9709/12/M/J/13 Questions

10. A student mixed 25cm³ of 0.10moldm^–3 sodium hydroxide solution with 25cm³ of 0.10moldm^–3 hydrochloric acid and noted a temperature rise of 2.5°C. What is the enthalpy change of the reaction per mole of NaOH?

A –209 kJ mol^–1

B –104.5 kJ mol^–1

C –209 J mol^–1

D –522.5 J mol^–1

Reference: 9701/12/M/J/13 Q10

Solution:

Answer: A

Q = mcΔT, where m is mass of solution.

Q = 50 x 4.18 x 2.5 = 522.5 J mol^-1

ΔH = Q/n, where n is the no. of moles of NaOH.

No. of moles of NaOH = MV = 0.10 x 25/1000 = 0.002500 mol

ΔH = 522.5/0.0025 = 209 000 J mol^-1

Since the temperature increases, it is an exothermic reaction. Thus, ΔH = -209 000 J mol^-1 = -209 kJ mol^-1

14. In which row of the table are all statements comparing the compounds of calcium and barium

correct?

	solubility of calcium hydroxide	solubility of barium hydroxide	thermal stability of calcium carbonate	thermal stability of barium carbonate
A	higher	lower	higher	lower
B	higher	lower	lower	higher
C	lower	higher	higher	lower
D	lower	higher	lower	higher

Reference: 9701/12/M/J/13 Q14

Solution:

Answer: D

Going down Group II (magnesium, calcium, strontium, barium), the solubility of hydroxides increases and the thermal stability of carbonates increases.

Thus,

(a) solubility of calcium hydroxide < solubility of barium hydroxide

(b) thermal stability of calcium carbonate < thermal stability of barium carbonate

25. The compound shown is menthol, a naturally-occurring alcohol found in peppermint oil.

When menthol is heated with concentrated sulfuric acid it reacts. The products that form include

compound T.

What could be the structure of compound T?

Reference: 9701/12/M/J/13 Q25

Solution:

Answer: D

Concentrated sulphuric acid is a dehydrating agent. Thus, menthol undergoes dehydration to form an alkene. Thus, B is incorrect.

H is removed from the adjacent carbon atom together with the –OH group. There are two ways:

D is the only compound that is formed based on the diagram above. Thus, D is the answer. 

31.  In which pairs do both species have the same number of unpaired p electrons?

1 Al^2– and O⁺

2 N and Cl²⁺

3 C and Cl⁺

Reference: 9701/12/M/J/13 Q31

Solution:

Answer: B (1 and 2 only correct)

Statement 1:

Al^2- has 15 electrons. The electronic configuration is 1s²2s²2p⁶3s²3p³.  

O⁺ has 15 electrons. The electronic configuration is 1s²2s²2p⁶3s²3p³.  

Looking at the outer electrons only,

both Al^2- and O⁺ have three unpaired p electrons. Thus, statement 1 is correct.

Statement 2:

N has 15 electrons. The electronic configuration is 1s²2s²2p⁶3s²3p³.  

Cl²⁺ has 15 electrons. The electronic configuration is 1s²2s²2p⁶3s²3p³.  

Looking at the outer electrons only,

both N and Cl²⁺ have three unpaired p electrons. Thus, statement 2 is also correct.

Statement 3:

C has 14 electrons. The electronic configuration is 1s²2s²2p⁶3s²3p².  

Cl⁺ has 16 electrons. The electronic configuration is 1s²2s²2p⁶3s²3p⁴.  

Looking at the outer electrons only,

C has two unpaired p electrons but Cl⁺ has one unpaired p electron. Thus, statement 3 is false.

40. An organic compound, Z, will react with calcium metal to produce a salt with the empirical formula CaC₄H₆O₄.

What could be the identity of Z?

1 ethanoic acid

2 butanedioic acid

3 methylpropanedioic acid

Reference: 9701/12/M/J/13 Q40

Solution:

Answer: D (1 only is correct)

Statement 1: Ethanoic acid is CH₃COOH.

Ca + 2CH₃COOH à (CH₃COO)₂Ca + H₂

The empirical formula for the salt is CaC₄H₆O₄. Thus, statement 1 is true.

Statement 2: Butanedioic acid, HO₂CCH₂CH₂COOH.

Ca + HO₂CCH₂CH₂COOH à (OOCCH₂CH₂COO)Ca + H₂

The empirical formula for the salt is CaC₄H₄O₄. Thus, statement 2 is false.

Statement 3: Methylpropanedioic acid, HO₂CCH(CH₃)CO₂H

Ca + HO₂CCH(CH₃)CO₂H à (OOCCH(CH₃)COO)Ca + H₂

The empirical formula for the salt is CaC₄H₄O₄. Thus statement 3 is false.

9709/11/M/J/13 Questions

11. A solution of Sn²⁺ ions will reduce an acidified solution of MnO₄^– ions to Mn²⁺ ions. The Sn²⁺ ions are oxidised to Sn⁴⁺ ions in this reaction.

How many moles of Mn²⁺ ions are formed when a solution containing 9.5 g of SnCl₂ (Mr: 190) is added to an excess of acidified KMnO₄ solution?

A 0.010

B 0.020

C 0.050

D 0.125

Reference: 9709/11/M/J/13 Q11

Solution:

Answer: B

Sn²⁺ is oxidised. Thus, the half equation is: Sn²⁺ à Sn⁴⁺ + 2e^-

MnO₄^– is reduced. Thus, the half equation is: MnO₄^- + 8H⁺ + 5e^- à Mn²⁺ + 4H₂O

Hence, the overall ionic equation is: 5Sn²⁺ + 2MnO₄^- + 16H⁺ à 5Sn⁴⁺ + 2Mn²⁺ + 8H₂O

SnCl₂ produces one mole of Sn²⁺ and two moles of Cl^-.

SnCl₂ à  Sn²⁺ + 2Cl^-     

No. of moles of SnCl₂ = 9.5/190 = 0.05 mol. There is 0.05 mol of Sn²⁺ in 0.05 mol of SnCl₂. 

Since 5 mol of Sn²⁺ produces 2 mol of Mn²⁺, hence 0.05 mol of Sn²⁺ produces (0.05 x 2/5) = 0.020 mol of Mn²⁺. 

21. Lactic acid (2-hydroxypropanoic acid), CH₃CH(OH)CO₂H, is found in sour milk. Which reaction could occur with lactic acid?

A CH₃CH(OH)CO₂H + CH₃OH → CH3CH(OCH₃)CO₂H + H₂O

B CH₃CH(OH)CO₂H + HCO₂H → CH3CH(O₂CH)CO₂H + H₂O

C CH₃CH(OH)CO₂H + NaHCO₃ → CH3CH(ONa)CO₂H + H₂O + CO₂

D CH₃CH(OH)CO₂H + Cl₂ → CH3CH(Cl)CO₂H + HOCl

Reference: 9709/11/M/J/13 Q21

Solution: 

Answer: B

From the diagram above, lactic acid has two functional groups: secondary alcohol and carboxylic acid.
A is incorrect because methanol (CH₃OH) reacts with the carboxylic acid group, not the alcohol group. The correct answer should be CH₃CH(OH)CO₂CH₃.
C is incorrect because sodium hydrogen carbonate (NaHCO₃) reacts with the carboxylic acid group, not the alcohol group. The correct answer should be CH₃CH(OH)CO₂Na. (Remember that alcohol group can react with sodium, Na ONLY.)
D is incorrect because none of the two functional groups reacts with Cl₂.
B is the correct answer. 

25. Pentane, C₅H₁₂, is reacted with chlorine in the presence of ultraviolet light. A compound R is found in the products. R has molecular formula C₅H₁₀Cl₂. Each molecule of R contains one chiral carbon atom.

Which two atoms of the pentane chain could be bonded to chlorine atoms in this isomer?
A 1 and 3
B 1 and 5
C 2 and 3
D 2 and 4

Source: 9701/11/M/J/2013 Q25

Solution:
Answer: A

Having a chiral centre means that the organic compound shows optical isomerism, in which one carbon atom have four different groups of atoms attached to it, producing mirror images which do not superimpose. 

B is incorrect because the compound does not have any chiral centres.

C and D are incorrect because the compound has TWO chiral centres.
A is the correct answer. 

26. 2.30 g of ethanol were mixed with an excess of aqueous acidified potassium dichromate (VI). The reaction mixture was then boiled under reflux for once hour. The desired organic product was then collected by distillation. The yield product was 60.0%.

What mass of product was collected?
A 1.32 g
B 1.38 g
C 1.80 g
D 3.20 g

Reference: 9701/11/M/J/13 Q26

Solution:
Answer: C

Ethanol, C₂H₅OH (Mr: 46) undergoes oxidation to produce ethanoic acid, CH₃COOH (Mr: 60).
No. of mol of C₂H₅OH: 2.30/46 = 0.0500 mol
No. of mol of C₂H₅OH that forms product = 0.0500 x 60% = 0.0300 mol
Mass of CH₃COOH produced = 0.0300 x 60 = 1.80 g

. 

28. The structural formula of a compound X is shown below.

What is the name of the compound X and how does its boiling point compare with that of butanoic acid?

	Name of X	Boiling point of X
A	methyl propanoate	higher
B	methyl propanoate	lower
C	propyl methanoate	higher
D	propyl methanoate	lower

Reference: 9701/11/M/J/13 Q28

Solution:
Answer: D

Compound X is propyl methanoate. Remember the naming of esters is given by the part that shows the alkyl group (propyl) first, followed by the part that shows the acid (methanoate) group.

Compound X is lower than butanoic acid because butanoic acid can form hydrogen bonds. Thus, more heat is needed to overcome the forces of attraction. The boiling point of butanoic acid is higher.  

32. The Group II metals have higher melting points than the Group I metals. Which factors could contribute towards the higher melting points? 
1 There are smaller interatomic distances in the metallic lattices of the Group II metals. 
2 More electrons are available from each Group II metal atom for bonding the atom into the metallic lattice. 
3 Group II metals have a higher first ionisation energy than the corresponding Group I metal. 

Reference: 9701/11/M/J/13 Q32

Solution:
Answer: B (1 and 2 only are correct)

Statement 1 is correct. Interatomic distance refers to the distance between nuclei of neighbouring atoms. Since Group II metals are +2, the effective nuclear charge is higher. The attraction between nuclei and electrons is greater. Thus, more energy is needed to overcome these forces of attraction. 
Statement 2 is correct because Group II metals are +2. Hence, it can contribute to have more delocalised electrons, resulting in stronger metallic bond compared to Group I metals.
Statement 3 is false because having a higher first ionisation energy would make an atom less able to contribute to the ‘sea of delocalised electrons’ , causing a decrease in melting point. 

35. Solids W, X, Y and Z are compounds of two different Group II metals. Some of their applications are described below. 

Compound W is used as a refractory lining material in kilns.
Compound X is used as a building material. It can also be heated in a kiln to form compound Y. When Y is hydrated, it forms compound Z which is used agriculturally to treat soils. 

Which statements about these compounds are correct? 
1 More acid is neutralised by 1g of W than by 1g of X. 
2 The metallic element in W reacts with water more quickly than the metallic element in Y. 
3 Adding Z to a soil decreases the pH of the soil. 

Reference: 9701/11/M/J/13 Q35

Solution:
Answer: D (1 only)

W is magnesium oxide, MgO. X is calcium carbonate, CaCO₃. Y is calcium oxide, CaO. Z is calcium hydroxide, Ca(OH)₂.

Statement 2 is false because Mg does not react readily with water compare to Ca.
Statement 3 is false because Ca(OH)₂ neutralises the acid, thus the pH of soil increases, not decreases.
Statement 1 is true because MgO is a base but CaCO₃ is not.