CHEMISTRY PRACTICALS NECO 2017 QUESTION AND ANSWER

GN TEAM
 CHEMISTRY PRACTICALS NECO 2017

  1. A is a solution containing 6.30gdm-3 of impure ethanedioic acid. B is 0.100moldm-3 sodium hydroxide solution. Put A into the burette and titrate it against 20.0cm3 or 25.0cm3 portions of B using phenolphthalein as indicator. Repeat the titration three more times and record your results in the table below;

Volume of pipette = 25.0cm3 

TITRATION

ROUGH(cm3)

1ST(cm3)

2ND(cm3)

3RD(cm3)
FINAL BURETTE READING

21.20

30.60

25.70

22.80
INITIAL BURETTE READING

0.00

10.00

5.00

2.00
VOLUME OF ACID USED

21.20

20.60

20.70

20.80
 Calculate the average volume of A = (1st + 2nd + 3rd)/3

= (20.60+20.70+20.80)/3

= 62.10/3 = 20.70cm3 

From your results and information provided above, calculate

Concentration of A in moldm-3 

Eqn : H2C2O4 + 2NaOH —— Na2C2O4 + 2H2O

  Mole ratio of A : B = 1 : 2.

Using (CaVa/CbVb) = (Na/Nb)

(Ca*20.70/0.100) = ½

Ca = (0.10025)/(20.702)= 2.50/41.40= 0.0603864

Ca = 0.0604moldm-3

Concentration of solution A in gdm-3

Molar Mass of A{H2C2O4} = (12) + (122) +(16*4)

       = 2 + 24 +64

       = 90g/mol

Conc of A in gdm-3 = Conc of A in moldm-3 * Molar mass

        =0.0604*90

        = 5.44 gdm-3

Percentage purity of A

% purity of A = (Concq of pure/ Conc of Impure)*100

= (5.44/6.30)*100

= 86.35%

Hence, % impurity = 100 – 86.35

% impurity = 13.65%

Volume of solution A that would neutralize a solution containing 0.005 mole of NaOH

25cm3 of 0.100 moldm-3 NaOH contain (25/1000)*0.100 = 0.0025 mole

If 20.70cm3 of A neutralizes 0.0025 mole of NaOH,

09.005 mole of NaOH will be X

=(20.70*0.005)/0.0025 = 41.4cm3  
C         i.        What would be the color of methyl orange indicator in soln. B?

Ans: Yellow or Orange

ii. Give the reason why methyl orange is not suitable for this

Ans: Ethanedioic acid is a weak acid while sodium hydroxide is a strong 

       base. Methyl orange is not suitable because the endpoint will not 

       coincide with the pH of indicator.

S/N

TEST

OBSERVATIONS

INFERENCE
 2ai.

X + Heat

Effervescence occurs, a reddish brown gas with irritating smell evolved

NO2 gas from NO3 
ii.

X + Water

It dissolved to give a clear solution

X is a soluble salt
b

Divide the solution into 2 portions

i.

1st portion + NaOH in drop

White ppt formed

Pb2+ or Ca2+

In excess

Soluble in excess

Pb2+ present
ii.

2nd portion + NH3 in drops

White ppt formed

Pb2+ or Al3+

In excess

Ppt insoluble in excess

Pb2+ 
iii.

Solution from b(ii) above + dil. HCl

White ppt insoluble

Pb2+ present
Ci.

Y + Conc. HNO3

Yellow color is formed

Protein present
ii.

Solution from Ci. Above + Conc. NH3

Yellow colour remained 

Protein is confirmed

[6/12, 10:47] Master ExM: No3
3ai)

i) SO2 change the colour of acidified K2Cr2O7 solution from Orange to green when it is passed through it by changing to Cr2(SO4)3

solution 
ii) when zinc dust is added to CuSO4 solution, the solution turn from Blue to white

3aii)

i)Due to the formation of Cr2(SO4)3 solution 
ii)Due to the formation of white ZnSO4  since zinc is higher in electrochemical series
3b)

DRAW and label a diagram for a set up that can be used to separate  suspension of chalk dust in water
(check above images for the drawing)
S/N

TEST

OBSERVATIONS

INFERENCE‎

————————————‎

 2ai.

TEST‎

X + 5cm distilled water, divided the solution into 4 portion

OBSERVATION ‎

Effervescence occurs, a reddish brown gas with irritating smell evolved

INFRENCE ‎

NO2 gas from NO3 ‎

_____________________________‎
bi)‎‎

TEST‎‎

1st portion + NaOH in drop

OBSERVATION ‎

White ppt formed

INFRENCE ‎

Pb2+ or Ca2+

______________________________‎

TEST‎

In excess

OBSERVATION ‎

Soluble in excess

INFRENCE ‎

Pb2+ present

_______________________________‎
bii)

TEST‎

2nd portion + NH3 in drops & dilute NH3 solution 

OBSERVATION ‎

White ppt formed

INFRENCE ‎

Pb2+ or Al3+

_________________________________‎
TEST‎

In excess

OBSERVATION ‎

Ppt insoluble in excess

INFRENCE ‎

Pb2+ 


biii)

TEST‎

3rd portion + dil. HCl in drop

OBSERVATION ‎

White ppt insoluble

INFRENCE ‎

Pb2+ present

_____________________________________‎
Ci)

TEST‎

Y + Conc. HNO3

OBSERVATION ‎

Yellow color is formed

INFRENCE ‎

Protein present


Cii)

TEST‎

Heat the produce formed in Ci. Above + Conc. NH3

OBSERVATION ‎

Yellow colour remained 

INFRENCE ‎

Protein is confirmed

________________________________________‎

Ciii)

TEST

Add Conc. NH3 to the product in cii

OBSERVATION 

Yellow colour remained 

INFRENCE 

Protein is present 

_______________________________________‎Note:make use of your school end poin: Note:
 Candidates are strongly advised to check which size of pipette was used by their school.

 We used 25cm³

If your school used 20cm³, all you need do is use it in place of 25cm³ anywhere we used it and calculate accordingly

Succeess and goodluck… 👍🏻💪🏽✅

.. 

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2017 NECO CHEMISTRY PRACTICAL SOLUTION 

NO1)

Note:
 Candidates are strongly advised to check which size of pipette was used by their school.

 We used 25cm³

If your school used 20cm³, all you need do is use it in place of 25cm³ anywhere we used it and calculate accordingly

NOTE:ALL MAKE USE OF YOUR SCHOOL ENDS POINT
Volume of Pipette used 25.0cm^3
Titration | Rough | First | Second | Third 

Final burette reading (cm^3) | 22.50 | 45.10 | 32.65 | 27.60

Intial burette reading (cm^3) | 0.00 | 22.50 | 10.00 | 5.00

Volume of A used (cm^3) | 22.50 | 22.60 | 22.65 | 22.70
1a) Average volume of A = 22.60 + 22.65 + 22.70/3
= 22.65cm^3
1bi) GIVEN: CB = 0.10moldm^-3

VB = 25cm^3

CA = ?

VA = 22.65cm^3
Equation of reaction:

H2C2O4(aq) + 2NaOH(aq) ====> Na2C2O4(aq) + 2H2O(l)

hence, nA = 1

Nb = 2
Using CAVA /CBVB = nA/nB

CA = CBVBnA/VanB

=0.10 × 25 × 1/22.65 × 2
CA = 0.055 moldm^-3
Concentration of A in moldm^-3 = 0.055

bii) Using: 

gram concentration = molar concentration × molar mass

conc. Of A in gdm^-3 = 0.055 × [(1 × 2) + (12 × 2) + (16 × 4)]

= 0.055 × 90

= 4.95gdm^-3

======================
No2 solution 
2a)

TEST

X + 5cm of distilled water, Divide the solution into 4 portion
OBSERVATION

Salt X is soluble in distilled water 
INFERENCE

Salt X is soluble salt

=======================

2bi)

Test

-To the first portion, Add dilute NaOH in drop

-then in excess
OBSERVATION

-White precipitate solution is formed

-The white precipitate is soluble 
INFERENCE

Zn^2+ Pb^2+ and AL^3+ are present

=====================

2bii)

TEST
-To the 2nd portion add dilute NH3 solution in drop

-then in excess 
OBSERVATION
White gelatinous precipitate is formed  which is insoluble in excesa
INFERENCE
Zn^2+,Pb^2+ and Al^3+ are present

======================

2biii) 

TEST

To the 3rd portion, Add dilute HCL in drops

-then in excess

-heat the solution
OBSERVATION

White  precipitate is formed which Dissolved on heating and re-appears on cooling 
INFERENCE
Pb^2+ confirmed

======================

2ci)

TEST

Sample Y + concentrated HNO3
(Cii) 

Heat the product formed in (ci) 
OBSERVATION

Dirty white curd is formed

===========

It turn to yellow colouration on heating

INFERENCE

Protein is present

2ciii)

TEST
Add concentrated NH3 to the product in (cii) 
OBSERVATION

The product in (Cii) turns to orange colour 
INFERENCE
Protein is present

=======================
No3
3ai)

i) SO2 change the colour of acidified K2Cr2O7 solution from Orange to green when it is passed through it by changing to Cr2(SO4)3

solution 
ii) when zinc dust is added to CuSO4 solution, the solution turn from Blue to white

3aii)

i)Due to the formation of Cr2(SO4)3 solution 
ii)Due to the formation of white ZnSO4  since zinc is higher in electrochemical series
3b)

DRAW and label a diagram for a set up that can be used to separate  suspension of chalk dust in water
(check above images for the diagram)

DONE
GOODLUCK

Succeess and goodluck… 👍🏻💪🏽✅

.. 

Lemme come and be going to put my phone in order. 🚷

Visit www.gistnaira.com

Follow us on

FB:GistNaira.com

Twitter:@GistNaira_Page

Youtube:Freenaija TV

IG:GistNaira

To subscribers for correct question and answer call/WhatsApp: 08065538205

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