CHEMISTRY PRACTICAL ANSWER NOW AVAILABLE

🇰.🇴.🇱.🇱.🇾. 🇧.🇷.🇮.🇬.🇭.🇹.✔: 2ai )1) 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

legendary…..

+++++++++++++++++++++++++++++++++++++++++++++++

3i) S02 changes the colour of acidified K2Cr2O7 solution from orage to green when it is passed through it by changing it to Cr2(So4)3 solution

ii) When Zinc dust is added to CuSo4 solution, the solution turns from blue to white.

iii) Due to the formation of Cr2(So403 solution.

iv) Due to the formation of white ZnSo4 since Zn is higher in the electrochemical series

3b) DRAW THE DIAGRAM BY YOURSELFTabulate
- Test -
Xn +10cm 3of H 2O and stirred
- Observation -
Xn dissolves to form a colourless solution
- Inference -
Xn is a soluble salt

2aii )
- Test -
First portion of Xn + NaOH ( aq ) in drops ,then
in excess
- Observation -
White precipitate which dissolves in excess to give a colourless solution
- Inference -
Zn2 +,Pb 2+, orAl3 + present

2aii )
- Test -
Second portion of Xn +NH3( aq ) in drops ,
then in excess
- Observation -
White precipitate in soluble in excess
- Inference -
Pb 2+orAl 3+ present

2aiii )
- Test -
Third portion of Xn +dil. HCl
- Observation -
White crystalline / chalky precipitate
- Inference -
Pb 2+confirmed

2bi )
- Test -
Yn + litmus
- Observation -
Red litmus paper turned blue
- Inference -
Yn is alkaline

2bii)
- Test -
Portion of Yn+ conc . HNO 3
- Observation -
Yellow colouration
- Inferrence -
Protein present
[6/12, 7:39 AM] 🇰.🇴.🇱.🇱.🇾. 🇧.🇷.🇮.🇬.🇭.🇹.✔: Chemistry - Practical - Answers

*Note*
*Use Your School End Point*

1a )
tabulate
Volumeofpipette / baseused = 25. 0cm 3
Indicator used - Phenolphthalein
Burette

Readings
Rough - First , Second ,Third
FinalBurette Reading ( cm 3) | 16. 00|
| 31. 70| 15 .65 | 31. 25|

InitialBurette Reading ( cm 3 ) - | 0. 00|
| 16. 00| | 0. 00| | 15. 65|

Volumeof acid used ( cm 3) | 16 .00 |
| 15. 70| | 15. 65| | 15. 60|

Average titre value
= 15 .70 +15. 60+15 .65 cm ^ 3/ 3= 15 .65 cm 3

1bi )
Concentration of B in moldm ^ - 3
H 2C 2O 4( aq ) +2 XOH ( aq )→ X 2C 2O 4 ( aq ) +2H 2O ( i ) CAVA = nA CBVB = nB
0. 08× 15 .65 = 1
CB × 25/ 2
CB = 2× 0 .08 × 15. 65/ 25 = 0 .100 moldm ^ - 3

1bii)
Molar mass of XOH
Concentration of B in gdm^ -3
= 1. 0× 1000 / 250 = 4gdm^ - 3
Therefore ,molar mass of XOH
= 4gdm ^ -3
0. 100moldm ^ - 3
= 40 gmol ^ - 1

1biii )
Relative atomic mass of X
XOH = 40
X +16 +1= 40
X +17 = 40
X = 40- 17
X = 23
[6/12, 7:39 AM] 🇰.🇴.🇱.🇱.🇾. 🇧.🇷.🇮.🇬.🇭.🇹.✔: 1(a).Volumeofpipette/baseused=25.0cm3

Indicatorused-Phenolphthalein

Burette
Readings

Rough ==> First ==>Second ==>Third

FinalBurette
Reading
(cm3)
16.00 31.70 15.65 31.25

InitialBurette
Reading
(cm3)
0.00 16.00 0.00 15.65

Volumeof
acidused
(cm3)
16.00 15.70 15.65 15.60

Averagetitrevalue=15.70+15.60+15.65cm3
3
=15.65cm3

(b).
(i).ConcentrationofBinmoldm-3:
H2C2O4(aq)+2XOH(aq)→ X2C2O4(aq)+2H2O(l)
CAVA=nA
CBVB=nB�
0.08×15.65=1
CB×25 2
CB=2×0.08×15.65
25
=0.100moldm-3

(ii).MolarmassofXOH:
ConcentrationofBingdm-3=1.0×1000
250
=4gdm-3
Therefore,molarmassofXOH=4gdm-3
0.100moldm-3
=40gmol-1

(iii).RelativeatomicmassofX:
XOH=40
X+16+1=40
X+17=40
X=40-17
X=23.