| |0.00| |0.00| |0.00|

Volume of A used; |25.00| |23.40| |23.50| |23.60|

(1ai)

Average volume of Acid used=23.40+23.50+23.60/3 =70.5/3=23.50cm3

(1aii)

This is because it is a reaction between a weak acid and strong base

(1aiii)

A funnel should be used while transferring the acid into the burette.

(1bi)

Conc of A in mol|dm³ =?

Conc of A in g|dm³ = 12.75g|dm³

Molar mass of A, NaHSO⁴ = 23+1+32+64

=120g|dm³

From.

Conc(mol/dm³) = conc(g/dm³)/molar mass

Conc =12.75/120

Conc =0.106mol/dm³

(1bii)

Con of B in mol/dm³ =?

Ca =0.106, Cb =? Va =23.50cm³, Vb =25.00cm³

na=1 , nb=1

From; CaVa/CbVb=na/nb

Cb=CaVanb/Vbna

Cb=0.10623501/25.001

Cb=2.5/25

Cb=0.1mol/dm³

(1biii)

Conc of B in g/dm³=?

Molar mass of B , NaOH=23+16+1=40g/mol

From; conc(g/dm³)=conc(mol/dm³)molar mass

=0.140

=4.0g/dm³

(1biv)

Mass of salt formed =?

Recall mole of NaHSO⁴=?

From; mole =con(mol/dm³)*vol(cm³)/1000

Mole=0.106*23.50/1000

Mole=0.0025mol of NaHSO⁴

By proportion 1mol of NaHSO⁴ produce 1mol of Na²SO⁴ 0.0025mol of NaHSO⁴ will produce X mol of Na²SO⁴ X=0.00251=0.0025mol of NaSO⁴

Molar mass of Na²SO⁴=(232) + 32 + (16*4)

=46+32+64

=142g/mol

Recall; mole =mole/molar mass

Mass =mole * molar mass

Mass =0.0025*142

Mass =0.36g of Na²SO⁴

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