Activity 1: Osmosis across an artificial semipermeable membrane 1. Three solutio

Question

Activity 1: Osmosis across an artificial semipermeable membrane 1. Three solutions will be prepared for you and placed in an artificial semipermeable membrane ("dialysis tubing"): 30% sucrose, 60% sucrose and 30% magnesium sulfate (MgSO4). 2. The dialysis tubing will be placed in a beaker of water. 3. Record the height of the column of water above the second string in cm every 10 minutes for 60 minutes. 4. Draw one graph of column height over time, representing the different solutions with different colors. Questions 1. Why did the columns rise? Describe the mechanísm. 2. Using your graph, calculate the osmosis rate for each solution (cm/time) at the start. Example: Osmotic Pressure Initial rate: 20.0cm - 2.0 cm 10.0 mi min 60 Time (mint Thone Inc 3. Which column rose the most? Why? Was that expected? Why, or why not? 4, what if 30% urea (Mw-60.06g/mol; nonionic) had been in the dialysis bag instead of 30% magnesium sulfate, would it have risen higher or lower or the same? Why?

Explanation / Answer

1.semipermiable membrane allows solvent flow against solute concentration gradient is known as Osmosis .

Rise in column level is observed  due to the flow of solvent through semipermiable membrane .

we have theories that describes the mechanism of osmosis

1 solute bombardment theory

2 solute attraction theory

3 water conc. diffusion theory

4 Obstruction mechanism

5 water thirst theory

6 solute attraction theory

we will discuss about water concentration solute diffusion theory

according to this theory water molecules can pass by both sides of membrane but as the concentration of water in pure water is more than in solution molecules will pass from high water concentration to low water concentration.

i.e. from low solute conc. to high solute concentration.

2.

to answer this question you need to perform experiment & note down rise in tubes wrt time. as described above the questions.

you can easily find rate by formula

rate of osmosis = rise in tube /time taken to rise

if you want calculate intermidiate rate of osmosis just find the rise in tube in that intermediate time period & divide it by time period your considering

for example

rate between time 40 -50 min =rise in tube in this period /time interval

rate between time 40 -50 min=(80-70)/10 = 1cm /min

you can also calculate

overall rate = total rise/total time

to find instantaneous rate draw tangent ti curve & find its slope at required point

slope of curve = instantaneous rate at a point

I hope this will help you !

3.

which column rose the most

after successfull completion of experiemt from data you can see which column is rose the most.

was that expected

this can be answered by our theorotical knowledge of osmosis

why or why not that was expected we will known from this knowledge

according to osmosis theory the column rise will be based on concentration of solutions

we have

30 % sucross 60% sucross & 30% MgSO4  

concentration of solution is measured as no. of molecules per litre of solution

in case of sucrose the 60% sucross will rise more

but while answering for MgSO4 you need to note that unlike sucross MgSO4 dissociates in solution

there for no. of molecules becomes twice the conc. of MgSO4

as MgSO4 dissociates in two ions Mg2+ & SO2-4

therefore it will act as 60% concentration .

so the rise in 60% sucross & 30% MgSO4 will be observed same that is more than 30% sucross.

Osmotic Pressure = inRT/V

where i = Van't Hoff's factor

i=no.of particles after association or dissociation/no.of particles before dissociation or assocition

for sucross i =1

& for MgSO4 i =2

now you can compare your result with expected rise.

4.

as urea not disssociates nor associates

i = 1

osmotic Pressure will be same as 30% sucross

therefore the column rise maximum in 60% sucross tubing will be observed.

in this case

Osmotic pressure of 30%sucross = 30% urea<60% sucross

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