Re: How can I write this experiment?

Dorea,
     The simplest way to determine the concentration of the solutions 
would be to weigh a known volume of the solutions, allow the water to boil 
away and weigh the solid sucrose at the end.  This gives the most accurate 
result for concentration. However, since you seem to want to solve the 
problem using diffusion and osmosis here is a simple way to do it. 

Osmotic pressure: I dont know how familiar you are with osmosis but let me 
explain some basics anyways.  Consider a U tube shown below containing two 
concentrations A and B of sucrose solution in its two arms. The volumes of 
the two solutions are "equal".  These solutions are separated by a 
semipermeable membrane.  This is a membrane usually made of cellophane 
that can allow water to pass through but not sucrose.  This is because the 
pores in the cellphane are too small for the big sucrose molecules to pass 
through.  Assume that con(A) < con(B) i.e. A has more water than B. It is 
more dilute.  Because of this some water will permeate through the 
cellophane to the B-side of the U-tube. This will result in an increase in 
the level of the liquid on the B-side of the two. Thus on simply looking 
at the liquid levels in the two arms after allowing sufficient time for 
equilibrium to be established between the two liquids, you can tell which 
is the more concentrated solution. In this case since the level of the B-
side increases, B is more concentrated than A.


        
| |    | |         
| |    | |
| |    | |
|_|    |_|
| |A   | |B
| |    | |
\ \____/ /
 \__||__/      
    
Semipermeable membrane = Cellophane. This is shown as two vertical limes 
at the bottom of the U-tube. 

| |    | |
| |    | |
| |    | |
| |A   |_|B
| |    | |
|_|    | |    
\ \____/ /
 \__||__/      
 
Having said that why does water diffuse from the less concentrated to the 
more concentrated solution? This seems to be against a concentration 
gradient!! This is because of a phenomena called osmotic pressure.  
You may be aware of the gas law,

PV=nRT
Now, if we divide both sides of the equation by V we get

    P = cRT

where c = n/V .  This is the concentration in moles/liter (M in your 
problem). R is the universal gas constant and is equal to 0.08205 liter 
atm K-1 mol-1. (Here K represents Kelvin (unit of temperature))
T is the temperature. Thus if you have a concentration of 0.2M then the 
osmotic pressure is 
       0.2*0.08205*298K = 4.9 atm.  

For a 0.4M solution, 
P = 0.4*0.08205*298K = 9.8 atm.  

So you see that as the concentration is double the pressure is also 
doubled. So if you have these two solutions separated by the semipermeable 
membrane, water from the more dilute solution will diffuse towards the 
more concentrated solution with the aim of diluting that until the 
concentration become equal in the two arms.  Since we started out with 
equal volumes of A and B solutions, as water diffuses to B from A, the 
level on the B-side will increase. So you can tell which solution is more 
concentrated.

Note: This increase in the level of B occurs only when the membrane is 
completely "impermeable" to sucrose and completely "permeable" to water. 
If both sucrose and water can pass through the membrane then the overall 
concentration of the solution will be 0.3M and the levels in the two arms 
will be the same because of the age old truth  "Liquids find their levels".

For your experiment, you need to make four measurements, one for each 
concentration. In each case you must have the sucrose solution in one arm 
and plain water in the other. The arms are then closed by pistons to which 
are attached pressure gages. As soon as the solution is introduced and the 
pistons are in place make sure that the pistons dont move up or down.  The 
level of water will go down and the sucrose solution will exert a pressure 
on the piston with which it is in contact. This will show up as a pressure 
readin on the gage. This is the osmotic pressure. From the pressure and 
knowing that you are carrying out the experiment at room temperature, you 
can figure out the concentration of the sucrose solution.

  You can simplify the experiment by just using a U tube with a 
semipermeable membrane and no pistons or pressure gages if you know the 
concentrations of the solutions before hand but do now know which flask 
contains which solution as follows. Use the following pairs of solutions 
on the two arms of the balance each time.

A and B
A and C
A and D
B and C 
C and D.

Thus you would have to do five separate runs. In the first case with the 
Utube containing A and B, if A is more concentrated than B then the level 
of A will increase. Make sure you have equal volumes of both solutions. 

In this way after the five experiments you can arrange A, B, C and D in 
order of increasing concentration and since you know the concentrations 
(0.2M, 0.4M, 0.6M and 0.8 M) you can label the flasks correctly. 

If you have any questions regarding 
(a) the theory of osmotic pressure
(b) the exact experimental setup
(c) the logic in what I have explained 

write to me. My email is gt0063a@acme.gatech.edu. I am giving you some 
references that you can look up in case you have a problem with what I 
have just explained.

(1) Ignacio Tinoco Jr, Kenneth Sauer and James C. Wang, Physical 
Chemistry: Principles and Applications in Biological Sciences, 2nd 
Edition, Prentice-Hall Inc., NJ, 1985
(2) H. Nabetani, M. Nakajima, A. Watanabe, S. Ikeda, S. Nakao, S. Kimura, 
Development of a new type of membrane osmometer, Journal of Chemical 
Engineering of Japan, V 25, N3, 269-274, 1992.
 The second reference is a scientific research journal. It will contain 
some details about the building of a membrane osmometer using cellophane 
for sucrose solutions. If you are planning on actually constructing one 
then you can find it in the article. You will also find a picture of the 
osmometer in the article.