Units of Concentration

Units of Concentration:

Problem 11.47:

What is the difference between a saturated and a super-saturated solution?
A saturated solution contains the maximum amount of a solute which can be contained in a solution at equilibrium. In this case, we have a stable system.
On the other hand, a super-saturated solution contains more solute than the solvent can contain (take) at equilibrium. In this case, the system is not at equilibrium and the slightest perturbation (such as scratching the flask with a stirring rod or introducing a "seed crystal") will cause the excess solute to precipitate. A super-saturated solution is sometimes called a system in metastable equilibrium. What we mean here is that the system is not changing but is subject to any perturbation which will cause it to go to true equilibrium this case, a saturated solution.

Problem 11.49:

How would you prepare each of the following solutions?

(a) 100 mL of a 155 ppm solution of urea, CH₄N₂O, in water. Here, we would dissolve 15.5 mg of urea in water. 100 mL of water is about 100 g. 1 ppm would be 0.0001g = 0.1 mg. Thus, 155 ppm would just be 15.5 mg.

(b) 100 mL of an aqueous solution whose K⁺ concentration is 0.075M. First, let us choose a potassium salt. The most common (hence, "obvious") choice would be KCl (MW = 74.551₃). In 100 mL this would be 0.0075 moles = 0.55₉ g. So, what we would do is weigh out this amount (actually, the best number to do here with an analytical balance would be 0.5591 g), place this in a 100 mL volumetric flask, and dilute with water to 100 mL.
Actually, a good volumetric flask would have its volume at 100.0 mL (or even better). A good analytical balance would give you the mass as 0.5591 g. Thus, you could actually say that you could have prepared this solution to have a concentration of 0.07500M!

Problem 11.51:

How would you prepare 165 mL of a 0.0268 m solution of benzoic acid (C₇H₆O₂) in chloroform (CHCl₃)?
The MW of benzoic acid is 122.1213₄ g/mol. 0.0268 moles of this would be 3.27₃ g.
Here is what the instructor's manual says:
"Dissolve 3.27 g of C₇H₆O₂ in 1.000 kg of CHCl₃ and take 165 mL of the solution."
Now this would work BUT

maybe chloroform is expensive (after all, you'd throw away the remaining solution!) and

chloroform is toxic.

So, throwing away the remainder is not only wasteful but poisons something or someone.

So, what is really better is to find out the mass of 165 mL of chloroform, dissolve our solute in that, and then take 165 mL of the resulting solution. In this case, the amount of waste would be much less! Let us now proceed along this better path.
The density of chloroform is 1.4832 g/mL at 25°C (I stole this number from a handbook). So, the mass of 165 mL of chloroform would be

Measuring out this volume is rather simple. All we need do is then calculate how much benzoic acid to weigh out (which would then be added to the chloroform). If we let w_A be the required amount, we can then solve for it. First, the molality of the given solution would be given by

Simple rearrangement and solving then gives the final result.

Adding this number of grams to the 165 mL of chloroform would give us considerably less waste! We only need to discard a very small amount (< 1 mL).

Problem 11.53:

Which of the following solutions has the higher molarity?

(a) 10 ppm KI in water or 10,000 ppb KBr in water? 10 ppm = 10,000 ppb (there are just three orders of magnitude separating these). So, we have 10 ppm of each salt. KBr has a lower MW (ca. 78.2) than does KI (ca. 166.0). Thus, for a given mass, there are more moles of KBr than of KI and KBr would have the higher molarity.

(b) 0.25 mass-% KCl in water or 0.25 mass-% citric acid (C₆H₈O₇) in water? These are the same mass-%. So we compare the MW's:

KCl: 74.551

Citric acid: 192.1235

Obviously, there are more moles of KCl for a given mass and KCl would have the higher molarity.

Problem 11.54:

What is the mass percent concentration of the following solutions?

(a) Dissolve 0.655 mol of citric acid, C₆H₈O₇, 1.00 kg of water. MW (citric acid): 192.1235. So, the mass of citric acid would be 0.655(192.1235) = 125.₈ g. Obviously, the mass of the solution would the be 1.12₅₈ kg and the mass percent would be just

(b) Dissolve 0.135 mg of KBr in 5.00 mL of water. Assume 1 mL of water to be 1 g. Then, 5.00 mL = 5000 mg and

for the mass-% of KBr. Note that we were a little sloppy with sig. figs. in the intermediate steps but did the right thing at the end!

(c) Dissolve 5.50 g of aspirin, C₉H₈O₄, in 145 g of dichloromethane, CH₂Cl₂. This is quite simple in that we don't have to fiddle with units! Anyway,

It's nice when the last part of a problem is the easiest!

Problem 11.55:

What is the molality of each solution prepared in Problem 11.54?

(a)	Dissolve 0.655 mol of citric acid, C₆H₈O₇, 1.00 kg of water.	Every problem set needs a "no-brainer"! In this case, with 0.655 moles of solute and 1.00 kg of solvent, it should be obvious that the concentration here is 0.655 m.
(b)	Dissolve 0.135 mg of KBr in 5.00 mL of water.	For this we need the MW of KBr, which is 119.002₃. Now, we shall convert to grams for all masses and we get the following equation and result:
(c)	Dissolve 5.50 g of aspirin, C₉H₈O₄, in 145 g of dichloromethane, CH₂Cl₂.	The MW of aspirin is 180.1574₂. From here, we just apply the formula given in part (b) just above to get Again, please note how easy these things are if you can do simple algebra!

Problem 11.56:

The so-called ozone layer in the earth's stratosphere has an average total pressure of 10 mm Hg (0.0132 atm). The partial pressure of ozone in the layer in the layer is about 1.2 x 10^-6 mm Hg (1.6 x 10^-9 atm). What is the concentration of ozone (in parts per million) assuming that the average molar mass of air is 29 g/mol?
First, we need the mole fraction of ozone. From Dalton's law of partial pressures, this is just

If we have one mole of air (this is the easiest assumption which works), we can then calculate the mass-% of ozone as follows:
Mass of a mole of air: 29g
MW of ozone (O₃): 47.9982
Mass of ozone in a mole of air: (1.2₃ x 10^-7)(47.9982) = 5.91 x 10^-6g
Given all this, we can then easily calculate the concentration of ozone in ppm:

Not too difficult, really! Note that ppm is just mass fraction times 10⁶.

Problem 11.57:

Persons are medically considered to have lead poisoning if they have a concentration of greater that 10 micrograms of lead per deciliter of blood. What is this concentration in parts per billion?
We need to make an assumption here. Blood is essentially a dilute solution of salt in water. Hence, let us assume that it has a density of ca. 1g/mL. So, a deciliter (dL) is 0.1L and would have a mass of 100g. Given this, it is quite easy to get the answer. We show the result now.

Here, to get ppb, all you need do is multiply the mass fraction by 10⁹.

Problem 11.60:

The density of a 16.0 mass-% solution of sulfuric acid in water is 1.1094 g/mL at 25.0°C. What is the molarity of the solution?
Let us assume that we have exactly 1 L of the solution. Then, it should be obvious that the masses of sulfuric acid and water are, respectively, 160g and 840g. We know that density is given by r = w/V. Finally, the MW of sulfuric acid is 98.078₅. Putting all this information together give us

Please note that I did insert 10³ mL/L into the numerator! Formulas are all well and good but you must be sure that you are watching your units!

+Problem 11.61: (This is an added problem from what was originally given in the syllabus for 2004. It is here in order to make the next problem clearer and to give you a little more practice in calculations.)

Ethylene glycol, C₂H₆O₂, is the principal constituent of automobile antifreeze. If the density of a 40.0 mass-% solution of ethylene glycol in water is 1.0514 g/mL at 20°C, what is the molarity?
We shall do this in a straight-forward fashion. Here, we shall try things a little differently than in the previous problem.

Assume a kg of solution. Then, the mass of ethylene glycol is 400g and the mass of water is 600g. (If you can't do this, what are you doing in college?)

The MW of ethylene glycol is 62.0678₄ g/mol.

Thus, you ought to be able to see that this is 6.44₅ moles!

If you can't figure this out, then maybe you should transfer to FSU.

The volume of a kg of solution is obviously V = 1000g/(1.0514g/mL) = 951.₁ mL = 0.951₁L.

So, carrying out a rather simple division, we get c = 6.44₅mol/0.951₁L = 6.77₆M.

The answer:
6.77₆M ethylene glycol.

Lots of folks like their answers in this form. Heaven forbid that we actually use any formulas or any of the math you might have learned! Do you think this way is any better than the right way of doing things?

Problem 11.62:

What is the molality of the 40.0 mass-% ethylene glycol solution used for automobile antifreeze (Problem 11.61)?
Here, just for fun, let us assume a 100g total mass. The reason for this is that things are in percentages and the "natural" unit for these is 100.
Here, we have 40.0g of solute and 100.0 - 40.0 = 60.0g = 0.0600kg of solvent. So, let us do the calculation post haste!

This is a rather concentrated solution. In cases such as this (with water as the solvent) it is quite common to see the molality larger than the molarity.

Problem 11.63:

What is the molality of the 16.0 mass-% solution of sulfuric acid in Problem 11.60?
This is quite easy. We shall just let things "flow"...

Problem 11.66:

A 0.944 M solution of glucose, C₆H₁₂O₆, in water has a density of 1.0624 g/mL at 20°C. What is the concentration of this solution in the following units?
Before we answer the questions below, let us do a few simple and relevant calculations.

First, the MW of glucose is 180.1558₈.

0.944 moles of this is 170.₁ g.

The mass of exactly l L of this solution is 1062.4 g.

Of this, we have 170.₁g of glucose (or 0.944 mol of same).

This means that we have 1062.4 - 170.₁ = 892.₃g of water.

This is, of course, 0.892₃ kg of water.

The MW of water is 18.0152₈. Thus the number of moles of water present here is 49.5₃.

Now, we are ready to do the problem! Let's get at it!

(a) Mole fraction We worked out the numbers of moles of each substituent above. In this case, the answer is easily calculated as

(b) Mass percent Mass percent is quite easy. We have 170.₁g of solute in 1062.4 g of solution (assuming a liter). Thus, the mass percent is just

(Note that the final answer has just 3 sig. figs.--as it should!)

(c) Molality Here, we have (for a liter of solution), 0.944 moles of solute and 0.8923 kg of solvent. Thus,

That's all there is to this! I hope that you found this easy! If not, please look at the write-up on concentration conversions in the "pHundamentals" part of this presentation!

(a)	100 mL of a 155 ppm solution of urea, CH₄N₂O, in water.	Here, we would dissolve 15.5 mg of urea in water. 100 mL of water is about 100 g. 1 ppm would be 0.0001g = 0.1 mg. Thus, 155 ppm would just be 15.5 mg.
(b)	100 mL of an aqueous solution whose K⁺ concentration is 0.075M.	First, let us choose a potassium salt. The most common (hence, "obvious") choice would be KCl (MW = 74.551₃). In 100 mL this would be 0.0075 moles = 0.55₉ g. So, what we would do is weigh out this amount (actually, the best number to do here with an analytical balance would be 0.5591 g), place this in a 100 mL volumetric flask, and dilute with water to 100 mL. Actually, a good volumetric flask would have its volume at 100.0 mL (or even better). A good analytical balance would give you the mass as 0.5591 g. Thus, you could actually say that you could have prepared this solution to have a concentration of 0.07500M!

(a)	10 ppm KI in water or 10,000 ppb KBr in water?	10 ppm = 10,000 ppb (there are just three orders of magnitude separating these). So, we have 10 ppm of each salt. KBr has a lower MW (ca. 78.2) than does KI (ca. 166.0). Thus, for a given mass, there are more moles of KBr than of KI and *KBr would have the higher molarity*.
(b)	0.25 mass-% KCl in water or 0.25 mass-% citric acid (C₆H₈O₇) in water?	These are the same mass-%. So we compare the MW's: KCl: 74.551 Citric acid: 192.1235 Obviously, there are more moles of KCl for a given mass and *KCl would have the higher molarity*.

(a)	Dissolve 0.655 mol of citric acid, C₆H₈O₇, 1.00 kg of water.	MW (citric acid): 192.1235. So, the mass of citric acid would be 0.655(192.1235) = 125.₈ g. Obviously, the mass of the solution would the be 1.12₅₈ kg and the mass percent would be just
(b)	Dissolve 0.135 mg of KBr in 5.00 mL of water.	Assume 1 mL of water to be 1 g. Then, 5.00 mL = 5000 mg and for the mass-% of KBr. Note that we were a little sloppy with sig. figs. in the intermediate steps but did the right thing at the end!
(c)	Dissolve 5.50 g of aspirin, C₉H₈O₄, in 145 g of dichloromethane, CH₂Cl₂.	This is quite simple in that we don't have to fiddle with units! Anyway, It's nice when the last part of a problem is the easiest!

(a)	Mole fraction	We worked out the numbers of moles of each substituent above. In this case, the answer is easily calculated as
(b)	Mass percent	Mass percent is quite easy. We have 170.₁g of solute in 1062.4 g of solution (assuming a liter). Thus, the mass percent is just (Note that the final answer has just 3 sig. figs.--as it should!)
(c)	Molality	Here, we have (for a liter of solution), 0.944 moles of solute and 0.8923 kg of solvent. Thus, That's all there is to this! I hope that you found this easy! If not, please look at the write-up on concentration conversions in the "pHundamentals" part of this presentation!