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.