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Paper 5 Tips !! :)

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waittt...u got me confused now...!!!...what do u mean?...that my method is wrong??...
I think that would be normal dilution. For serial dilution it must be that for 0.5mol/dm^3, you take 50cm^3 distilled water and 50cm^3 of the previously diluted solution (i.e. 1 mol/dm^3).
Guys guys calm down, there're two types of dilution: one's one in ten, which i'm assuming Gémeaux is talking about and one's the C1V1=C2V2.
Am i right Gémeaux?
 
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Heyy everyonee. MashaAllah all of you are doing a wonderful deed helping out so selflessly. God bless you guys. Could you help me out with this though.
Whenever we draw the graphs its not neccessary to start our axes with 0 right? I mean the graph becomes extremely awkward thennn
 
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Heyy everyonee. MashaAllah all of you are doing a wonderful deed helping out so selflessly. God bless you guys. Could you help me out with this though.
Whenever we draw the graphs its not neccessary to start our axes with 0 right? I mean the graph becomes extremely awkward thennn

It sometimes is, and sometimes isn't! Depends on what the relation is! Direct proportionality usually requires a true origin whereas others require a straight line. Trick is, visualize the 0,0 point in your head with the given relationship! Like if I make one reactant 0 will the other be 0? And base it off of that!
 
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Heyy everyonee. MashaAllah all of you are doing a wonderful deed helping out so selflessly. God bless you guys. Could you help me out with this though.
Whenever we draw the graphs its not neccessary to start our axes with 0 right? I mean the graph becomes extremely awkward thennn
I totally second that but no, it doesnt have to start from the origin. Although its better, but if your graph doesnt fit when u start from 0, then break it and make sure you take up atleast half the graph on either axis!!!
hope i helped:)
and Jazak Allahu Khair for your wonderful way of asking:)
 
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http://papers.xtremepapers.com/CIE/... AS Level/Chemistry (9701)/9701_s12_qp_51.pdf
Q1 part (c) :/ Can anyone tell me about how to make the ranges and how to calculate the molar concentrations?
Asslamu Alykum!
I'm going to explain it generally; try doing the question yourself then:)
You use the formula C1V1=C2V2.
The total volume of your solution is to be kept constant, say 50cm3. And you are to make dilutions of a given concentration, say 2.00 moldm3.
So you first decide which concentrations you want to make. marking schemes usually prefer a minimum of five concentrations, so:
1.00, 1.20, 1.40, 1.60, 1.80, and 2.00.
If you're going to make the 1.00moldm3 conc, look at the following steps:
C1V1=C2V2
C1= 2.00 (the concentraiton you're given.)
V1= x (what the volume of the new concentration will be)
C2= 1.00 (the conc you're planning to make)
V2= 50 (the total volume)
So, 2x= 1 x 50
x= 25
So now that your V2 is 25, to know what volume of water you must add to make it up to that conc, 5o-25= 25.
So your procedure to make 1.00 moldm3 from the given 2.00moldm3 would be to add 25cm3 of water into 25cm3 of acid.
Got it?
 
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Can some one list down examples of the reasons when the anomalous point is befor or after what reasons do we give??
it's different for different situations, just go through the marking schemes for different years, look at the qn , and the reason for anomalies they have posted
That's what i think at least, unless someone's got another way :)
 
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Asslamu Alykum!
I'm going to explain it generally; try doing the question yourself then:)
You use the formula C1V1=C2V2.
The total volume of your solution is to be kept constant, say 50cm3. And you are to make dilutions of a given concentration, say 2.00 moldm3.
So you first decide which concentrations you want to make. marking schemes usually prefer a minimum of five concentrations, so:
1.00, 1.20, 1.40, 1.60, 1.80, and 2.00.
If you're going to make the 1.00moldm3 conc, look at the following steps:
C1V1=C2V2
C1= 2.00 (the concentraiton you're given.)
V1= x (what the volume of the new concentration will be)
C2= 1.00 (the conc you're planning to make)
V2= 50 (the total volume)
So, 2x= 1 x 50
x= 25
So now that your V2 is 25, to know what volume of water you must add to make it up to that conc, 5o-25= 25.
So your procedure to make 1.00 moldm3 from the given 2.00moldm3 would be to add 25cm3 of water into 25cm3 of acid.
Got it?
Yes thank you :) But in the paper I have mentioned, they have asked for calculation of molar concentration of one of the ranges produced.How do we do that?
 
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Sorry to have confused you.
No, your method is not wrong but that is simple dilution. Serial dilution is slightly different.
no its alrite...i am just panicked you know a day before the exam...u don't need to be sorry...can u plzz elaborate what the difference is in serial dilution?
 
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me too stuck at this point :(
help required here

xhizors
Gémeaux
knowitall10

or any one?

JazakAllah :)
I remember doing this question orally with my teacher. I don'y precisely remeber the method but it's based on the following lines:
From the co-ordinates, you get, you have to relate it somehow to the equation given at the start of the question; you'll get a quadratic equation from which you find the value of x. i'm sorry, i'm too lazy to solve it for myself.. i was an idiot to rub it off from my paper!
 
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Can anyone please tell me about how to calculate soulbility when we have the mass of water and mass of solid in grams? And how do we calculate molar concentration/molarity?
 
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