ICSE Board
Class X Chemistry
Board Paper – 2011 Solution
SECTION I
(1)
i. Barium chloride
ii. Platinum
iii. Coke
iv. Acetylene gas
v. Bronze
(2) On addition of ammonium hydroxide in small quantity, a blue-coloured copper hydroxide
precipitate is formed. This copper hydroxide of light blue colour dissolves in excess of ammonium
hydroxide to yield a deep blue solution.
(3) Charring of sugar takes place. Sulphuric acid dehydrates sugar leaving behind carbon which is
black in colour.
(4) Brown fumes of nitrogen dioxide are produced. Copper reacts with concentrated nitric acid to
produce copper nitrate, water and nitrogen dioxide gas.
(5) Aluminium reacts with nitrogen to form aluminium nitride, which on addition of water forms
aluminium hydroxide with the evolution of ammonia which turns moist red litmus blue.
(6) Black copper oxide is reduced to brown/red copper with the release of carbon dioxide gas.
(7) Pure water is a non-electrolyte and does not form ions. On addition of sulphuric acid, water
being a polar covalent compound ionises and becomes an electrolyte. So, it is regarded as an
example of catalysis.
(8) Carbon can form a large number of compounds because of tetravalency and catenation.
(9) In insufficient supply of air, methane burns to produce carbon monoxide which is a toxic gas.
(10) In hydrogen chloride, chlorine has greater nuclear charge so it pulls the shared pair of
electrons towards itself more strongly. As a result, a slight negative charge is developed on the
chlorine atom and a slight positive charge is developed on the hydrogen atom. Thus, hydrogen
chloride can be termed a polar covalent compound.
(11) The oxidising power of elements depends on the tendency to gain electrons which
increases from left to right along a period due to increase in nuclear pull.
(12) Sharing
(13) High
(14) Nitrogen
(15) Increases
(16) Decreases
(17)
i. Gram molecular mass of SO2 = (1 × 32) + (2 × 16) = 64 g
Mass of 1 mole of SO2 = 64 g
Molar volume =22.4 litres
64 g of SO2 at STP will have volume = 22.4 litres
320 g of SO2 at STP will have volume = (320 × 22.4)/64 = 112 litres
Volume occupied by 320 g of SO2 at STP = 112 litres
ii. Gay-Lussac’s Law states “When gases react, they do so in volumes which bear a
simple ratio to one another and to the volume of the gaseous product, if all the
volumes are measured at the same temperature and pressure.”
iii. C3H8 + 5O2 →3CO2 + 4H2O
Molar mass of propane = 44
44 g of propane requires 5 × 22.4 litres of oxygen at STP.
8.8 g of propane requires
5  22.4
 8.8 = 22.4 litres
44
(18) D
(19) B
(20) A
(21) D
(22) B
(23) A
(24) D
(25) C
(26) B
(27) C
(28)
i. Na2S2O3 +2HCl 2NaCl +H2O+SO2 +S (yellowppt.)
ii. Ca(HCO3)2 + 2HCl → CaCl2 + 2H2O + 2CO2
iii. Na2SO3 + H2SO4 → Na2SO4 + H2O + SO2
iv. Pb(NO3)2 + 2NaCl → PbCl2 + 2NaNO3
v. Zn + 2NaOH Na2ZnO2 + H2
SECTION II
(29)
(i) Copper metal conducts electricity due to the flow of electrons, whereas the copper sulphate
solution conducts electricity due to the flow of ions.
(ii) Copper metal does not decompose by passage of electricity, whereas copper sulphate
solution decomposes.
(iii) Copper metal usually exists in the solid state but allows the passage of current in the
molten state only, whereas copper sulphate conducts electricity in the molten state or in
solution form.
(iv) Copper metal involves physical change and copper sulphate solution involves chemical
change.
(30)
i. D
ii. C
iii. E
v. A
v. F
vi. B
(31)
i. At the cathode, copper metal is deposited. At the anode, oxygen gas is evolved.
ii. The blue colour of copper sulphate solution becomes light and discharges at the end.
iii. At cathode:
Cu2+ + 2e- → Cu
At anode:
4OH- − 4e → 2H2O + O2
(32)
i. Aluminium
ii. Calcination: The process of heating an ore to a high temperature in the absence of
air is called calcination.
Roasting: The process of heating an ore in the presence of air is called roasting.
iii. Froth flotation process
iv. Iron haematite: Fe2O3
Aluminium bauxite Al2O3.2H2O
v. Pure alumina ( Al2O3 ), cryolite (Na3AlF6) and fluorspar ( CaF2 )
(33)
i. Ammonia
ii. Equation: 2NH4Cl + Ca(OH)2 → CaCl2 + 2H2O + 2NH3
iii. The gas is collected by downward displacement of air.
iv. Quick lime
v. Dense white fumes are seen when a glass rod dipped in conc. HCl is brought near the
mouth of the gas jar.
(34)
Element Relative
atomic
mass
%
Relative
compound number
of atoms
Simple
Ratio
H
1
2.13
2.13/1
2.13
Simple
whole
number
Ratio
2
C
12
12.67
12.67/12
1.05
1
Br
80
85.11
85.11/80
1.06
1
Empirical formula = CH2Br
n(Empirical formula mass of CH2Br) = Molecular mass (2 × VD)
n(12 + 2 + 80) = 94 × 2
n=2
Molecular formula = Empirical formula × 2
= (CH2Br) × 2
= C2H4Br2
(35)
i. 1022 atoms of sulphur
6.022 × 1023 atoms of sulphur will have mass = 32 g
1022 atoms of sulphur will have mass = (32 × 1022)/(6.022 × 1023) = 0.533 g
ii. 0.1 mole of carbon dioxide
1 mole of carbon dioxide will have mass = 44 g
0.1 mole of carbon dioxide will have mass = 4.4 g
(36) i.
ii. (a) To prevent back suction of water into the apparatus.
(b) Provides a larger surface area for dissolution of hydrogen chloride gas.
iii.
200  C
NaCl(s) + H2SO4(l)  NaHSO4(aq) + HCL(g)
Sodium Sulphuric
Sodium
Hydrogen
chloride acid
hydrogen
chloride
sulphate
100 C
2NaCl(s) + H2SO4(l)  Na2SO4(aq) + 2HCl(g)
Sodium Sulphuric
Sodium
Hydrogen
chloride acid
sulphate
chloride
(37)
i. Nickel
ii. Acetic acid
iii. Esterification
iv. CH2 CH2 2KOHCH2 CH2 2KBr
(Alc.)
Br
Br
v. Ethanol
(38)
Boil
 C2H5OHKCl
i. C2H5Cl KOH(aq) 
Cao
ii. CH3COONa+NaOH
CH4 +Na2CO3
∆
AcidicK Cr O O 
AcidicK Cr O O 
7
7
 CH3CHO+H2O 22 
 CH3COOH
iii. C2H5OH 22 
iv. CaC2+2H2O  Ca(OH)2+C2H2
v. 2C2H5OH + 2Na → 2C2H5ONa + H2
(39)
i. Step 1: Production of SO2 from sulphur.
S + O2 → SO2
Step 2: Purification of SO2. Air impurities such as dust and arsenic are removed;
otherwise, these will lead to poisoning of the catalyst.
Step 3: Catalytic oxidation of SO2 to SO3.
Catalyst – V2O5 (Vanadium pentoxide)
Temperature – 450°C–500°C
Pressure – 1–2 atmospheres
Reaction – 2SO2 + O2 → 2SO3
Step 4: Absorption of SO3 in 98% sulphuric acid to form oleum.
SO3 + H2SO4 → H2S2O7 (oleum)
Step 5: Dilution of oleum to obtain sulphuric acid of desired concentration.
H2S2O7 + H2O → 2H2SO4
ii. (A) Sulphuric acid when heated with sodium nitrate shows its non-volatile nature.
200 C
KNO3(S)
+ H2SO4(S)  KHSO4(aq)
+ HNO3(vap)
Potassium nitrate Sulphuric acid
Potassium hydrogen Nitric acid
(B) With carbon, it shows oxidising nature.
C(s)
+ 2H2SO4(aq)  2 H2O(l) + 2SO2 (g)
+ CO2(g)
Carbon Sulphuric acid Water
Sulphur dioxide Carbon dioxide
(40)
i.
The complete apparatus should be made of glass only.
ii.
At high temperature, nitric acid decomposes and the glass apparatus may get
damaged. The sodium formed at a higher temperature forms a hard crust which
sticks to the walls of the retort.
(41)
i. 8NH3 (g) + 3Cl2 (g) → N2 (g) + 6NH4Cl
ii. Fe(OH)3 + 3HNO3 (aq) → Fe(NO3)3 + 3H2O
iii. ZnO + 2KOH → Na2ZnO2 + H2O
(42)
i. Group – 13
Period – 3
ii.
iii. When crystals of washing soda are exposed to air, it loses its water of crystallisation
and the phenomenon is known as efflorescence.
(43)
i. (B) Neutralisation
ii. (E) Direct synthesis
iii. (D) Double decomposition
iv. (A) Simple displacement
v. (C) Decomposition by acid