Molarity & Stoichiometry Review

Scribe Post Author: Serene P.

11/8/13

To begin class today, Ms. Friedmann checked our previous night's homework assignments. Then, she informed us of our homework over the weekend.

1) Unit 4 Test has been moved to Wednesday!

-A study session will be held Wednesday morning at 7 am

- A review packet was given and a set of extra precipitation stoich problems (however, these are optional preparation items for the test)

2) Read sections 10.3 and 10.4 in your textbook and take notes in your journal-due Tuesday

-You need to know about Colligative Properties for the test, but you do not need to know how to calculate freezing point depression, boiling point elevation, or osmotic pressure.

3) WebAssign 10.3-due Monday night at 11:59 pm

During class, we also picked up a sheet titled, "Chemistry Scene Investigation~Trouble in the Chemistry Store Room." However, we were not able to get to this activity.

We went over the two homework sheets. Ms. Friedmann demonstrated how to do question 2a on the homework sheet titled 'Molarity.' Here is the work for the question:

2) What is the molarity of each ion present in aqueous solutions prepared by dissolving 15.00g of the following compounds in water to make 5.25 L of solution?

a) magnesium bromide?

15.00g MgBr2 x 1 mole MgBr2 = 0.08148 moles MgBr2

                        184.10g MgBr2   

0.08148 moles MgBr2   = 0.0155 M MgBr2 ←overall chemical molarity

5.25 L Solution        

Chemical concentration x number of particles

So…

0.0155 x 1M Mg2+ ions = 0.0155M

0.0155 x 2M Br- ions = 0.031 M

If you want the total ion concentration, do 0.0155 x 3 = 0.0465M

For question one of the ‘Molarity’ worksheet, the answer was 6.46M NH3. The work is shown below:

11.0g NH3 x 1 mole NH3  = 0.64554 moles NH3

                   17.04g NH3

100 mL x 1 L       = 0.1 L

            1000mL

0.64554 moles NH3  = 6.46 M NH3   

0.1 L NH3                     

       

On the ‘Concentration and Solution Stoichiometry’ sheet, the answer for question one should have been 0.519 M NaOH. To solve this question, you should have done the total number moles in solution divided by the total volume of the solution. The answer was a little closer to the higher molarity.

For question two on this worksheet, the answer was that both reactants were used up equally. There was not any limiting reactant because the amount that resulted from both Pb(NO3)2 and KI was 0.0875g PbI2.

The remaining answers for questions 2b, 2c, and 3 on the ‘Molarity’ worksheet will be found on the key that should be posted on moodle.

Ms. Friedmann also discussed the differences of molecular compounds and ionic compounds getting dissolved in water. She explained that the ionic compounds will be able to split into more particles when dissolved in water. This is because the water molecules will be able to attach to the ends of particles with opposing charges. An example is MgBr2.

Ionic compounds are able to conduct electricity as well.

However, molecular compounds in water will act differently. They will not conduct electricity and will not break apart. One particle of the compound will become one particle in solution. However, from being a solid, the compound will be referred to as aqueous because water gets in between molecules of the compound. It will separate molecules of the compound, not ions. Ionic compounds will separate into ions when in solution.

We also learned how colligative properties depend on the number of particles in solution. Water boils at a higher temperature when particles are added. Particles with charges that hold onto water molecules keep them from escaping into a gaseous state. Therefore, you must increase the heat energy to reach the boiling point of the water(boiling point elevation). If there is a lot of solute present, the water can boil at a temperature higher than a 100 degrees.

At the end of class, we had some fun with Georgia’s large hot pack.

The next scribe post author will be Colin S.