CAPE Chemistry Unit 1 May/June Module 1 Solutions

Added: 2026-05-19

[00:00:07]Welcome back to Chem Exam explained where the aim is chemistry clarity exam mastery. In this video we will be looking at Cape Chemistry paper 2 unit 1 module 1 2025.

[00:00:20]Let's go. Question one part A. The sketch in figure one below shows the first ionization energy of the eight elements of period 3. So here you see your graph with the eight elements across period 3 and our first question is to state two factors which influence the value of the first ionization energy. So here you are given three you can use any two one nuclear charge two atomic radi and three shielding or screening. Part two, explain how the ionization energy data provides evidence for the idea of subshells using appropriate elements from the sketch in figure two. So we'll be looking at these two anomalies here between magnesium and aluminium and phosphorus and sulfur. So in explaining the trend we'll see that the general trend is an increase in the ionization energy and that is consistent with an increase in the nuclear charge and a decrease in the atomic radi. The deviations include magnesium to aluminium. We'll discuss that one first.

[00:01:35]So the ionization of aluminium is lower than that of magnesium due to the aluminium outer electrons and this is in the 3p subshell which is higher in energy and less tightly bound than the magnesium 3s subshell. For phosphorus to sulfur, the ionization energy of sulfur is lower than that of phosphorus due to the electron electron repulsion of the fourth electron being placed in the first 3p orbital. These anomalies provide evidence that electrons occupy different subshells. Example 3s and 3p.

[00:02:17]Part B. Table one shows successive ionization energies for element D. We can see here that we are given the number of electrons removed which is 12.

[00:02:29]And we have our ionization energy. But we are now supposed to convert the ionization energy in KJ per mole to log base 10 of the ionization energies. And so once you get your calculator, you'll put in log base 10 of 690 and you'll get 2.84.

[00:02:50]You'll do the same for the remainder of the ionization energies for the electrons removed. That would complete our table. And now we are to use the data in the table to plot a graph. And that graph is on page seven. And you have to plot the log base 10 of the ionization energies against the number of electrons removed. And of course we were given the first plot. So let's get to our graph. Here you can see that we have on the x-axis the number of electrons removed and we have 12 electrons removed and on the y-axis we have log base 10 of the ionization energies. You can see that the two electrons here with the highest ionization energy are the ones closest to the nucleus. And so these would be your 1 s2 electrons.

[00:03:49]And this is your 1s orbital or subshell.

[00:03:53]Okay. The energy level of course is n equ= 1. You'll see that the eight electrons here now represent a subshell going farther away from the nucleus. And so we'll have 2 s2 and 2p6.

[00:04:11]And these electrons are in the subshell further away from the nucleus. And they would now require a lower IE when these electrons are removed. and the two. So these are the eight electrons right here. These two electrons here would now be the electrons on the outer subshell are the farthest from the nucleus. And now we have our two electrons that require the least amount of energy to remove them. And so from that we can write the configuration to be 1 s2 2 s2 2 p6 and 3 s2.

[00:04:53]So we can go back to our question. So part three we are to explain the shape of the graph. So the first two electrons are farthest from the nucleus and these are your 3s2 electrons and they belong to the same subshell or the same energy level. Okay. So once you remove those first two the these are the ones that require the least amount of energy to remove them. So now we're going to remove the third electrons which will take us into the p subshell which of course now will show a jump in the ionization energy suggesting that we are moving closer to the nucleus which is giving us evidence for shells. And so once we remove the third to the 10th electrons, we are now going to have another jump as we move to the subshell closest to the nucleus. And so we'll read for part three. The first two electrons are farthest from the nucleus in the same energy level or shell. The third 10th 10th electrons are in the lower energy level which is closer to the nucleus. The last two electrons are closest to the nucleus in a different energy level. Part four, state the electronic configuration of element D based on the shape of the graph. So that was explained earlier and we got 1 s2 2 s2 2 p6 3 s2 and this element belong to group two since it has two electrons on the outer shell. Part C. Substance S is a solid with a low melting point. A student is asked to determine the melting point of a sample of substance S by using the following apparatus and materials available in the school laboratory. A thermometer, capillary tubes sealed at one end, 500 mil beaker, bouncing burner, tripod, gauze, and tap water. See part one. outline fully the steps involved in the procedures that the students would follow in order to obtain the melting point of substance S.

[00:07:05]So the first thing we'll do is crush the solid sample and place it in the capillary tube. We'll then attach the capillary tube to a thermometer. We're then going to heat the capillary tube with a thermometer slowly in the water bath and observe the the temperature when the solid just start to melt. After that we will record the melting point. C part two. Based on the melting point data obtained, the student concludes that sulfur is a covealent compound.

[00:07:36]Further analysis of substance S reveals that it is non-polar. So the first thing we should note here being that S is covealent and non-polar, it means that it will have a low melting point. So low MP and of course it will be soluble in organic solvent but not in water. So the question want us to state whether substance S will dissolve in water or tetrachloromthane and tetrachloromthane is an organic solvent. So substance S will not dissolve in water because it is non-polar but it will dissolve in the tetrachloromthane because a non-polar solvent will dissolve a non-polar solute.

[00:08:21]C part three. It was determined that substance S is a solid with a low melting point. Describe the type of forces of attraction that exist between the molecules of substance S. Well, the fact that it is non-polar would mean that it has weak vanderwalls forces due to the temporary induced dipole dipole forces. Part D. Use the kinetic theory to explain each of the following. D.

[00:08:49]Part one, the change from a solid to a liquid state. Solid particles gain energy and start to vibrate faster overcoming the forces of attraction and the solid particles changes to liquid.

[00:09:03]Deep part two the nature of the liquid state. The liquid particles are not closely packed and are held together by weak intermolecular forces. The particles in the liquid state have more kinetic energy than in the solid state and the particles are able to move or slide past each other. D part three. The change from liquid to the gaseous state.

[00:09:29]The particles in the liquid state gain enough energy to overcome the forces of attraction and move far apart into the gaseous state. Part E. A gas syringe contains 18.2 2 cm cq of air at 55°.

[00:09:46]A volatile liquid with a mass of 0.185 g is injected into the syringe. The volume of gas in the syringe then becomes 54.5 cm cq at 55° and 1.01 * 10 5th pascal. E part one.

[00:10:05]Calculate the molar mass of the liquid given the molar gas constant to be equal to 8.314 JW per kelvin per mole. So the first thing we'll do is we would write down the information given and we have the mass to be 0.185.

[00:10:24]We have the volume which is the initial volume to be 18.2 cm cq and we have the final volume to be 54.5 cm cq. We were given the temperature to be 55° C which we must convert to Kelvin and we know the pressure to be 1.01 * 10 5th Pascal.

[00:10:45]So the first thing we do is to calculate the volume change and that is the final volume minus the initial volume and we get our volume change to be 36.3 cm cq after which we convert that to me cube using the conversion table. We then use our ideal gas equation to make mole the subject of the formula and then plug in the information that we have for pressure, volume, the molar gas constant and the temperature and calculate the moles. After we know the moles, we then use moles equal mass over mass and make molar mass the formula plugging in our mass and our moles that was calculated.

[00:11:30]And then we get our answer to be 137 g per mole. It can be worked out another way using the ideal gas equation PV= NRT. We are then going to make N be equal to mass over molar mass.

[00:11:50]And of course, we're now going to make molar mass the subject of the formula.

[00:11:55]So we'll multiply PV * the molar mass.

[00:11:59]That is equal to mass * RT. So molar mass is equal to M RT over P V. This is V volume. And so now if you do your calculation your molar mass should be very close or equal to 137 g per mole. So we can work it out more than one way using the ideal gas equation. E part two suggest which measurement used in the calculation of the molar mass is subject to the greatest error. And you can see that because we have our volatile liquid as it is evaporating the mass would change giving us various mass change and so that will give us the greatest error.

[00:12:52]This ends our question three module 3 for unit one paper 2 2025. Thank you.