cambridge chemistry challenge lower 6th - c3l6 · cambridge chemistry challenge lower 6th june 2012...
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Cambridge Chemistry Challenge Lower 6th
June 2012
Marking scheme for teachers(please also read the additional instructions)
C L63
Page 1
mark
p2 p3 p4 p5 p6 p7 p8 Total
8 9 3 5 11 14 6010
leaveblank1(a) (i) oxidation state of As
(ii) equation for production of As2O3
(b) Structure of H3AsO3 geometry around As:
(ii)
(i)
(c) (i)
Equation for formation of H3AsO3:
(iii) Concentration of H3AsO3:
Mass of As2O3 in 100 cm3:
Page 2
+3 (or 3)
2
1
1
1
2
1
1
Page total8
2FeAsS + 5O2 Fe2O3 + As2O3 + 2SO2
As2O3 + 3H2O 2H3AsO3
pyramidal
[ accept a bond anglebetween 90° and 108°]
(give a mark if planarbut correct bonding)
number of moles of As2O3 in 1 dm3 = 20.6 / 197.84 = 0.104
from equation, 1 mol of oxide forms 2 mol acid
so concentration of acid = 0.208 mol dm–3
RMM of As2O3 = (2 × 74.92) + (3 ×16.00) = 197.84
20.6 g × 10–30 in 1 dm3
∴ mass in 100 cm3 = 2.06 × 10–30 g
[ note this is much less than the mass of a single protonso there are no molecules of As2O3 in the glass ]
OHOH
HO
As
leaveblank1(c) (ii) volume needed for a fatal arsenic dose:
answer expressed as a fraction of the Earth’s volume:
1(d) concentrations of HCl in mol dm–3
(i) D2 (ii) D4 (iii) D6
(iv) D8 (v) D10
(iii) number of bottles to contain one atom:
Page 3
3
3
3
2.06 × 10–29 g of As2O3 in 1 dm3
so 0.1 g contained in (0.1 / 2.06 × 10–29 ) dm3
= 4.854 × 1027 dm3
= 4.854 × 1024 m3
Volume of Earth = 1.08 × 1012 km3
= 1.08 × 1012 × 109 m3 = 1.08 × 1021 m3
so volume needed is (4.854 × 1024 / 1.08 × 1021) times the Earth
= 4500 times the volume of the Earth.
2.06 × 10–29 g of As2O3 in 1 dm3
∴ number of atoms of As in 1 dm3
= NA ×( 2.06 × 10–29 × 2) / 197.84
=1.254 × 10–7
volume required for 1 atom = 1 / 1.254 × 10–7 dm3
= 7.97 × 106 dm3 = 7.97 × 109 cm3
∴ number of bottles needed
= 7.97 × 109 / 28 = 285 million
[ 3 marks if all correct; −1 for each wrong answer down to zero ]
[ 2 marks if all correct; 1 mark if a factor of 1000 out ]
10–2 10–4 10–6
10–8 10–10
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leaveblank1(e) (i) pH of pure water:
(ii) pH of solutions:
(f) concentration of conc. HCl
pH of conc. HCl
Page 4
[H+] for pure water:
stock solution: pH = D2 pH =
D4 pH = D6 pH =
D8 pH = D10 pH =
4
2
7
3
1
½ ½
½ ½
2.00.0
4.0 6.0
7.0 7.0
10–7 mol dm–3
38 g = 38 / 36.5 mol in 100 g of solution
100 g of solution = 100 / 1.2 cm3
∴ mols of HCl in 1 dm3 =
(38 / 36.5) × (1.2 / 100) × 1000 = 12.5
concentration = 12.5 mol dm–3
− log10 (12.5) = −1.1
[ note the acid solution cannot becomealkaline on diluting with water ! ]
[ 3 marks if correct final answer; partial marks for any key steps]
RMM of HCl = 1.008 + 35.45 = 36.458
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leaveblank1(g) concentration of pure water:
(h) mass of acid:
Page 5
2
1
RMM of H2O = (2 × 1.008) + 16.00 = 18.016
1 dm3 water ≡ 1.0 kg = 1000 / 18.016 mol
concentration = 55.5 mol dm–3
[ Note this is more dense than a neutron starwhich essentially consists of touching nucleons.
An additional idea might be to calculate thedensity of a single proton. ]
[H+] = 1018 mol dm−3
so 1 dm3 has a mass of 1018 × 1 g
5 cm3 has a mass of
(1018 / 1000) × 5 g = 5 × 1015 g
Page total3
Page 6
+OOO OHHO
O
O
O
OH
OH
O
OO OHH
OH O
(i)
2 (a)
(ii)
(iii)
(iv)
(v)
(vi)
(vii)
O
OH
OH
(viii)
(ix)
(x)
C6H5
C6H5
C6H5
C6H5
C6H5
C6H5 C6H5 C6H5
NH2NH
N
H2N
O
O
OH
N
OC2H5
C2H5OH
O
H
O
OH OH
O
+
ox. red. hyd.
ox. red. hyd.
ox. red. hyd.
ox. red. hyd.
ox. red. hyd.
ox. red. hyd.
ox. red. hyd.
ox. red. hyd.
ox. red. hyd.
ox. red. hyd.
[ half mark per answer ]
Page total5
leaveblank
Page 7
(i)2(b)
C (174)
D (58)
G (176)
A (46) B (74)
H (74) J (60)
E (46) F (46)
(ii)
(iii)
(iv)
K (92) L (44) M (17)
2
1
3
5
[ answers can be either way round ]
[ allow this alternative ]
H OH
O
HO
NH
NH
O CO2H O
NH2
NH
NH
O
OO
HO NH2
H2N
HO
H
O
O
H2N
O
NH2H2N
HO
OHHO
O HO
H
OHOH
O
OC
HO OH
O
H2NNH
HHO
O
O
NH
HH
O[ allow thisalternative ]
Page total11
½ ½
leaveblank
Page 8
2(c)
W
S
R
T
U V
X
Z
Y
2
2
2
2
4
2
OHH
O
O
OH
O
O
O
O O
O
O
O
O
HO
OHOO
O
Page total14