CHEMISTRY PRACTICAL ANSWERS:
NOTE Make sure u ask your
chemistry teacher to tell you the titre value to
be used.
Note that (cm)^3
Means raise to the power of 3
(1)
volume of pippete(VB)=25c m
tabulate
burette readings:rough( cm)^3,1st(cm)^3 ,2nd(cm)^3
final readings:15.60, 15.40,15.20
initial readings:0.00,0 .0
average titre=(15.40+15 .20)/ 2=15.30cm^(3)
(1b)
cava/cavb= na/nb
0.05*15.30/ (cb*25.00)=1/2
2(0.05*15.30)=1 9cb*25.00)
cb=(1.53*25.00) /25.0
cb=0.0612moldm^ -3
(1bii)molarmass of KOH=39+16=56g/ mole
(1biii)relative mass of k=39g
(2) tabulate
(2)Test:
(i)Solid C+H20
(2bi)1st portion + NaOH(aq)+ heat
(2bii)2nd portion+NaoH in drops and then in excess
(2c)3rd portion+Bacl2+d il Hcl
(2)observation:
(2a)c dissolves to form a colourless solution which turns blue litmus paper red
(3a)
aqueus ion(ii) tetra oxo sulphate can be converted into an aqueous solution of Magnesium tetra oxo sulphate (VI) byreacting it with magnesium metal. since magnesium is above ion in electrochemical series.
i.e Displacement reaction
equation
(ii) FeSo4+MG+=>MgSO 4+Fe
(3bi)some compound are recrysdtasllize d after preparation because they are soluble in water at devated temperature so as the temperature is allow to decrease the crystal of the compond are formed. some compounds are also seperated from the unwanted product by recrystallizati on
(3bii)
-dissolution of mixture in hot water
-filter the hot slution
-allowing the filtrate to cool
-refiltration of the filtrate after crystallization
(3c)
-hydrogen sulphide
-sulphur(iv) oxide
(3cii)the colour changes from yellow to green.
QUANTITATIVE ANALYSIS
TEST
1. Sample C + Heat
2. Sample C + distilled water
3. Sodium hydroxide portion + NAOHag
4. 2nd portion + NH3(ag)
5. 3rd portion + Baruim Chloride solution +
dilute hydrochloric acid
OBSERVATION
1. There was some water droplet Of the
upper part of test tube.
2. Sample C dissolve change blue litmus
paper to red,
It is cool to touch.
3. There was a dirty green precipitate in
soluble in excess NAOHag
4. Dirty green precipitate.
5. There was a white ppt insoluble in excess
Bacl2
White ppt is insoluble in dilute hydrochloric
acid
INFERENCE
1. Water of crystallization present in sample
C.
2. It is acidic due to salt hydrolysis.
3. Iron (ii) ion, fe2+ present.
4. Iron (ii) ion, fe2+ present.
5. So32-� or tetraoxosulphat e(vi) ion, So42-�
or trioxocarbonate (iv) ion Co32-,
tetraoxosulphat e(vi)ion So42- is confirmed
1. Solution A is 0.049mole of dilute
tetraooxosuipha te(vi) acid dissolved in
500cm3 of solution.
Solution B is 0.100mol/dm3 solution of a
base.
a. Titrate solution A against B using methyl
orange as the indicator and calculate the
average titre
b. With ur result and the information
provided calculate Concentration of solution A in mol/dm3 and
B in g/dm3
Mole ratio of acid to base and hence
Suggest name and formula for the base
Write a balanced equation of reaction
between solution A & B
Titre Value = 0.25
Note-: School titre value might be differ from
naijahero' tk own. Make sure u ask your
chemistry teacher to tell you the titre value to
be used.
SOLUTION
Concentration of solution A in mol/dm3
and B in g/dm3. Solution A IS 0.049mole in
500cm3?
Mole of A is in 10000
Mole of A = 0.049 x 1000cm3/500cm3
=0.090mol
Solution A is 0.98mol/dm3
Conc. in mol/dm3 of A = Conc. in g/dm3/
Molar mass
No of moles of A =Molar Conc. of A x Volume
= 0.098mol/dm3 x 12.25cm3/1000
=0.0012005
Molar mass of solution A = H2So4=
(1x2)+(32x1)+(1 0x4)
= 2+32+64
=98g/mol
Conc. of mol/dm3 = Conc. in g/dm3/Molar
mass
Conc. in g/dm3 = Conc. in mol/dm3 x Molar
mass
=0.098x98
=9.604 g/dm3
Mole ratio of acid to base
Na/Nb = 0.0012005/ 0.0025
= 0.4802 approximately 0.50 or �
No of moles of B = Conc. of B x Vol in dm-3
= 0.100 x 25/100dm3
=0.0025moles
H2S04 + 2NH4OH => (NH4)2 SO4 + 2H20
CaVa/CbVb = � this can be base
So, the base can be ammonia solution or
ammonium hydroxide or
H2S04 + 2Na0H => Na2S04 + 2H20
CaVa/CbVb = �
So the base can be NH4OH or NH3� or NaoH
or KOH
Therefore the balanced equation is:-
2Na0H + H2S04 => Na2S04 + 2H20
QUESTION 3
In the usual qualitative analysis why the
litmus paper is first moistened?
In volumetric analysis, why is burette used
and not measuring cylinder?
Give a test in the laboratory to (i) Distinguish
btw ALNO3 and Mg(N03)2 (ii) Confirm btw
ZnC03(aq) and Al2(C03)3(aq)
NOTE Make sure u ask your
chemistry teacher to tell you the titre value to
be used.
Note that (cm)^3
Means raise to the power of 3
(1)
volume of pippete(VB)=25c m
tabulate
burette readings:rough( cm)^3,1st(cm)^3 ,2nd(cm)^3
final readings:15.60, 15.40,15.20
initial readings:0.00,0 .0
average titre=(15.40+15 .20)/ 2=15.30cm^(3)
(1b)
cava/cavb= na/nb
0.05*15.30/ (cb*25.00)=1/2
2(0.05*15.30)=1 9cb*25.00)
cb=(1.53*25.00) /25.0
cb=0.0612moldm^ -3
(1bii)molarmass of KOH=39+16=56g/ mole
(1biii)relative mass of k=39g
(2) tabulate
(2)Test:
(i)Solid C+H20
(2bi)1st portion + NaOH(aq)+ heat
(2bii)2nd portion+NaoH in drops and then in excess
(2c)3rd portion+Bacl2+d il Hcl
(2)observation:
(2a)c dissolves to form a colourless solution which turns blue litmus paper red
(3a)
aqueus ion(ii) tetra oxo sulphate can be converted into an aqueous solution of Magnesium tetra oxo sulphate (VI) byreacting it with magnesium metal. since magnesium is above ion in electrochemical series.
i.e Displacement reaction
equation
(ii) FeSo4+MG+=>MgSO 4+Fe
(3bi)some compound are recrysdtasllize d after preparation because they are soluble in water at devated temperature so as the temperature is allow to decrease the crystal of the compond are formed. some compounds are also seperated from the unwanted product by recrystallizati on
(3bii)
-dissolution of mixture in hot water
-filter the hot slution
-allowing the filtrate to cool
-refiltration of the filtrate after crystallization
(3c)
-hydrogen sulphide
-sulphur(iv) oxide
(3cii)the colour changes from yellow to green.
QUANTITATIVE ANALYSIS
TEST
1. Sample C + Heat
2. Sample C + distilled water
3. Sodium hydroxide portion + NAOHag
4. 2nd portion + NH3(ag)
5. 3rd portion + Baruim Chloride solution +
dilute hydrochloric acid
OBSERVATION
1. There was some water droplet Of the
upper part of test tube.
2. Sample C dissolve change blue litmus
paper to red,
It is cool to touch.
3. There was a dirty green precipitate in
soluble in excess NAOHag
4. Dirty green precipitate.
5. There was a white ppt insoluble in excess
Bacl2
White ppt is insoluble in dilute hydrochloric
acid
INFERENCE
1. Water of crystallization present in sample
C.
2. It is acidic due to salt hydrolysis.
3. Iron (ii) ion, fe2+ present.
4. Iron (ii) ion, fe2+ present.
5. So32-� or tetraoxosulphat e(vi) ion, So42-�
or trioxocarbonate (iv) ion Co32-,
tetraoxosulphat e(vi)ion So42- is confirmed
1. Solution A is 0.049mole of dilute
tetraooxosuipha te(vi) acid dissolved in
500cm3 of solution.
Solution B is 0.100mol/dm3 solution of a
base.
a. Titrate solution A against B using methyl
orange as the indicator and calculate the
average titre
b. With ur result and the information
provided calculate Concentration of solution A in mol/dm3 and
B in g/dm3
Mole ratio of acid to base and hence
Suggest name and formula for the base
Write a balanced equation of reaction
between solution A & B
Titre Value = 0.25
Note-: School titre value might be differ from
naijahero' tk own. Make sure u ask your
chemistry teacher to tell you the titre value to
be used.
SOLUTION
Concentration of solution A in mol/dm3
and B in g/dm3. Solution A IS 0.049mole in
500cm3?
Mole of A is in 10000
Mole of A = 0.049 x 1000cm3/500cm3
=0.090mol
Solution A is 0.98mol/dm3
Conc. in mol/dm3 of A = Conc. in g/dm3/
Molar mass
No of moles of A =Molar Conc. of A x Volume
= 0.098mol/dm3 x 12.25cm3/1000
=0.0012005
Molar mass of solution A = H2So4=
(1x2)+(32x1)+(1 0x4)
= 2+32+64
=98g/mol
Conc. of mol/dm3 = Conc. in g/dm3/Molar
mass
Conc. in g/dm3 = Conc. in mol/dm3 x Molar
mass
=0.098x98
=9.604 g/dm3
Mole ratio of acid to base
Na/Nb = 0.0012005/ 0.0025
= 0.4802 approximately 0.50 or �
No of moles of B = Conc. of B x Vol in dm-3
= 0.100 x 25/100dm3
=0.0025moles
H2S04 + 2NH4OH => (NH4)2 SO4 + 2H20
CaVa/CbVb = � this can be base
So, the base can be ammonia solution or
ammonium hydroxide or
H2S04 + 2Na0H => Na2S04 + 2H20
CaVa/CbVb = �
So the base can be NH4OH or NH3� or NaoH
or KOH
Therefore the balanced equation is:-
2Na0H + H2S04 => Na2S04 + 2H20
QUESTION 3
In the usual qualitative analysis why the
litmus paper is first moistened?
In volumetric analysis, why is burette used
and not measuring cylinder?
Give a test in the laboratory to (i) Distinguish
btw ALNO3 and Mg(N03)2 (ii) Confirm btw
ZnC03(aq) and Al2(C03)3(aq)
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