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Job 3: Reporter
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Colligative Properties Lab

Today in chemistry we did a lab that tested the colligative properties. In the lab we boiled a solution of water and an unknown substance to its boiling point. We used four beakers that were each filled with about 50 ml of water. The first beaker was the control group so we didn't add any of the substance. In the second beaker we added about 5 grams of the substance and for the third and fourth beakers, we added about 10 and 15 grams respectively. While the solutions boiled we placed a thermometer inside the beaker to find its boiling point. As expected, the solution's boiling point got higher with a higher concentration of the substance. For the first calculation, you will want to use this equation:

T(b)=i x m x k(b)

You are supposed to solve for molality here. The ionic solid has 2 ions so i would equal 2. m would equal the molality and k(b) is a constant, which would be 0.51 for water. The T(b) is the Delta H that comes from the data in the lab. The homework is to finish up the post lab which is due on friday. Also there are 3 WebAssigns due on friday so keep up with them. The next scribe is Lauren C.

Colligative Properties

Today in chemistry we kicked off class with a lecture regarding colligative properties. There are 4 types of colligative properties and they are:
1.Vapor Pressure Lowering
2. Boiling Point Elevation
3. Melting Point Depression
4. Osmotic Pressure
Basically, the relations are simple. The more solute added, the more the vapor pressure decreases. The more solute added, the more the boiling point raises. The more solute added, the lower the melting point becomes. We also got a few formulas to accompany the above.

To calculate the change in boiling point(Blogger isn't letting me use equation editor, get notes off moodle to really see this):

Change in Boiling Point=molal boiling point elevation constant x molality
Traingle T subscript b= K subscript b * m

To calculate freezing point depression, you use the same formula except the constant is different. The K subscript b is repaced with K subscript f.

The van't Hoff factor basically mens that if you have a mole of NaCl, and the ions dissociate when in water, you will get one mole of Na+ ions and one mole of Cl- ions. The van't Hoff factor is i so simply modify the previous equations by multiply i to the end like so:

Triangle T subscript f= K subscript f * m * i

That pretty much sums it up for new material covered today. Liebs demonstrated these new concepts with one demo involving 4 of us as water molecules, showing that the more solute added (balloons), the more interactions we had to make with it (touching each ion), which made it harder for us water molecules to turn into our gaseous state. Also, we took some club soda, stuck it into some ice and upon opening the glass, thus releasing the CO2 which had raised the freezing point, the club soda immediately froze.

Liebs also discussed some issues with the faulty Webassign that was due today, and said he would make corrections. I know it probably frustated myself as much as did to some of you (honestly, comment if you remembered how to use sig figs before this unit started).
That's about all though. Homework tonight is to do all the worksheets you have been given this unit that are due on the day of the test, as well as the 3 remaining webassigns also due on the day of the test.

Justin J., come on down, you are the next contestant on the price is right.

Stoichiometry Returns

Today in 7th period chem we did some stoichiometry with solutions.

That was one problem that we did. Basically you can use mole ratios from a balanced equation, molar masses, and now you can use the molarities to get what you want.

Now we know the mole ratio is related in several ways:

-PV=nRT

-molar mass

-M=moles/L

We can use stoichiometry for pure substances, gases, and now solutions.

If anyone is still having trouble, I would recommend going through the notes from last semester on Stoichiometry, or watch this video:

http://www.youtube.com/watch?v=uiEc5LcsGIg

(my blogger never lets me put the actual video in... :( )

The next scribe will be...... Nicole C

Hey Guys! So today in chem class we went over notes on molarity and molality dilution.

The first concept that we talked about was the concentration of solute. That is, the amount of solute in a solution is given by its concentration. There are four units if concentration.

The first is molartiy and you can compute it by this equation, molarity (M) = moles of solute/liters of solution. This equation comes in hand when you are trying to figure out the mass of a substance needed to make a certain amount of solution at a specific molarity, or to simply find the molarity of a solution.

The second unit in with you can compute the concentration of a solution is molality, be sure to not get this and molarity confused for the sound very similar. With this unit the equation is molality (m) of solution moles of solute/kilograms of solvent.

The third unit in with you can compute the concentration of a solution is percent by mass. The equation of this unit is % by mass = (grams of solute/grams of solution) X 100.

The fourth and final way of compute the concentration of a solution is by finding the mole fraction. To find this you use the equation mole fraction (x) = moles of solute/ moles of solvent + moles of solution. For this you want to remember to have your final answer be in decimals.

The second concept we discussed was molarity and dilution. This concept talked about diluting a solution and this is simply adding more solvent, in our case water, to the solution. This will dilute the solution but remember that the amount of solute does not change and because of this, it is really easy to calculate the new molarity. All you do is take the information you have (at least three points) and plug it into the equation, MaVa=MbVb.

That was all for today’s class. Homework for Monday is to finish your lab and do the webassigns also remember to keep up with your work sheets.

The next scribe is …………………….. Rachel M.

Solubility Curve Lab

Today in 7th Period Chem we did yet another lab! This one dealt with measuring the saturation temperatures for six difference solution concentrates to construct a solubility curve.

The lab itself wasn't too complicated. Each lab group had 6 test tubes: 3 for Series 1 and 3 for Series 2. The first step was to mass the empty test tubes and record that in the lab table. Next, we added potassium nitrate to each test tube: each test tube had a specific amount of potassium nitrate it needed to be added [the lab direction sheet explains the specifics]. Once that was done, we massed the test tubes with the potassium nitrate and added that in the data table. The next step was to add 20 drops of distilled water into each test tube. Then -you guessed it- we massed the test tube plus potassium nitrate plus water....and added that into our data table. Finally, we placed the test tubes in a hot water bath, letting the potassium nitrate dissolve, then put the test tubes in an ice water bath. This cooled the solution which enabled it to crystallize. We took note of the temperature the instant crystallization occurred and recorded that into our lab table.

With all this information, there were some things that needed to be calculated in the lab. The calculations were quite simple- you just needed to use your data.

1. Calculate the mass of potassium nitrate and the mass of water in each solution

For the mass of potassium nitrate, you take the mass of test tube and KNO3 and subtract it by the mass of the empty test tube. So essentially, the second column of your data table minus the first. You have to do that for all 6 of the test tubes, so it gets pretty tedious.

For the mass of water, it's the same idea. Subtract the mass of test tube plus KNO3 and water by the mass of test tube and KNO3, which is really column 3 in your data table minus column 2.

2. Calculate the ratio of the mass of potassium nitrate to the mass of water for each solution

Here, you take the values you got from part one and set them to a ratio. For each of the 6 test tubes, divide the mass of potassium nitrate by the mass of water. That will be your ratio.

3. Multiply each ratio by 100 to get the solubility in g/100g of solvent

This one speaks for itself. Multiply what you got in in part two by 100 and you're set.

So that's the gist of the lab and the calculations. After you do that, all you have to do is place ALL of the results in a results table and plot a graph of solubility of potassium nitrate!

This lab is due Monday.

Other than that, we have a Webassign due tomorrow as well as the usual worksheets that will be due at the end of the unit.

That's all I've got for today. The next scribe will be.............................................NIRALI P! Good luck :)

Baby Monkey Going Backwards on a Pig

What a great day in 7th period chem. Arguably the best day of the year. It started out with a discussion about the english mini field trip. Summarized to be "the english teachers needed three periods to assign a project that we all knew we had." Once all of our shenanigans ended, the real chemistry began. We discussed the concept that pressure and temperature both effect solubility.

Container A is under normal pressure and therefore has a normal solubility. Container B is under high pressure and therefore has a higher solubility.

As this graph shows, when the temperature increases, so does the solubility. It is the opposite effect for gases in solution. Ce2(SO4)3 is a hipster.

We then discussed saturation. When the solubility is even with the line on the graphs, the solution is saturated. When there is less solubility than possible, the solution is under saturated. Super saturation comes when a solution is saturated at one temperature, and then is cooled down. This video shows what happens when a super saturated solution is agitated.

This is a video of a baby monkey riding backwards on a pig.

Remember to do the lab. And keep up with your webassigns.

Nicole C can do the next post. Your welcome Nicole.