Colligative Properties Lab

Today in Chemistry we did the Colligative Properties Lab. The goal of this lab was to use boiling point elevation data to identify an unknown salt. We began by labeling four 100 mL beakers A,B,C, and D and recording their masses in our super neat data tables. Next we filled each of the beakers about half full with distilled water. We recorded the mass of the beaker and water in our data table. We then placed the beakers full of water on the hot plate and heated them to about 85°C at which point we removed them from the hot plate. We determined the boiling point of the water in beaker A by noting the plateau on our lab pro. We recorded this number and proceeded to add about 5.0 grams of the unknown ionic solid to beaker B. We placed it on the hot plate and recorded its boiling point in our data table. We then added about 10.0 grams of the unknown solid to beaker C. We placed it on the hot plate and recorded its boiling point once again. Finally we added 15.0 grams of the unknown solid to beaker D, placed it on the hot plate, and recorded its boiling point in our data tables.

As a part of the data/calculations we were asked to calculate the molality of each solution and were given the molal boiling point elevation constant: 0.51°C kg/mol. To calculate the molality of each solution, we used the equation ΔT_b = K_b · m · i. To calculate change in temperature, we subtracted the boiling point of beaker A from beaker B, beaker B from beaker C and so on. We were told the ionic sold has 2 ions, so i=2. We plugged these values into the equation and solved for m to get a value for molality.

To calculate the moles of solute in beakers B, C, and D, we used the molality equation which says that molality = mol solute / kg solvent. We have the value for molality and to find the kg of solvent we simply subtracted the mass of the beaker from the mass of the beaker and water and divided the answer by 1000 to obtain the value in kilograms. To find the moles of solute, we multiplied the molality by kg solvent.

To find the molar mass of the solute in beakers B,C, and D we divided the number of grams of solute in each mixture by the moles of solute obtained in the previous step. We added 5.0 grams to beaker B, 10.0 grams to beaker C, and 15.0 grams to beaker D. These values were divided by the moles solute to obtain the molar mass of the solutes. To calculate the average of the molar masses, we just added them together and divided by three.

The conclusion asks you to decide the formula for the unknown solute, so choose the formula with the molar mass closest to your average. The choices are NaCl, KI, NaNO3 or NaBr. Support your claim with evidence and then calculate the percent error:

| actual value – theoretical value | x 100 %
theoretical value

The lab is due Friday and so are all of the worksheets and Webassigns. Study for the Test Friday!

The next scribe will be Emilio I!

Title

Today in the class of chemistry, we discussed colligative properties, which are properties that focus on the amount of particles in a solution and not their actual identities. There are 4 colligative properties, but we only need to know 3 of them: vapor pressure lowering, boiling point elevation, and freezing point depression. Osmotic pressure is the fourth. Vapor pressure change can be expressed by multiplying the vapor pressure of the solvent by the mole fraction of the solute. A change in boiling point can be calculated by subtracting the BP of the solvent from the BP of the solution. For freezing, do the opposite and subtract the FP of the solution from the FP of the solvent. We also had a demo where a bottle of unopened club soda was placed into a beaker of ice and salt in order to change its freezing point. After a few minutes, Mr. Lieberman took out the soda and opened it, causing it to instantly freeze and have the carbonation come bubbling out. That's about it for Tuesday March 22 in chem. Do the worksheets and webassigns for Friday and study for the test! The next scribe will be Zoe S

Stoichiometry Expanded

We first turned in our solubility labs from Friday. When then began our discussion, extending off the solubility problems. (Also we have a test on friday) Everything was mostly a review from stoichiometry 1st semester except for molarity and molality.

When then preceded to solve this problem for practice:

How much calcium carbonate will be precipitated by adding 25.0 ml of calcium chloride to 25.0ml of 56 M potassium?

I left out the ml to l conversation factor and the 1 to 1 mole conversation for the second equation above.

We also did the first problem of the worksheet.

Remember to keep doing those web assigns and worksheets for preparation for the test on friday!
The next scribe will be Ben T !

Molarity vs Molality vs Molasses

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.

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.