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A LEVEL CHEMISTRY - PRACTICALS

Measuring Mass:

Using the Electronic Balance:

• Before measuring any mass, press the “tare” button to make sure that the intial reading is zero.
• Make sure that the electronic balance is clean with no residues on it. Blow lightly if you feel there are any.
• The precision of the electronic balance is up to 2 decimal places e.g. 13.45 g.
• Units: Electronic balance measure in grams, so always write units (“g”, “/g”, or “grams”) beside every reading.
• All recordings should be made in a single table. Otherwise your 1 Mark will be on stake.
• As already stated, the headings and their units have separate marks. Write mass/g or mass (g) but m/g or mass g are not allowed. Full name of quantity must be used.

Below is a table that may be used for recording Masses:

	Exp. 1	Exp. 2
Mass of empty test-tube / g
Mass of empty test-tube + Sample X / g
Mass of test-tube + residual Sample X / g
Mass of sample X used / g

Measuring Temperature:

There’re only a few tips relating to this measurement. First, always stir the mixture before recording the temperature. Second,  when measuring temperature, make sure that the thermometer is not taken out from the solution.

A LEVEL CHEMISTRY - PRACTICALS - TITRATION

Burette can be started from any point. The first titre must be from about 2­6 cm³ because we don’t know how much

volume will be needed. but rest of them can be started from any value which allows you to measure the volume. Let’s say that rough titre was 23 cm³ you can start the next titration from 15 cm³ because you know 15+23=38 which is inside the 50 cm³ range of burette. But if the rough titre is 38 cm³ and you start from 15 cm³ then you will be in trouble.

The two best titre must be consecutive as well as within 0.1 cm³. If first titre was 23.40, second was 23.10 and third was 23.50 then first and second can not be best titres. the reason is obvious. the second titre was more accurate than first and third.

If first two titres are within 0.1 cm³ then there is no need for third titration. If you think that any reading is not correct because it is out of pattern or too far away from line of graph circle that point on graph, again take that reading on the same table but on new row, and then plot new point.

All readings on table must be consistent. All your burette readings must be given to the nearest 0.05 cm³. For example if your volume comes out to be 24.1 cm³, then don’t write it as 24.1 cm³ rather write it as 24.10 cm³.

Always beware of endpoints. Whenever you know the end point is near, add the solution in burette drop by drop and close tap when the end point is reached without overly-titrating. 

All readings should be in a single organized table. 1 MARK would be deducted otherwise. An example of a good table for recording readings while performing titrations is given below (you may memorize this table!) :

	1	2
Initial Burette Reading / cm3
Final Burette Reading / cm3
Titre / cm3
Best Results (✓)

In the table, the headings and their units have separate marks. Write Temperature/C or temperature (C) but T/C or temperature C are not allowed. Full name of quantity must be used. Writing V for volume might be acceptable but ‘T’ is not acceptable as it may have been used for temperature or for time.

A LEVEL CHEMISTRY - ENTROPY

Short analysis on entropy to help predict the signs (+/-) of entropy change for a reaction.

ENTROPY PROBLEMS

Determine if the entropy change will be positive or negative for the following reactions:

A) (NH₄)₂Cr₂O₇(s) → Cr₂O₃(s) + 4 H₂O(l) + CO₂(g)

B) 2 H₂(g) + O₂(g) → 2 H₂O(g)

C) PCl₅ → PCl₃ + Cl₂(g)

SOLUTION

Entropy of a reaction refers to the positional probabilities for each reactant. An atom in gas phase has more options for position than the same atom in a solid phase. This is why gases have more entropy than solids.

In reactions, the positional probabilities must be compared for all the reactants to the products produced.

If the reaction involves only gases, the entropy is related to the total number of moles on either side of the reaction. A decrease in the number of moles on the product side means lower entropy. An increase in the number of moles on the product side means higher entropy.

If the reaction involves multiple phases, the production of a gas typically increases the entropy much more than any increase in moles of a liquid or solid.

Reaction A

(NH₄)₂Cr₂O₇(s) → Cr₂O₃(s) + 4 H₂O(l) + CO₂(g)

The reactant side contains only one mole where the product side has six moles produced.

The was also a gas produced. The change in entropy will be positive.

Reaction B

2 H₂(g) + O₂(g) → 2 H₂O(g)

There are 3 moles on the reactant side and only 2 on the product side. The change in entropy will be negative.

Reaction C

PCl₅ → PCl₃ + Cl₂(g)

There are more moles on the product side than on the reactant side, therefore the change in entropy will be positive.

ANSWER:

Reactions A and C will have positive changes in entropy.
Reaction B will have negative changes in entropy.

A LEVEL CHEMISTRY

TIPS FOR CHEMISTRY PAPER 5

When asked to draw a diagram, ALWAYS mention the volume of the apparatus being used. The one most often used is gas syringe. It is better to limit the volume of syringe to less than 500 cm3 . Also, the volume of a commonly used small test tube is around 16 cm3 and that of a boiling tube is about 25 cm3, volume of beaker 250cm3, conical flask 100cm3 and burette 50cm3.

 We always have to keep the volume of the apparatus we are using while choosing the volume of the solution for the planning. Saying that we are going to place 50cm3 of aqueous HCl in a test tube is of course not going to please the examiners.However, we can use greater volume if we state that the volume of the beaker is 400cm3 or 500 cm3 or so on. So, stating the volume of the apparatus is very very important

Always while measuring volume of solutions, use burette or pipette because they have low PERCENTAGE error. The question of percentage errors are usually asked in P5. The percentage error becomes very unacceptable if we are measuring SMALL volumes of solution or small masses. So, a 3dp balance is much better for measuring masses than a 2dp balance as it would have much less % error when small masses are being measured

While heating crystals strongly, do not use a simple test tube or any apparatus with sharp edges as they are liable to crack at the sharp edges. For very strong heating to a constant mass, a crucible placed on a pipe clay triangle is appropriate. And of course to measure the mass of the crucible + solid, you would have to place it on a balance. But we wouldn't want to fry our poor balance by placing a very hot crucible on top of it. So, allow the crucible to cool for a few minutes by placing it on a heat mat.

When using a magnesium ribbon or any dirty surface, clean it with SANDPAPER

How to prepare crystals (which is also linked with solubility)
So, we have to prepare a saturated solution first:
1. Take a fixed volume of water in a beaker of appropriate volume
2. Add the crystal to the water and stir continuously. You have to allow some time for the crystal to dissolve as it is an equilibrium process
3. After 5 min of stirring , if no solid crystals appear, add further mass of crystal
4. And repeat the process until solid appears in the beaker
5. Filter the solution using a filter paper and funnel so that the saturated solution is collected in a beaker(whose mass has been measured previously) beneath the funnel
Now we have a saturated solution in a beaker.

How to get the crystals:
1. We have to place the beaker in a warm water bath. We could use a burner as well but there is a risk of overheating the solution which could decompose the crystal. If the heat is appropriate, the water of the solution should evaporate and we should have dry crystals ready.

How to measure solubility:
6. Measure the mass of beaker + solution from step 5
7. Subtract the mass of beaker from the mass in 6 to get the mass of saturated solution
8. And evaporate as shown above to get the mass of crystals.
8. Measure the mass of crystal + beaker
9. Subtract mass of beaker from mass in 8 to get the mass of crystals
10. Subtract the mass of crystals from the mass of saturated solution to get the mass of water in the solution
11. I have assumed that all the masses are in grams. So, to get the solubility: Mass of crystal x 100/Mass of water

When you have to remove moisture from :
1. Surface:
Wash the surface with a stream of propanone. The water gets dissolved in the propanone and repeat it multiple times. Then gently heat the surface to evaporate the propanone from the surface

2. Vapour:
Use dessicants such as:
1. ANHYDROUS sulphuric acid
2. ANHYDROUS calcium chloride
3. Silica gel
You have to pass the vapour from the beaker containing the dessicant

Also, a useful property of soda lime is that it absorbs BOTH water vapour and carbon dioxide

We are also regularly asked to measure enthalpy changes
We use a plastic cup and thermometer for this purpose, however this has many disadvantages (asked regularly) and here are some of them with the required measures:

1. Heat loss to the surroundings from the beaker:
To avoid this:
a. Cover the plastic cup with a lid
b. Place the cup in a beaker. The air in the beaker acts a good insulator.
c. Use multiple cups so as to thicken the lateral layer of plastic

2. Instability of the cup
a. Place the cup in a glass beaker

3. For exothermic reactions, spray of the solution very likely
a. Use a large beaker to carry out the experiment rather than the small plastic cup (which has a small volume)
b. Put a lid on the top of the beaker (this only MINIMISES the spray, doesn't prevent it totally)

4. When we are heating a volume of water in glass beaker, there are two cases of heat loss to consider:
a. Heat loss from the beaker of water
b. Heat loss from the burner heating the volume of water

Titration is accurate because:
1. Standard solution of acid/base is used
2. we obtain concordant titres
3. % error in pipette and burette is very small
4. The end point of a titration is sharp

We do not need to measure the mass of a reagent or volume of reagent that is in excess

If the percentage difference between the measured value and the true value is more than the maximum apparatus error, this means the ex foperimenter's technique needs modification. If not, the error could be considered to be entirely due to the intrinsic error or the apparatus used

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