What Are Hypertonic, Hypotonic, and Isotonic Solutions in Dialysis Experiments?

  • Thread starter Thread starter DespicableMe
  • Start date Start date
AI Thread Summary
Hypertonic solutions contain a higher concentration of solute compared to another solution, leading to water moving out of the dialysis tubing. In contrast, hypotonic solutions have a lower solute concentration, causing water to move into the tubing. The discussion emphasizes that water is the key factor in determining hypertonic and hypotonic conditions, as it is the only substance that can diffuse through the dialysis membrane. Additionally, the impact of temperature on the rate of diffusion was mentioned, with higher temperatures increasing particle movement. Understanding these concepts is crucial for accurately interpreting dialysis experiments.
DespicableMe
Messages
40
Reaction score
0

Homework Statement



Purpose: To use dialysis tubing and two different solutions, water and a concentrated starch solution, to demonstrate hypertonic, hypotonic and isotonic solutions.

I'm confused as to which situation is hyper or hypotonic:

The Attempt at a Solution



If the water is placed INSIDE the tube and the starch solution is in the beaker, that would be a HYPERTONIC situation because water concentration inside is higher?

If the water is placed OUTSIDE the tube and starch is placed INSIDE the tube, that would be a HYPOTONIC situation because water concentration inside is lower (than in the beaker)?

We only talk about hypo- and hypertonic situations in terms of the WATER solution because it is the only solution out of the two used that can diffuse in and out of the dialysis tube, right?

There was also one section where it asked us to repeat the HYPERTONIC situation at a different temperature to look for the RATE of transport.

We did not have enough time to complete this part so I was wondering if there is any insight on that.

Thanks : )
 
Physics news on Phys.org
Hypertonic = more solute = less water

Particles move faster at higher temperatures, and diffusion is the movement of particles...
 
Thread 'Confusion regarding a chemical kinetics problem'
TL;DR Summary: cannot find out error in solution proposed. [![question with rate laws][1]][1] Now the rate law for the reaction (i.e reaction rate) can be written as: $$ R= k[N_2O_5] $$ my main question is, WHAT is this reaction equal to? what I mean here is, whether $$k[N_2O_5]= -d[N_2O_5]/dt$$ or is it $$k[N_2O_5]= -1/2 \frac{d}{dt} [N_2O_5] $$ ? The latter seems to be more apt, as the reaction rate must be -1/2 (disappearance rate of N2O5), which adheres to the stoichiometry of the...
Back
Top