CH 101/Keys 6
From WolfWikis
| CH 101 |
|---|
| Exercises 1 |
| Exercises 2 |
| Exercises 3 |
| Exercises 4 |
| Exercises 5 |
| Exercises 6 |
| Exercises 7 |
| Exercises 8 |
| Exercises 9 |
| Exercises 10 |
| Exercises 11 |
| Exercises 12 |
| Exercises 13 |
| Exercises 14 |
| Exercises 15 |
|
Back to CH 101 |
Contents |
Exercises
Exercise 1
- How many grams of alcohol (CH3CH2OH) must we add to 1 liter of water to obtain a solution of 1 [mmol/l]?
- We want 1 [l] of 1 [mmol/l] concentration. Multiplying those two we see that we need 1*1 [
l][mmol/l]= 1 [mmol] of alcohol. - To convert that to grams we need the molar mass: M = 2*12+6*1+16= 46 [g/mol] = 46 [mg/mmol]
- To get rid of the mmol we need to multiply. We get:
- 1*46 [
mmol][mg/mmol] = 46 [mg] = 0.046 [g] of alcohol.
- Notice that I carefully made use of the double bookkeeping and canceled units to make sure I did not accidentally divide where I should multiply. (Try doing it the wrong way around and see what undesirable units you get!)
Exercise 2
- The concentration of oxygen [O2] in a blood sample of 300ml is 0.0023 [mol/l]. How many grams of oxygen does the sample contain?
- Given is [O2]= 0.0023 [mol/l]= 0.0023 [mmol/ml]
- If we multiply we find: 300* 0.0023 [
ml][mmol/ml]= 0.69 [mmol] of O2 in the blood sample. - To go to grams we need the molar mass of 2*16= 32 [g/mol]= 32 [mg/mmol]
- To get rid of the milimoles we need to multiply:
- 0.69*32 [
mmol] [mg/mmol] = 22 [mg] = 0.022 [g] of oxygen in the blood sample.
Exercise 3: dilution
- To 500 ml of a solution with a molarity of 0.1 [mol/l] is added more solvent until the volume is 2 l. What is the resulting concentration?
- Dilution is a source of many, many mistakes. Concentration is a compound quantity that involves both the quantity of the solute and the volume of the solution. Concentrations change if either are tampered with. If your tea is too sweet, try adding some water!
- We can use the bookkeeping to keep track of things though.
- First let us see how much solute we have, because that is the quantity that does not change.
- The old concentration is 0.1 [mol/l]=0.1 [mmol/ml]. Multiplying with the old volume makes the ml's drop out
- 500*0.1 [
ml][mmol/ml] = 50 [mmol] of solute.
- All we need to do now is put that in the new volume by dividing"
- 50 / 2 [mmol]/[l]= 25 [mmol/l] = 0.025 [mol/l]
- As you can see the final concentration is lower by a factor of old/new = 0.1/0.025 = 4. This factor is known as a dilution factor. We could have guessed that I suppose, because the volume went up from 0.5 to 2 liters and that is a factor of 4 as well.
Exercise 4: adding solutions
- 40 ml of an aqueous solution of 1-propanol with a concentration of [C3H8O] = 0.021 [mol/l] is added to 60 ml of a 1-propanol solution with [C3H8O] = 0.0105 [mol/l]. The solutions are thoroughly mixed. What is [C3H8O] in the resulting solution?
- Adding solutions to each other leads to even more errors than just diluting. Again good bookkeeping is essential!!!
- First note what does not change: the quantity of the solute(s). Atoms do not go up in smoke just because you throw them together!
- So let's see how much solute we have in A. We did that in the previous problem. It involves multiplying the concentration with the volume to get the quantity of solute in moles.
- 40 ml * 0.021 [mmol/ml] = 0.84 [mmol]
- In solution B we have:
- 60 ml * 0.0105 [mmol/ml] = 0.63 [mmol]
- As the solutions contain the same solute, we can add these numbers to arrive at the amount of solute in the final mixture: 0.84+0.63 = 1.47 [mmol]
- Now we need the final volume. For dilute solutions we can just add the volumes together: 60+40=100 ml
- The concentration is then found by dividing: [C3H8O] = 1.47/100 [mmol]/[ml]= 0.0147 [mmole/ml] = 0.0147 [mol/l].
- As you can see this concentration is somewhere in between the two original concentrations. In fact it is a weighted average of the two. (The weights are 40:60).
- It is a good thing that the problem says that there is good mixing. Without that we cannot really compute anything because the concentrations will vary in your liquid in unpredictable ways,if you do not mix well.
Exercise 5: adding solutions
- 40 ml of an aqueous solution of 1-propanol with a concentration of [C3H8O] = 0.021 [mol/l] is added to 60 ml of a butanol solution with [C4H10O] = 0.0105 [mol/l]. The solutions are thoroughly mixed. What is [C3H8O] and [C4H10O] in the resulting solution?
- This looks a lot like the previous problem but it is really quite different, because we have two different solutes. The point is that the propanol concentration in the second solution is zero and the butanol concentration in the first likewise.
- We could rewrite the problem as:
| Solution | Volume | [C3H8O] | [C4H10O] |
|---|---|---|---|
| A | 40 ml | 0.021 [mol/l] | 0 [mol/l] |
| B | 60 ml | 0 [mol/l] | 0.0105 [mol/l] |
- This means that the problem has more in common with exercise 3 (the dilution problem) than with 4, because as the propanol in A is concerned, solution B really is just so much more solvent. The same holds for the butanol in B.
- One could say this is an exercise in mutual dilution.
- We could solve it by bookkeeping (do try), but we can also look at the dilution factors. The final volume is 60+40=100 ml.
- The propanol gets diluted by a factor of 40/100, so its concentration drops to 0.021*40/100 = 0.0082 [mol/l]
- The butanol goes from 60 to 100 ml, so its concentration drops from 0.0105 to 0.0105*60/100 = 0.0063 [mol/l]
Execise 6: evaporation
- 100 ml of a solution with a concentration of 0.1 [mol/l] is exposed to evaporation until the volume is only 75 ml. The solvent is volatile, the solute is not. What is the resulting concentration?
- This is the opposite process of dilution. Instead of adding solvent, we get rid of some.
- Again we can do rigorous bookkeeping, but we can also use the opposite of the dilution factor: the concentration factor. The ratio of the old and the new volumes gives us this dimensionless factor: 100/75 = 4/3 = 1.3333
- The solution should go stronger by this factor, so it goes from 0.1 [mol/l] to 1.3333*0.1= 0.13333 [mol/l]
- Notice that we need to be sure that only one of the components (the solvent in this case) evaporates, so that the amount of solute remains the same.
- What we have really done is saying:
- new concentration / old concentration = (
solute/volume)new / (solute/volume)old = old volume / new volume = concentration factor.
- If both solvent and solute evaporate, we cannot cancel the solute quantities. In fact, we do not know what happened, because one may have disappeared more quickly than the other....
Exercise 7
- Myoglobin is a protein found in the muscle tissues of many vertebrates. Its molar mass is 16,700 Daltons. The globular protein is soluble in water and contains an active center consisting of a heme molecule in the center of which is a single iron atom. What is the iron concentration in 5 ml of a solution containing 5 g of myoglobin? What is the mass of the iron present in this solution?
- Lots of interesting but pretty irrelevant information. That happens a lot in science. You really need to become an expert at picking out the juicy tidbits that lead you to the answer...
- Because the myoglobin molecule contains only one iron there are as many iron atoms as there are myoglobin molecules. This means that [My] and [Fe] must be the same.
- We know the molar mass for My, so we can compute the number of moles of protein (and thus Fe) in the solution. It is 5 /16,700 [
g]/[g/mol]= 0.299 [mmol] - The concentration is found by dividing by the volume: [My] = [Fe]= 0.299 /5 = 0.06 [mmol/ml] = 0.06 [mol/l]
- The mass of the iron requires a look at the periodic table to find the atomic mass of this element 55.8 [g/mol]= 55.8 [mg/mmol]. Thus 0.299*55.8 [mg/mmol].[mmol] gives 16.7 [mg] of iron.