On Friday we talked about the idea of molarity and molality. Molarity (M) is = to the moles of solute/liters of solution. Molality is not as common, as shown by the fact it is not in some dictionary. Molality is = to the mol solute / kilograms of solvent.
We then continued to solve some example problems of calculating concentrations. Take a look at your notes for some extra practice.
Molarity and Dilution
At the end we introduced the concept of Molarity and dilution, and established this formula
m1v1 = m2v2.
Here is a video walking through the conversation of Molality to Molarity.
Here doesn't sound very happy, but he is pretty concise. Also remember to do your webassigns and homework. Also molasses has no link to chemistry if you were curious,but they did come to mind.
A peek inside the everyday happenings of our classroom. This is an interactive learning environment for students and parents in my Honors Chemistry 173 class. This ongoing dialogue is as rich as YOU make it. Visit often and post your comments freely.
Saturday, March 19, 2011
Thursday, March 17, 2011
Heat, Cool and Repeat
Today we started off class by turning in our Solution Formation lab. (Mr. Lieberman also wants to know where his door stopper is...if anyone knows or finds it)
The rest of class was working on our new Solubility Curve lab (due Monday).
This took up the whole classtime and we even had to split up the work amongst our groups to get the job done faster. Essentially, we had two "series" to run, each with the same procedures but with different masses for the materials in the three test tubes.
In the first session, test tube A had to contain 0.45 to 0.50 grams of potassium nitrate (KNO3), test tube B had 0.70 to 0.80 grams, and C had to have 1.20 to 1.30 grams.
I massed each of these tubes and then with the KNO3. Then I massed each test tube with water added. From there, I placed each test tube into boiling water and stirred them gently (well actually, prodded at them more) to make the KNO3 dissolve faster. Once it was completely dissolved, I first took out test tube C and placed it in a beaker of iced water to make it crystallize. The temperature at which test tube C started turning white was recorded. This is the same as the saturation temperature. What was done to test tube C was then done for the other test tubes.
The second series (that was being tested at the same time) had test tube A massed at 0.35 to 0.40 grams, B with 0.90 to 1.00 grams and C with 1.50 to 1.60 grams. My partners then repeated the process I explained above.
<------(crystallized KNO3)
There are some calculations and a graph to do on the computer to go with this lab, along with the conclusion.
(There is also a Webassign due at 1:00 Friday.. just in case)
The next scribe will be.....Austin W.
The rest of class was working on our new Solubility Curve lab (due Monday).
This took up the whole classtime and we even had to split up the work amongst our groups to get the job done faster. Essentially, we had two "series" to run, each with the same procedures but with different masses for the materials in the three test tubes.
In the first session, test tube A had to contain 0.45 to 0.50 grams of potassium nitrate (KNO3), test tube B had 0.70 to 0.80 grams, and C had to have 1.20 to 1.30 grams.
I massed each of these tubes and then with the KNO3. Then I massed each test tube with water added. From there, I placed each test tube into boiling water and stirred them gently (well actually, prodded at them more) to make the KNO3 dissolve faster. Once it was completely dissolved, I first took out test tube C and placed it in a beaker of iced water to make it crystallize. The temperature at which test tube C started turning white was recorded. This is the same as the saturation temperature. What was done to test tube C was then done for the other test tubes.
The second series (that was being tested at the same time) had test tube A massed at 0.35 to 0.40 grams, B with 0.90 to 1.00 grams and C with 1.50 to 1.60 grams. My partners then repeated the process I explained above.
<------(crystallized KNO3)
There are some calculations and a graph to do on the computer to go with this lab, along with the conclusion.
(There is also a Webassign due at 1:00 Friday.. just in case)
The next scribe will be.....Austin W.
Wednesday, March 16, 2011
Factors Affecting Solubility
I'm filling in for Michelle T.
Today was a notes day. We explored how certain conditions, specifically temperature and pressure, impacted the solubility of solutions. Each of these conditions affects their solute differently based on its state of matter; a solid, gas, or liquid.
An increase in temperature for solids and liquids results in an increase of solubility, as the intensified intermolecular motion allows for a more thorough interaction between the solute and solvent. However, higher temperatures in gases results in molecules being released to the atmosphere instead of contained within the solvent, resulting in lower solubility.
Pressure only changes the solubility of gases. Solids and liquids already have their molecules tightly packed so that an increase in pressure would not change the structure of the solute enough to change solubility. Gases are more prone to be altered by pressure because their molecules are more spread out. The increase in pressure forces the molecules of gas down into the solvent, increasing solubility.
We discussed real world examples describing the relationship between pressure and solubility. Soda is kept under high pressure in its can. When that pressure is released when the can is opened, pressure no longer forces the CO2 molecules in to the soda. The CO2 rises to the top in the form of bubbles and thus, carbonation.
Here's some scientific photos to help explain:
BEFORE:
AFTER:
It seems as though a rapid decrease in solubility also has the side-effect of a spontaneous age, gender, and race change.
I guess the next scribe is Michelle T., to fill her duties from today.
Tuesday, March 15, 2011
Solution Formation Lab
Today was late arrival and we had shortened periods, which means we spent the whole period working with our lab groups on the solution formation lab. The goal of the lab was to determine the effect of crystal size, temperature, and degree of mixing on the solution formation. Each group had to develop their own procedures to test the three variables for a solution of copper sulfate disolved in water.
For our groups procedures, we kept all of the variables the same except for the one we were testing, in order to make sure the that change in results is because of the change on the variable that we controlled. For example, the first part of the lab was to determine how the size of the crystal effected the solution formation. By testing three different sized crystals of copper sulfate at the same temperature, our group got the conclusion that the larger the crystal, the longer it takes for it to dissolve.
So, that is pretty much it for our shortened class.
For our groups procedures, we kept all of the variables the same except for the one we were testing, in order to make sure the that change in results is because of the change on the variable that we controlled. For example, the first part of the lab was to determine how the size of the crystal effected the solution formation. By testing three different sized crystals of copper sulfate at the same temperature, our group got the conclusion that the larger the crystal, the longer it takes for it to dissolve.
The second part of the lab was to test the effect of the temperature of the solvent on the formation. Our group used water at three different temperatures and the same amount of copper sulfate, and by dissolving copper sulfate in all of them, we discovered that if the temperature is higher, the crystals will dissolve faster.
The last part of the lab was to test the effect of the degree of mixing on the formation the of the solution. To test this, our group mixed the solution with a stirring rod at different speeds but the same amount of copper sulfate and at the same water temperature. By keeping time, our group found that the faster you stir, the faster the crystals will dissolve.
So, that is pretty much it for our shortened class.
The conclusion part of the lab is for homework and it is due on Thursday.
The next scribe will be Michelle T.
The next scribe will be Michelle T.
Monday, March 14, 2011
Let Chemistry Absorb You. Ha.
Alright, boys and girls, this here's a chemistry post and it's about to get real.
Mr. Lieberman reviewed the test with us, like a boss, and gave us two points because of an error and because of leniency. Then he reviewed with us the basic principles of solutions.
- A solute is the substance that is being dissolved
- A solvent is the liquid in which the solute is dissolved
- Solute dissolves in solvent
- Aqueous = solution with water as solvent
Easy, right?
A saturated solution is a solution where the solute has dissolved into it at a maximum. No more solute can be dissolved in this saturated state.
In the notes that Mr. Lieberman explained in class, there is a diagram of NaCl dissolving in H2O showing the driving forces that cause the dissolving of this solute, NaCl, into this solvent, H2O (most of the solvents in this unit will be H2O and all of them will be liquids! Awesome!). In the diagram, the H2O molecules, which are polar, attach themselves to the Na+ or Cl- atoms according to polarity. So an H2O molecule's negative pole will attach to the positive Na atom. And an H2O molecule's positive pole will attach to the negative Cl atom. This will pull the crystal-like structures of NaCl apart. This is a demonstration of an ionic solute dissolving by dissociation into its ions. There are two more types: Covalent solutes dissolving by H-bonding to water and covalent solutes dissolving by London dispersion forces (LDF).
This process is carried out instantaneously, it cannot be viewed through a microscope or observed at all, for that matter.
Furthermore, there are three stages to this same solution process explained in further detail here.
- Primarily, there is the separation of a solute, and in order of this to happen, the solute's molecules must surpass their intermolecular forces (IMF) and it requires energy, making it endothermic.
- Secondly, the separation of a solvent occurs when the solvent overcomes its intermolecular forces. This also requires energy, also making it endothermic.
- Thirdly, the interaction of these two substances occurs. An attractive bond forms between the solvent and solute molecules and this releases energy, making it exothermic.
Phenominal! Now we understand the heat exchanges that occur within this process, let's continue!
We continued with the in-class notes and took a look at the factors affecting the solubility.
We know that Like dissolves like, which means that molecules with the same type of intermolecular forces will dissolve in eachother. E.g. dipole-dipole, Hydrogen bonds, and LDF.
Cool, huh?
Tell you what's not cool, that the raise of temperature in these solutions causes more collisions which allows easier access into these crystal structures, allowing for further saturation of a solution.
Finally, there is pressure. Solids and liquids are hardly affected by pressure changes in relation to solutions, but gas, under higher pressure, will have a higher solubility.
Well done, Captain, you have successfully acquired knowledge of solutions and can continue on your path to success. Enjoy your good HEALTH:
The next scribe is Becky N. (Rebecca N.)
Labels:
Emilio I,
HEALTH,
Mr. Lieberman,
solutions,
Unit 10
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