01
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02
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03
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04
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05
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06
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07
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08
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09
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10
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D
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C
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A
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B
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C
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A
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B
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D
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C
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C
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11
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12
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13
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14
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15
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16
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17
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18
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19
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20
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D
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C
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C
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A
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D
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C
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C
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A
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D
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A
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21
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22
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23
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24
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25
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26
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27
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28
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29
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30
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A
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C
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A
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D
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B
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D
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B
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B
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D
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D
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31
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32
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33
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34
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35
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36
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37
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38
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39
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40
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C
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C
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C
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A
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C
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A
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B
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A
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C
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B
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Q1 [Experimental methods] – D
Ammonia is an alkaline gas so calcium oxide must be used as the drying agent (ammonia will react with sulfuric acid). Ammonia is less dense than air so it can be collected via upward delivery.
Q2 [Chemical analysis] – C
The correct option is essentially the test for carbon dioxide with limewater (calcium hydroxide). The white ppt formed is calcium carbonate.
Q3 [Kinetic particle theory] – A
When a perfume is detected across a room, it must be due to the diffusion which caused the scent to travel.
Rate of diffusion of a gas is inversely proportional to its relative molecular mass. Hence perfume X would reach the opposite side in a shorter time.
Q4 [Atomic Structure] – B
All four atoms contain 8 neutrons.
Q5 [Atomic Structure] – C
The ion has 13 protons and 10 electrons → Group III
The ion (on losing its valence electrons) has 2 completely-filled electron shells → atom originally has 3 filled electron shells → Period 3
Q6 [Chemical Bonding] – A
Based on the formula of the compound, the ions present are X+ and Y2–.
Hence each atom of X gives away one electron only
Q7 [Chemical Bonding] – B
Covalent molecules are formed from two non-metal atoms i.e. carbon and sulfur.
Q8 [Chemical Bonding] – D
1 is wrong: ionic compounds conduct electricity through mobile ions (not electrons)
2 and 3 are correct.
Q9 [Chemical Calculations] – C
No. of moles of X = 12/24 = 0.5 mol
1 mole of X weighs 56 g (molar mass)
Molar mass of C4H8 = 56 g
Q10 [Chemical Calculations] – C
Find the reacting ratio between what is given (nitrogen) and what you want to find (nitric acid)
1 mole of nitrogen produces 2 moles of nitric acid.
0.5 mole of nitrogen produces 1 moles of nitric acid.
Q11 [Electrolysis] – D
No. of moles of copper = 16/64 = 0.25 mol
Each Cu2+ requires 2 electrons to be discharged.
No. of moles of electrons = 0.25 × 2 = 0.5 mol
No. of moles of titanium = 6/48 = 0.125 mol
Each titanium ion would have taken in 0.5/0.125 = 4 electrons
Charge on titanium ion = 4+
Q12 [Electrolysis] – C
Lead ions are attracted to the negative electrode and are reduced (gain electrons)
Chloride ions are attracted to the positive electrode and are oxidised (lose electrons)
Q13 [Electrolysis] – C
In the electrolysis of aqueous sodium chloride:
Ions present at positive electrode: chloride and hydroxide → hydroxide ions oxidised to oxygen gas
Ions present at negative electrode: sodium and hydrogen ions → hydrogen ions reduced to hydrogen gas
In the electrolysis of concentrated aqueous sodium chloride, the high concentration of chloride ions cause it to be discharged at the positive electrode instead. Chlorine gas is produced.
Q14 [Energy Changes] – A
Bonds broken = 2 × H–H + 1 × O=O = 2(436) + 498
Bonds formed = 4 × O–H = 4(463)
Enthalpy change of reaction = bonds broken – bonds formed
= 1370 – 1852 = –482 kJ/ mol
Q15 [Energy Changes] – D
From the table, the increase in enthalpy for each successive member is 600-700 kJ/ mol.
Hence to obtain -3900 kJ/ mol from -2640 kJ/ mol (difference = 1260 kJ/ mol), the alcohol must have 2 more carbons than butanol.
Given the general formula of alcohols (CnH2n+1OH), the formula of an alcohol with 6 carbons would be C6H13OH i.e. 6 carbon atoms and 14 hydrogen atoms.
Q16 [Speed of reaction] – C
1 is correct. Vol of CO2 produced = 400 cm3
Moles of CO2 produced = 400/24000 = 0.0167 mol
Moles of CaCO3 = 0.0167 mol
2 is wrong: rate of reaction is reflected by the slope of the product-time curve i.e. decreasing.
3 is correct: no further production of CO2 after 15 mins (volume remained constant)
Q17 [Redox] – C
1: Mn (+2) → Mn (+3) [oxidised]
2: Mn (+4) → Mn (+2) [reduced]
3: Mn (+7) → Mn (+2) [reduced]
Q18 [Redox] – A
Since charge on sulfate ion is 2–, charge on chromium ion must be 3+
→ oxidation state of chromium in chromium sulfate is +3.
Q19 [Acids/ Bases/ Salts] – D
1 is wrong: With the same concentration of acids, a strong acid will produce a higher concentration of hydrogen ions than a weak acid → pH will be lower (not higher)
2 is correct.
3 is wrong: UI is green in a solution of pH 5.
4 is correct. Hydrogen ions are reduced by the metals to form hydrogen gas.
Q20 [Acids/ Bases/ Salts + Ammonia] – A
This question is easier solved by elimination: B, C and D are obviously wrong.
In the formation of ammonia from nitrogen and hydrogen, it is mentioned that the reaction is reversible and that the forward reaction is exothermic.
Hence when temperature is raised, the reverse reaction will occur i.e. ammonia decompose to nitrogen and hydrogen.
Q21 [Acids/ Bases/ Salts] – A
Lead(II) chloride is insoluble.
Q22 [Acids/ Bases/ Salts] – C
Again, this question is easier solved by elimination: A, B and D are obviously wrong.
A: basic oxide is an oxide of a metal
B: acidic oxide is an oxide of a non-metal → both are non-metals → covalent bonds
D: basic oxide is an oxide of a metal → ionic compound → solids
C: the amphoteric oxides in syllabus are oxides of Zn, Al and Pb which are metals.
Q23 [Periodic Table] – A
Group I: reactivity increases down Group Group VII: reactivity decreases down Group
For the most violent reaction, select the most reactive Group I and VII element i.e. caesium and fluorine.
Q24 [Periodic Table] – D
QCl3: Q has +3 oxidation state
QSO4: Q has +2 oxidation state
Since Q exhibits variable oxidation states, it must be a transition element.
Q25 [Periodic Table] – B
Hydrogen can donate an electron to form hydrogen ion or share an electron to a non-metal to form a covalent bond.
Group I metals react with water to give the hydroxide and hydrogen gas.
Q26 [Qualitative analysis] – D
Green solution: Iron(II) ions present
Yellow solution obtained on adding chlorine: iron(II) oxidised to iron(III) by chlorine (oxidising agent)
On adding aq ammonia, green ppt – Fe(OH)2, yellow ppt – Fe(OH)3
Add nitric acid followed by barium nitrate → white ppt → sulfate ions present.
Q27 [Periodic Table] – B
Q and R formed at positive electrode → Q and R forms negative ions i.e. non-metals
P and S conduct electricity → P and S are metals
P has a higher mp than S → melting point of metals increases across period and decreases down Group
Q displaces R from solution → Q is more reactive than R (reactivity of halogens decreases down Group i.e. Q is above R)
Q28 [Metals] – B
A (wrong): alloys are mixtures
C (wrong): both aluminium and copper ions are positively charged
D (wrong): an impure substance have lower mp than the pure substance
Q29 [Metals] – D
When heating a hydrated salt, the water of crystallization is removed as water vapour.
Since reactive metals form stable compounds, sodium carbonate is stable and will not decompose to form carbon dioxide.
Q30 [Metals] – D
Zinc oxide cannot be reduced to zinc by hydrogen gas.
Q31 [Metals] – C
A (wrong): haematite contains iron(III) oxide and not iron(II) oxide
B (wrong): limestone is added to remove acidic impurities and not as a catalyst
D (wrong): the main impurity in iron formed is carbon
Q32 [Metals + Chemical Calculations] – C
M + 2HCl → MCl2 + H2
Since the reacting ratio is the same for all the metals, the metal present in the greatest no. of moles will produce the greatest volume of gas.
Given the same mass of metal, the metal with the smallest molar mass will be present in the greatest no. of moles (since no. of moles = mass/ molar mass)
Q33 [Air] – C
In the catalytic converter, oxides of nitrogen are removed by reduction by carbon monoxide into nitrogen and carbon monoxide.
Q34 [Metals] – A
Q35 [Alkenes] – C
The reaction described is hydrogenation; hence X must be an alkene.
Since X reacts with only one mole of hydrogen gas, it contains only one C=C i.e. general formula CnH2n.
Q36 [Macromolecules] – A
In addition polymerisation, the alkene and polymer formed have the same empirical formula i.e. CH2.
Q37 [Alkanes] – B
C10H22 → x C2H6 + y C2H4 + z C3H6
Substitute the values of x, y and z
Correct combination will have same total number of each atoms on both sides (atom conservation).
Q38 [Alkenes] – A
Reaction of alkene with chlorine is an addition reaction → only one product is formed.
Q39 [Alcohols] – C
When alcohols are oxidized, the carbon attached to the –OH group is converted to –COOH.
So methanol will form methanoic acid, HCOOH (no CH3 group)
Q40 [Carboxylic acids] – B
An ester is formed from the condensation reaction where one molecule of water is removed i.e. ester will contain 2 less hydrogen atoms than the total no. from the carboxylic acid and alcohol combined.
No. of hydrogen atoms = (y + 1) + (2n+1 + 1) – 2 = y + 2n + 1
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