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## Neco 2023 Chemistry Practical Answers

Posted By on July 9.
Found In: Neco, Past Questions & Solution
Source:

### Neco 2023 Chemistry Practical Answers

(1a)
[TABULATE]

Burette Reading (cm³) | Rough | Titration I | Titration II
Final Reading (cm³) | 21.50 | 43.00 | 22.00
Initial Reading (cm³) | 0.00 | 21.50 | 00.00
Volume of A used (cm³) | 21.50 | 21.50 | 22.00

(1ai)
Average Volume of acid used = 21.50 + 21.50/2 = 21.50cm³

(1aii)
Solution B[Sodium trioxocarbonate (iv)] feels soapy

(1bi)
Conc. B in mol/dm³ =?
Ca= 0.04, Va= 21.50cm³, na= 1
Cb= ? , Vb= 25.00cm³ nb= 1

CaVa/CbVb = na/nb = 0.04×21.50/Cb×25.00 = 1/1
Cb= 0.04×21.50/25.00 = 0.86/25.00
Cb= 0.0344mol/dm³

(1bii)
Conc. of B in g/dm³ = Conc. of B in mol/dm³ × molar mass of B

Molar mass of B(NaCO₃)= (23×2) + 12 + (16×3) = 106

:. Conc. of B in g/dm³ = 0.0344×106 = 3.6g/dm³

(1biii)
1 mole of Na₂Co₃ = 1mole of Na₂So₄

:. 0.0344 mole of Na₂Co₃ = 0.0344 mole of Na₂So₄
№ of mole= mass/ molar mass

Molar mass of Na₂So₄= (23×2)+32+(16×4)=142
:. 0.0344 = mass/142
Mass=142×0.0344=4.9g

(1biv)
1 mole of Na₂Co₃= 1 mole of CO₂

0.0344 mole of Na₂Co₃ = 0.0344 mole of CO₂
:. № of mole = volume/ molar volume = 0.0344 = volume of CO₂/22400cm³
=771cm³
==================================

(2ai)
[TABULATE]

=TEST=
C + heat

=OBSERVATION=
Formation of yellow flame

=INFERENCE=
Na⁺ present

(2aii)
=TEST=
C + distilled water + shake

=OBSERVATION=

=INFERENCE=
C is a soluble in water

(2aiii)
=TEST=
Solution C + litmus paper

=OBSERVATION=
It changes moist red litmus to blue and no effect blue

=INFERENCE=
Solution C is alkaline

(2bi)
=TEST=
1st portion of solution C + barium chloride

=OBSERVATION=
White precipitate formed

=INFERENCE=
CO₃²⁻ , SO₄²⁻ , S²⁻ or SO₃²⁻

(2bii)
=TEST=
Add dilute HCl to product in (bi)

=OBSERVATION=
White precipitate dissolved

=INFERENCE=
CO₃ , SO₃²⁻ or S₃²⁻

(2biii)
=TEST=
To the second portion add phenolphthalein

=OBSERVATION=
Purple colour observed

=INFERENCE=
C is alkaline
=======================
(3ai) Sulphur (iv) oxide and hydrogen sulphide.

(3aii) Liberation of hydrogen gas with pop sound and formation of green iron (ii) chloride solution

(3aiii) Spatula, beaker, weighing balance, standard volumetric flask

(3bi) Solution turns red

(3bii) Turns from brown to green

(3biii) A colourless, odourless, acidic gas liberated which turns blue litmus paper red and turn lime water milky.

 COMMENT SECTION
 COMMENT SECTION

## Neco 2023 Chemistry Practical Answers

Posted By on July 9.
Found In: Neco, Past Questions & Solution
Source:

### Neco 2023 Chemistry Practical Answers

(1a)
[TABULATE]

Burette Reading (cm³) | Rough | Titration I | Titration II
Final Reading (cm³) | 21.50 | 43.00 | 22.00
Initial Reading (cm³) | 0.00 | 21.50 | 00.00
Volume of A used (cm³) | 21.50 | 21.50 | 22.00

(1ai)
Average Volume of acid used = 21.50 + 21.50/2 = 21.50cm³

(1aii)
Solution B[Sodium trioxocarbonate (iv)] feels soapy

(1bi)
Conc. B in mol/dm³ =?
Ca= 0.04, Va= 21.50cm³, na= 1
Cb= ? , Vb= 25.00cm³ nb= 1

CaVa/CbVb = na/nb = 0.04×21.50/Cb×25.00 = 1/1
Cb= 0.04×21.50/25.00 = 0.86/25.00
Cb= 0.0344mol/dm³

(1bii)
Conc. of B in g/dm³ = Conc. of B in mol/dm³ × molar mass of B

Molar mass of B(NaCO₃)= (23×2) + 12 + (16×3) = 106

:. Conc. of B in g/dm³ = 0.0344×106 = 3.6g/dm³

(1biii)
1 mole of Na₂Co₃ = 1mole of Na₂So₄

:. 0.0344 mole of Na₂Co₃ = 0.0344 mole of Na₂So₄
№ of mole= mass/ molar mass

Molar mass of Na₂So₄= (23×2)+32+(16×4)=142
:. 0.0344 = mass/142
Mass=142×0.0344=4.9g

(1biv)
1 mole of Na₂Co₃= 1 mole of CO₂

0.0344 mole of Na₂Co₃ = 0.0344 mole of CO₂
:. № of mole = volume/ molar volume = 0.0344 = volume of CO₂/22400cm³
=771cm³
==================================

(2ai)
[TABULATE]

=TEST=
C + heat

=OBSERVATION=
Formation of yellow flame

=INFERENCE=
Na⁺ present

(2aii)
=TEST=
C + distilled water + shake

=OBSERVATION=

=INFERENCE=
C is a soluble in water

(2aiii)
=TEST=
Solution C + litmus paper

=OBSERVATION=
It changes moist red litmus to blue and no effect blue

=INFERENCE=
Solution C is alkaline

(2bi)
=TEST=
1st portion of solution C + barium chloride

=OBSERVATION=
White precipitate formed

=INFERENCE=
CO₃²⁻ , SO₄²⁻ , S²⁻ or SO₃²⁻

(2bii)
=TEST=
Add dilute HCl to product in (bi)

=OBSERVATION=
White precipitate dissolved

=INFERENCE=
CO₃ , SO₃²⁻ or S₃²⁻

(2biii)
=TEST=
To the second portion add phenolphthalein

=OBSERVATION=
Purple colour observed

=INFERENCE=
C is alkaline
=======================
(3ai) Sulphur (iv) oxide and hydrogen sulphide.

(3aii) Liberation of hydrogen gas with pop sound and formation of green iron (ii) chloride solution

(3aiii) Spatula, beaker, weighing balance, standard volumetric flask

(3bi) Solution turns red

(3bii) Turns from brown to green

(3biii) A colourless, odourless, acidic gas liberated which turns blue litmus paper red and turn lime water milky.