CAPE Chemistry June 2021 Unit 1 Paper 2 - Past Paper Solutions

 Global Chemistry Lessons

Call or WhatsApp 739-2656 to join our class

L.P. #307 S.M.R. Cunupia

Notes relevant to each question follow the solutions.

Question 1 Notes:

1(e)    It could be that CXC has obtained permission to use past exam questions from         

          different examining bodies. Here is the source of this         

          one: https://capechemistry.blogspot.com/2021/12/source-of-cape-chemistry-june-2021-u1.html

Source Of CAPE Chemistry June 2021 U1 P2 Q1(e)

 Global Chemistry Lessons

Call or WhatsApp 739-2656 to join our class

L.P. #307 S.M.R. Cunupia

 

The question below was seen before, in a 2005 paper from another examining body.

Its mark scheme follows. The only significant difference is "NaoH" in part (iv) of the CAPE exam vs. "NaOH" in the corresponding part of the original question.

CAPE Chemistry June 2009 U1 P1 - Answers And Explanations

 Global Chemistry Lessons

Call or WhatsApp 739-2656 to join our class

L.P. #307 S.M.R. Cunupia

 

Download Paper Here

1. B

2. D

3. A

4. D

5. C

6. D

7. B

8. C

9. C

10. B

11. C

12. D

13. C

14. B

15. D

16. A

17. B

18. C

19. B

20. A

21. D

22. B

23. A

24. C

25. B

26. A

27. D

28. C

29. B

30. B - Detailed solution below.

31. A

32. B

33. D

34. B

35. C

36. A

37. D

38. C

39. D

40. A

41. D

42. A

43. D

44. C

45. B

Question # 30 - Detailed Solution

CAPE Chemistry Past Paper Answer - 2016 Unit 2 Paper 2 Question 5

Global Chemistry Lessons

Call or WhatsApp 739-2656 to join our class

L.P. #307 S.M.R. Cunupia

5(a)(i)

As applied to completely miscible binary systems, this states that the partial pressure of a constituent of a binary mixture at any given temperature is equal to the normal vapor pressure of that constituent at the stated temperature multiplied by the mole fraction of the constituent in the mixture.

5(a)(ii)

1. The interactions between components are similar to those in pure components.

2. There is no volume change on mixing the components.

5(b)(i)

An azeotropic mixture is one which boils or distils without change in composition, and in general it has a boiling point higher or lower than that of any of its pure constituents.

5(b)(ii)

An azeotrope is not a compound because its composition varies with pressure.

5(b)(iii)

Mixture X boils at a temperature T1 and at equilibrium gives off a vapor of composition Y richer in the more volatile component A. Continued distillation produces the azeotrope as the distillate while pure B is the residue.

5(c)

At equilibrium, mass of compound in water = (5-x) g and mass of compound in solvent = x g. 

Partition coefficient = concentration in water/concentration in ether, therefore  ((5-x)/100)/(x/25) = 0.2 i.e. x = 2.8 g.

Unit 1 Module 1 SS 2 Notes

Global Chemistry Lessons

Call or WhatsApp 739-2656 to join our class

2.1

State the various forces of attraction between particles.

Ionic bonds, covalent bonds, metallic bonds, van der Waals' forces.

Both bonding (intramolecular) forces and intermolecular forces arise from electrostatic attractions between opposite charges. Bonding forces are due to the attraction between cations and anions (ionic bonding), nuclei and electron pairs (covalent bonding), or metal cations and delocalized valence electrons (metallic bonding). Intermolecular forces, on the other hand, are due to the attraction between molecules as a result of partial charges, or the attraction between ions and molecules. The two types of forces differ in magnitude, and Coulomb's law explains why:

• Bonding forces are relatively strong because they involve larger charges that are closer together.
• Intermolecular forces are relatively weak because they typically involve smaller charges that are farther apart.

Ion-Dipole Forces

When an ion and a nearby polar molecule (dipole) attract each other, an ion-dipole force results. The most important example takes place when an ionic compound dissolves in water. The ions become separated because the attractions between the ions and the oppositely charged poles of the H2O molecules overcome the attractions between the ions themselves.

Dipole-Dipole Forces

When polar molecules lie near one another, as in liquids and solids, their partial charges act as tiny electric fields that orient them and give rise to dipole-dipole forces: the positive pole of one molecule attracts the negative pole of another (diagram below).

Polar molecules and dipole-dipole forces. In a solid or a liquid, the polar molecules are close enough for the partially positive pole of one molecule to attract the partially negative pole of a nearby molecule. The orientation is more orderly in the solid (left) than in the liquid (right) because, at the lower temperatures required for freezing, the average kinetic energy of the particles is lower. (Interparticle spaces are increased for clarity.)

For molecular compounds of approximately the same size and molar mass, the greater the dipole moment, the greater the dipole-dipole forces between the molecules are, and so the more energy it takes to separate them. Consider the boiling points of the compounds in the next diagram. Methyl chloride, for instance, has a smaller dipole moment than acetaldehyde, so less energy is needed to overcome the dipole-dipole forces between its molecules and it boils at a lower temperature.

Dipole moment and boiling point. For compounds of similar molar mass, the boiling point increases with increasing dipole moment. (Note the increasing color intensities in the electron density models.) The greater dipole moment creates stronger dipole-dipole forces, which require higher temperatures to overcome.

The Hydrogen Bond

A special type of dipole-dipole force arises between molecules that have an H atom bonded to a small, highly electronegative atom with lone electron pairs. The most important atoms that fit this description are N,0, and F. The H-N, H-O,
and H- F bonds are very polar, so electron density is withdrawn from H. As a result, the partially positive H of one molecule is attracted to the partially negative lone pair on the N, 0, or F of another molecule, and a hydrogen bond (H bond) forms. Thus, the atom sequence that allows an H bond (dotted line) to form is -B:····H-A-, where both A and B are N, O, or F. Three examples are

The small sizes of N, O, and F are essential to H bonding for two reasons:

1. It makes these atoms so electronegative that their covalently bonded H is
    highly positive.


2. It allows the lone pair on the other N, O, or F to come close to the H.

The Significance of Hydrogen Bonding

Hydrogen bonding has a profound impact in many systems. Here we'll examine one major effect on physical properties and preview its enormous importance in biological systems.

Why Warm Water Freezes Faster Than Cold

Nearly 50 years ago, Erasto B. Mpemba and Denis G. Osborne reported that if samples of water at 90 °C and 25 °C are cooled, the one starting at 90 °C begins freezing first. Many explanations for the “Mpemba effect” have been proposed, including ones based on evaporation, temperature gradients, impurities, and dissolved gases.

In warm water, weak hydrogen bonds break (top, red squiggles), leaving fragments

that easily reorganize into an ice lattice (bottom), a new study says.

A new computational study suggests that the effect arises from the liquid’s hydrogen bond network (J. Chem. Theory Comput. 2016, DOI: 10.1021/acs.jctc.6b00735). Southern Methodist University’s Dieter Cremer and colleagues investigated clusters of 50 and 1,000 water molecules, characterizing the types and strengths of the clusters’ 350 and more than 1 million hydrogen bonds, respectively. In (H2O)1,000 , raising the temperature from 10 °C to 90 °C led to fewer hydrogen bonds, as weaker, predominately electrostatic bonds broke.

That left behind cluster fragments with strong hydrogen bonds with more covalent character and proportionately more “dangling” or terminal hydrogen bonds. That hydrogen bond combination enables the fragments to easily reorganize and form the hexagonal lattice of ice.

Apart from learning what the name of this effect is and why it occurs, you can now answer two unit 1 past paper questions with the knowledge that:

1. Hydrogen bonds are largely electrostatic in nature.

2. Each water molecule forms four hydrogen bonds (from diagram).