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Chemo P5 tips

H

HarisShahzad

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When investigating how the boiling point of a solution changes with concentration, it is
convenient to represent the concentrations of the solute as a molality.

The molality of a solution is defined as the number of moles of a solute
dissolved in one kilogram of water e.g. a one molal solution has one mole of
solute dissolved in one kilogram of water.

In addition to the standard apparatus present in a laboratory you are provided with the
following materials.
100 g of distilled/deionised water (you should take particular note of this limited
supply of water)
solid potassium chloride, KCl
Give a step-by-step description of how you would
(i) prepare a series of solutions of potassium chloride that can be used in the apparatus
you have shown in (c) to give sufficient data to plot a graph as in (a)(ii),
(ii) show how you would calculate the molality of one of these solutions.
[Ar: K, 39.1; Cl, 35.5]
 
H

HarisShahzad

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Design a two-part laboratory experiment to investigate your prediction in (a).
Part 1 – solubility of cerium(IV) sulphate in water
In addition to the standard apparatus present in a laboratory you are provided with the
following materials.
60 cm3 distilled water
solid cerium(IV) sulphate, Ce(SO4)2
Give a step-by-step description of how you would
(i) prepare a saturated solution using all of the 60 cm3 of distilled water,
(ii) control the variable given in (b) (iii),
(iii) separate the saturated solution from undissolved solid,
(iv) obtain the mass of cerium(IV) sulphate and mass of water in the saturated
solution,
(v) calculate the solubility of cerium(IV) sulphate from the experimental results.



Part 2 – solubility of cerium(IV) sulphate in sulphuric acid
In addition to the standard apparatus present in a laboratory you are provided with the
following materials.
5 mol dm–3 sulphuric acid
distilled water
solid cerium(IV) sulphate
Give a step-by-step description of how you would prepare a range of solutions of
sulphuric acid of different concentrations. These solutions could then be used to
investigate the solubility of cerium(IV) sulphate in different concentrations of sulphuric
acid. See section (e).
Your plan should include details of the following.
(i) the number and concentrations of the solutions to be prepared
(ii) the volumes of acid and water used to prepare the solutions
(iii) the apparatus and method you would use in their preparation
 
N

nealDSA

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Hey Everybody !! I've A Question .. In Planning And Designing Question Of Oct/Nov & May/June - 2006 There Are Question Wherein We Have To Determine Which Of The Compound Is Being Used Through Calculation :%) ??!!!

Pls Look At Ques - 2 Of Both The Attachments .. Q2 - e ) For The Jun 06 Paper & Q2 - c ) For The November Paper

PLS HELP PEOPLE :Bravo:
 
A

abrraza

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zeebujha said:
When asked "Which equipment should be changed so that the accuracy is improved the most?", always choose the equipment which has the highest percentage error in the experiment.
Click to expand...
can we have precentage errors approx. of the most used appratusss..
 
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MukeshG93

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Hey Zeebu! Know me? Lol, I finally made an account, I figured enough procrastination had been done and I need to get down to business. Tomorrow's Mechanics paper will be no sweat but I am still a little iffy about P5. I read all of the posts on this thread and I doubt roughly about 23% (I actually counted and divided by total) of the tips would not come in handy. I mean, not to be too critical but, sandpapers, measuring the enthalpy change of combustion and what not, I don't know, it seems pretty unrealistic to me, do you really expect such things to come in the exam. If so, my house is tall enough for me to crack some bones if I jump off of the terrace, hence sparing me the terror of the pending unpredictable examination. Nevertheless, good points, I enjoyed reading through and I learned a lot. Ask me if I can help in any way. Cheers! :D
 
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MukeshG93

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abrraza said:
zeebujha said:
When asked "Which equipment should be changed so that the accuracy is improved the most?", always choose the equipment which has the highest percentage error in the experiment.
Click to expand...
can we have precentage errors approx. of the most used appratusss..
Click to expand...

Percentage error = (least value that can be read)/(value of measurement) x 100
So, the percentage error is less if the least quantity that can be read is smaller, and it is even less if the quantity you are measuring is bigger. So, for bigger measurements you can use instruments that don't measure so finely, but for smaller measurements please go for pipettes, micrometre screw gauges and digital balances calibrated to 3 decimal places. :)
PROTIP: If the instrument is used to measure two points on the measure, i.e. like a metre rule, the percentage error is x2.
 
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abrraza

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MukeshG93 said:
Hey Zeebu! Know me? Lol, I finally made an account, I figured enough procrastination had been done and I need to get down to business. Tomorrow's Mechanics paper will be no sweat but I am still a little iffy about P5. I read all of the posts on this thread and I doubt roughly about 23% (I actually counted and divided by total) of the tips would not come in handy. I mean, not to be too critical but, sandpapers, measuring the enthalpy change of combustion and what not, I don't know, it seems pretty unrealistic to me, do you really expect such things to come in the exam. If so, my house is tall enough for me to crack some bones if I jump off of the terrace, hence sparing me the terror of the pending unpredictable examination. Nevertheless, good points, I enjoyed reading through and I learned a lot. Ask me if I can help in any way. Cheers! :D
Click to expand...

tell something about enthalpy change of neutralization plzzz
 
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MukeshG93

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abrraza said:
tell something about enthalpy change of neutralization plzzz
Click to expand...

Enthalpy change of neutralisation, basically, is the energy lost (usually gained) when 1 mol of acid neutralizes 1 mol of base. It is different for every acid and every base. If you are to find the enthalpy change of neutralisation of something, use the standard plastic cup in beaker apparatus with a known amount of the substance in about 90% aqueous solution (this ensures that the specific heat capacity does not change much and remains close to that of water). Then, mix the appropriate amount of neutralising substance, preferably in excess and stir with the thermometer and record the maximum temperature. Use the change in Q = mcT and divide Q by the number of mols of the substance used and VIOLA, you got it! :D
 
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MukeshG93

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nealDSA said:
Can Someone PLS EXPLAIN HOW TO USE THE FORMULA " C1V1=C2V2 " w.r.t Some Data ?? :%) :oops:
Click to expand...

Dude, firstly, DO NOT PANIC! This is just an exam and usually you just need like 17 marks out of 30 to get an A so you will always be in the safe zone AS LONG AS YOU DO NOT PROCRASTINATE (which is hardly of the matter here because the exam's tomorrow). Anyways, getting to the point, when you mix a fraction of a solution of known concentration with water, the concentration decreases. Now, this decrease in concentration, or should I say, the final concentration is determined by how much of the solution you mix with how much water. If you had taken Biology, this task would have been so much easier. Anyways, the detour to this is: Percentage of original solution in the final solution = Percentage of original concentration in the final concentration

So, if you have an original solution of concentration 1.0 mol/dm3 and you are to make a final solution of 0.5 mol/dm3, here is how you do it. 0.5 mol/dm3 is 50% of the original concentration. So, the final solution should include 50% of the original solution and consequently 50% distilled water. Hence, if the final solution is supposed to be, say, 100 cm3, make 50% of it out of the original solution (50 cm3) and 50% of it distilled water (50 cm3). Similarly, if you are to make something like 0.6 mol/dm3 (Which is 60%), you use 60% of the original solution (60 cm3) and 40% of the original solution (40%). Hope you are good with math!) :D
 
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abrraza

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MukeshG93 said:
abrraza said:
tell something about enthalpy change of neutralization plzzz
Click to expand...

. Use the change in Q = mcT and divide Q by the number of mols of the substance used and VIOLA, you got it! :D
Click to expand...
what do u mean by .. ''no of moles of substance used? :%)
 
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