PAPER 5

Year Paper Question Remark2010 (MJ)

MJ(51) 2 (d)3 (e), (f)

MJ(52) 1 a(i), c, d2010 (ON)

ON(51) 1, 2(F)

2015(ON) ON(51) 2 (e)i, (f)ON(53) 2

2010 M/J(51) 2 (d)

I'm sure you could easily draw the graphs.....

The tricky thing is.... what do they mean?

We are told.... "The graph you have drawn for ethanol and cyclohexane is typical of some non-ideal mixtures of liquids." This means there is an imbalance between the energy of bonds breaking in pure ethanol and pure cyclohexane, compared to the bonds between ethanol and cyclohexane.

We can see that wnen ethanol is introduced into the cyclohexane, the boiling point goes down. This means LESS ENERGY is NEEDED to beak the conds in cyclohexane. This must mean that the bonds in the mixture become weaker, the molecues (ALL the molecules) do not bond as strongly as before. This means that we have replaced stronger bonds in the pure state with weaker bonds in the mixture. Going from strong bonds to weaker bonds must be an endothermic process (imagine if we made stronger bonds in the mixture, that would have given out energy. So if we do wht opposite, i.e. make weaker bonds that must absorb energy). Ethanol is preventing the cyclohexane molecules from engaging in London dispersion forces amongst themselves as much, and the cyclohexane is disrupting the ethanol molecules h-drogen bonding with each other.

2010 M/J(51) 3 (e), (f)

mass of mass of crucible mass of crucible and lid mass of

crucible and lid + magnesium mass of magnesium

and lid + magnesium oxide magnesium oxide

student / g / g / g / g / g1 25.37 26.62 27.50 1.25 2.132 25.18 27.01 28.19 1.83 3.013 25.44 27.73 29.19 2.29 3.754 25.26 27.71 24.96 2.45 -0.305 25.39 28.11 29.84 2.72 4.456 25.04 27.89 28.54 2.85 3.507 25.13 28.08 29.93 2.95 4.80

Whenever 'error in data' is being quizzed, make a statement about what the difference between that data point and the rest of the data.

Student 6's mass is lower relative to the other data points. That means some mass was lost during the experiment. It could have been that the lid was left off when heating the Mg so a lot of small particles of MgO (the white smoke) was lost, hence mass of MgO is lower than it should have been.

e) A LOT of mass was lost, over 2g !!! The mass was negative! It is probable that the student left the lid off when doing the final weighing.

Some N compound must have formed. The high temperatures at which Mg burns in oxygen can supply enough energy for N2(g) in the atmosphere to react too. Mg(NN)2 was probably formed (in small but detectable amounts !!). This means the compound formed isn't purely MgO so the method wasn't really that accurate. Confidence in the method is questionable..

2010 M/J(52) 1 a(i), c, d

a (i) We are told.... "When the liquid contains a dissolved solid (solute), the vapor pressure above the liquid is reduced." Hence it is reasonable to assume the vapor pressure will reduce further when more KCl(aq) is dissolved

2010 M/J(52) c

Suitable apparatus would be that which is used in distillation. Here's a reasonable diagram from https://www.tes.com/teaching-resource/distillation-6168252 . If the liquid was a volatile liquid a different boiling point technique could have been used (which is in mentioned in the Planning exercised document I'm still working ion).

The heat should be a heating mantle with temperature control and NOT a Bunsen burner which will defy the questions instruction NOT to allow overheating.

The boiling point is recorded from reading the thermometer when a constant steady continuous set of distillate condensate come off.

2010 M/J(52) d

You could do this the usual way - split the water into 15ml (which will give you 6 solutions and a little bit left over). In each 15ml, dissolve different amounts of salt. Say 1g, 2g, 3g, 4g, 5g, 6g. The molality can be calculated. moles of KCl = x/74.6. If this amount of moles is dissolved in 15ml, then if we had enough of these solutions ot make 1 L (which would require 1000/15 of these solutions) then the molality would be x/74.6 x 1000/15 where x is the mass of KCl used. Put one of the solution in the round bottomed flask and heat until boiling, and record it's temperature. Wash round bottomed flask with ordinary water and allow to dry. Repeat for each solution.

OR you could use a large amount of water e.g. 80g of water and make a solution of a certain modality. You could reuse the distilled water to make other solutions, as the distilled water should be pure.

2010 O/N) (51) 1

The whole question guys??????

Pre-amble...

Assuming 1L volumes (to make the math and discussion easier), in 0.10 mol dm-3 lead nitrate solution, there is 0.1 moles of Pb. It will require 0.2 moles of NaCl2 to form the maximum amount of PbCl2(s)

Or,

Equal volumes of a 0.20 mol dm-3 solution of NaCl with 0.10 mol dm-3 Pb(NO3)2 solution will give all the Pb <should precipitate

(a) The precipitate will increase as more moles of NaCl is added. Due to the 1 Pb(NO3)2 to two NaCl ration, half the number of moles of PbCl2 form for each (fraction of a) mole of NaCl added.

(b) I am really unsure what exactly this question is asking for due to the extremely brief and vague nature. I guess as you are able to keep adding NaCl, that it's not NaCl that's limiting, but the amount of moles of Pb in the lead nitrate solution.

(c) Interdependent and dependent variables are really really common Paper5 questions. The dependent one is the RESPONS to some Input/initial conditions you set-up. So dependent variable is the amount of Pb nitrate formed, the independent one is the vol (hence moles) of NaCl(aq) added. We are dealing with a sparingly soluble salt i.e. an eqm system so we need to fix the temperature.

Mr of Pb(NO3)2 = 207 + ( 14+16x3) x 2 = 331 g.

In 1 L of solution you need 33.1g to make a 0.1 mol dm-3 solution.in 250 cm3 you need (33.1/4) g = 8.275 g. There will be 8.275/331 moles of lead nitrate in the solution. Weight out that mass and using a clean funnel transfer into a clean 250cm3 volumetric flask. Rinse weighting container and transfer washings into the flask. Shake. Add up the 250ml mark and stopper the flask. Invert several times holding the stopper in place.

Pipette 20cm3 of the lead nitrate solution into a small clean beaker. This will contain 0.025 * (20/250), i.e. 0.002 moles of Pb(NO3)2 and will require 0.004 miles of NaCl. As concentration of NaCl is 0.20 mol dm3 the volume needed to extract all precipitate should be vol = moles / concentration = 0.004 / 0.20 = 0.02 L of NaCl = 20cm3 (use a burette to dispense the NaCl solution). Filter the ppt formed on a pre-weighed filter-paper , wash the filtered ppt ith a small amount of ice cold water, dry in a warm oven. Find the mass of dry ppt and filter paper. Subtract mass of filter paper.

Mass of

Mass of filter paper Mass of

Vol of Vol of filter paper

and dry ppt dry ppt

Pb(NO3)2 (aq) NaCl(aq) /g /g /g

Expt used / cm3 used / cm3 (A) (B) (B) - (A)

2 20 5 3 20 10 4 20 15 5 20 20 6 20 25

Shaded row = the one expected to give max ppt for min vol of NaCl(aq)

(f) One can ensure It’s thoroughly dried when cycles drying the ppt and weighing gives a constant mass.

2010 O/N) (51) 2(F)

The number of 'units' in NaCl is two. Whereas in a molecule of glucose, there is only one. Doubling the amount of particles, doubles the effect. This bit of physical chemistry is called 'colligative properties'

2015 O/N) (51) 2 (e)i

Common with enthalpy experiments, heat loss is always an issue. In this experiment very little precautions are done to prevent heat loss. E.g. use a lid.Results are more inaccurate as they reach their maximum because the rate of heat loss (and cumulative heat loss) is greatest. Heat flow is greater between larger heat differentials.

2015 O/N) (51) 2 (f)

The curve should be lower. Ethanoic acid is a weak acid. Enthalpy is required to break the O-H bond in the weak acid, hence less energy in total is lost.

2015 O/N (53) 2 - The whole question again???

DO NOT USE “square brackets, [ ], as they mean concentration. In Kp we don’t use concentration values, but partial pressures instead.

[NO2(g)]2 / mol2 dm–

6 [N2O4(g)] / mol dm–

3NO2(g)] / mol dm–

3 ( raw values ) to 3sf0.900 0.0729 0.0053144100 0.005310.800 0.0687 0.0047196900 0.004720.700 0.0643 0.0041344900 0.004130.600 0.0595 0.0035402500 0.003540.500 0.0548 0.0030030400 0.003000.400 0.0486 0.0023619600 0.002360.300 0.0390 0.0015210000 0.001520.200 0.0344 0.0011833600 0.001180.100 0.0243 0.0005904900 0.000590

Choose points ON THE LINE OF BEST FIT. Best to use at a minimum of 75% of the line between the points. Graph obviously should start from zero, but it’s not a “point” - I’m lazy to tell the computer to start from there. You can usually use 0,0 as a starting position for measuring gradients.

I’m not going to calcualte the value here, but you divide the difference of the y-axis squared term points, by the corresponding difference of the x-axis N2O4 term.

You can tell from the graph that the point at 0.300 is obviously anomalous. The partial concentration of NO2 is lower than expected. What would cause it to lower? Equilibrium hadn’t yet been reached? Condition of the reaction acting on the equilibriom may have changed therefore a new eqm value may have resulted

If T was raised, you can usr the LeChatelier trick to find out what happenss to the system, although it is NOT a correct explanation of why it happens., it merely gives you the right direction.

If T increases, the “eqm” will shift to oppose the increase. It will move to the endothermic side, hence [N2O2(g)] will decrease. Hence at eqm, the numerator must be smaller and/or denominator bigger, so the vaue of the gradient must decrease, hence draw a curve with lower gradient.

Only temperature changes the value of the equilibrium constant. On increasing the pressure of the eq-m mixture, the system hs moved away from eqm as the partial pressure of the NO2^2 increases a lot. Hnce the eqm tries to get back to equlibrium.

By the way, when at eqm the temperature changes. The system is also no longer at eqm and the eqm shifts to try and return to equilibrium BUT it tries to get back to a NEW EQM VALUE.

Last part….

Each mole of NO2 that formed came from half a mole of N2O4. So 0.0729 moles NO2 came from 0.0729/2 moles of N2O4. And there is 0.9 moles of unreacted N2O4 left at eqm, so the initial amount must be (0.0729/2) + 0.9