Last minute questions...

----Question----

I can't figure out the ppm question when it asks what is the concentration in ppm of a solution made by dissolving 14.18mg of dinitrotoluene (182.13g/mol) in 4.00L of water. I thought you took .01481g/182.13 than you took that answer divided by 4kg and than take that times 1 million but its not working out for me.

----Answer----

"ppm" is (by convention) usually thought of as a mass-mass unit, so it's part of the mass fraction family.

ppm (dinitrotoluene, DNT) = [ {mass DNT} / {total mass of sample} ] * 10^6 =

[ {0.01418g DNT} / {4000g + 0.01418g} ] * 10^6 = 3.55ppm DNT

NOTE: because the volume of water only known to 3 sig figs and is so large compared to the mass of DNT, the denominator simplifies to just 4000g.

----Question----

You don't need to know molar mass of a substance to find what: molarity, molality, normality, mole fraction, or mass percent?

----Answer----

The only reason you would need the molar mass would be to calculate moles of the substance, so anything that includes moles will require the molar mass. Because mass percent is just a ratio of masses, you can calculate it without knowing the molar mass of the substances involved. This is related to the previous problem above.

Questions...

A few email questions came in. I may be able to answer a few more of these tonight, but by about 8 or 9pm I'll be offline until morning, so if you have questions, get them in sooner rather than later.

--Question--------------

I just had a question for the freezing point depression problems..for example on the Spring 2009 exam 1b #13, as far as the calculations where do you get the #mols particles/mols ? In that problem it says (2 mols particles/mol LiNO3) how do you get that part?

--Answer--------------

Quite a few people are having trouble with this. The number of particles per solute (the van't Hoff Factor) is a measure of how many pieces each formula unit of solute breaks into when it dissolves. For molecular solutes, the solute formula is a single piece in solution; for example, when a sugar molecule dissolves in water, it's still a single sugar molecule, it doesn't break down into individual carbon, hydrogen and oxygen atoms. When ionic solutes dissolve in water, they break down into the ions that make up the formula. In the LiNO3 example given above, when LiNO3 dissolves in water it does NOT float around as LiNO3 units in solution, it breaks down into lithium ions in solution and nitrate ions in solution, so for every 1 LiNO3 unit, there are 2 particles in solution.

--Question--------------

I am not really understanding the formula for the ppm problems, for ex. (Winter 2006, Exam 1), you have times 10^6, how did you get to the sixth power?

--Answer--------------

10^6 is a million. When you're converting a fraction (mass fraction or volume fraction) to ppm {or ppm(v)}, you have to multiply by a million, 10^6.

--Question--------------

I wanted to make sure I did my work right for the mol fraction questions. If the question asks, what is the mol fraction of sugar in a saturated aqueous sugar solution at 25 C. (solubility of sugar in water is 211.4 g/ 100 mL.

Would it be 211.4 - 180.1548 (total grams of sugar) / 180.1548 = .1744

--Answer--------------

This is essentially a unit conversion problem. The given solubility means that 211.4g of sugar will dissolve in 100mL of water. The mols of sugar is:

211.4g / 180.1548g/mol = 1.173mols sugar

Mols of water in the system:

(100mL)(1g/1mL) / 18.015g/mol = 5.55mols water

So the mol fraction of sugar in this solution is:

(mols of sugar) / (total mols) = (1.173mols sugar) / (1.173mols sugar + 5.55mols water) = 0.1745

--Question--------------

I know we did our lab on molar mass today, but I am not sure how to calculate question #16 on exam 1a (spring 2009). It has atm added to it.

16. A newly discovered protein has been isolated from seeds of a tropical plant and needs to be characterized. A total of 0.137g of this protein was dissolved in enough water to produce 2.00mL of solution. At 31.68°C the osmotic pressure produced by the solution was 0.134atm. What is the molar mass of the protein? (20pts)

--Answer--------------

In lab we were looking at one colligative property (freezing point depression) to determine the molar mass of an unknown. This problem is looking at a different colligative property (osmotic pressure) to determine the mass of an unknown. The expression for osmotic pressure is:

P = MRTi

OK, we have to make an assumption here, we're going to assume that the protein is a single particle in solution, which makes the van't Hoff Factor (i) equal to 1. After that, we can start plugging in the info from the problem.

0.134atm = M(0.08206 L.atm/mol.K)(31.68+273.15K)(1)

M = 0.0053569 mols protein/L solution

We've made 2.00mL of solution so:

(0.0053569 mols protein/L solution)(0.00200L solution) = 0.000010714mols of protein

So the molar mass of the protein is:

(0.137g protein)/(0.000010714mols of protein) = 12800g/mol

Exam #1 next Friday

We've gone through Chapters 10 (Gases) and 11(Solids and Liquids), we'll look at Chapter 15 (Solutions) next. The first exam is next Friday, let me know when you have questions, I'll post answers here.