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NAROK FORM FOUR DISTRICT MOCK – 2007
233/3
CHEMISTRY
PAPER 3
MARKING SCHEME.
1. Table I :
School value - 1mk 8cm3 – 9cm3 1mk
Decimal / Accuracy 0.1 ( 1mk) Max ( 2mks)
Penalise for unrealistic values
a) Average volume 8.5 + 8.5 + 8.5 ½ = 8.5 cm3 ½ penalise for missing units 3
b) Moles of solution A 8.5 x 0.25 ½ = 0.002125 ½ moles 1000
c)i) Moles of HCl in 25cm3 of solution D: HCl + NaOH NaCl + H2O
1 : 1 ½
Moles of HCl in 25cm3 = 0.002125 ½ ( Mole ratio) ii) Moles of HCl on 100cm3 of solution D 25 0.002125 100x 0.002125 ½ = 0.0085 moles ½ 25 iii) Moles of HCl in 100cm3 Solution B:
0.5 moles 1000cm3
0.5 x 100cm3 = 0.05moles ½ 1000 iv) Moles reacting with x2CO3
X2CO3 + 2HCl 2xCl + CO2 + H2O 1 : 2 Moles of HCl reacting = 0.05 – 0.0085 ½ = 0.0415 moles X2CO3 ½
d) mole ratio 1 : 2 ½
Moles of X2CO3 = 0.0415 2 = 0.02075 Moles that reacted
e) i) Moles of X2CO3 = 0.02075
0.02075 2.2 g
1 mole x 2.2 ½ = 106.024g ½ 0.02075 = ~ 106 g ½
ii) RAM of x X2CO3 = 2x + 60 = 106 ½ 2x = 46 ½ 2x 2
x = 23 ½
2. Table II
Initial
Final
DT.C
13.0C
9.0C
- 4.0C
a) H = Mc T ½
= mass = 20g ( 1 g/cm3 ) ½
C = 4.2
DT = 4.0C = 20 x 4.2 x 4 = + 336J ½ Deny ½ for lack of +ve sign
b) Mole of KCl RFM = 39 + 35.5 = 74.5 ½
2 = 0.02685 moles 74.5
0.02685 336 1 x 336 = 12514 J Mol-1 ½ 0.02685 ½ 12514 1000
= + 12.514 KJ Mol-1 Insist on units as KJ Mol-1 J-1 Hsoln = Hhyd + Hlattice ½
12.514 = ½ Hhyd + 690 ½ = -677.5 KJ mol-1 ½ deny ½ mark for of sign (-ve)
3.
OBSERVATION
INFERENCES
a)i) Colourless
233/3
CHEMISTRY
PAPER 3
MARKING SCHEME.
1. Table I :
School value - 1mk 8cm3 – 9cm3 1mk
Decimal / Accuracy 0.1 ( 1mk) Max ( 2mks)
Penalise for unrealistic values
a) Average volume 8.5 + 8.5 + 8.5 ½ = 8.5 cm3 ½ penalise for missing units 3
b) Moles of solution A 8.5 x 0.25 ½ = 0.002125 ½ moles 1000
c)i) Moles of HCl in 25cm3 of solution D: HCl + NaOH NaCl + H2O
1 : 1 ½
Moles of HCl in 25cm3 = 0.002125 ½ ( Mole ratio) ii) Moles of HCl on 100cm3 of solution D 25 0.002125 100x 0.002125 ½ = 0.0085 moles ½ 25 iii) Moles of HCl in 100cm3 Solution B:
0.5 moles 1000cm3
0.5 x 100cm3 = 0.05moles ½ 1000 iv) Moles reacting with x2CO3
X2CO3 + 2HCl 2xCl + CO2 + H2O 1 : 2 Moles of HCl reacting = 0.05 – 0.0085 ½ = 0.0415 moles X2CO3 ½
d) mole ratio 1 : 2 ½
Moles of X2CO3 = 0.0415 2 = 0.02075 Moles that reacted
e) i) Moles of X2CO3 = 0.02075
0.02075 2.2 g
1 mole x 2.2 ½ = 106.024g ½ 0.02075 = ~ 106 g ½
ii) RAM of x X2CO3 = 2x + 60 = 106 ½ 2x = 46 ½ 2x 2
x = 23 ½
2. Table II
Initial
Final
DT.C
13.0C
9.0C
- 4.0C
a) H = Mc T ½
= mass = 20g ( 1 g/cm3 ) ½
C = 4.2
DT = 4.0C = 20 x 4.2 x 4 = + 336J ½ Deny ½ for lack of +ve sign
b) Mole of KCl RFM = 39 + 35.5 = 74.5 ½
2 = 0.02685 moles 74.5
0.02685 336 1 x 336 = 12514 J Mol-1 ½ 0.02685 ½ 12514 1000
= + 12.514 KJ Mol-1 Insist on units as KJ Mol-1 J-1 Hsoln = Hhyd + Hlattice ½
12.514 = ½ Hhyd + 690 ½ = -677.5 KJ mol-1 ½ deny ½ mark for of sign (-ve)
3.
OBSERVATION
INFERENCES
a)i) Colourless