Sample Questions: (As usual, I left out questions that were too similar to ones that were already included)
1. (JUNE 2000: 416): From the table below, which has the lowest concentration?
|
Solution |
Quantity of Salt in Solution |
Volume of Solution |
|
1 |
2.0 |
500 mL |
|
2 |
1.0 |
1.5 L |
|
3 |
3.0 |
2.0 L |
|
4 |
6.0 |
3.0 L |
A Solution 1
B Solution 2
C Solution 3
D. Solution 4
2. (JUNE 2000: 416): In the lab, you are given
Explain how you would prepare 100 mL of a 5 g/L salt solution by making a dilution. Show calculations and list all steps involved in making the solution.
Answer:
0.100 L (5g/L) = 0.5 g
1. Weigh out 0.5 g of salt
2. Dissolve in a beaker containing less than 100 mL of water.
3. Carefully transfer to 100 ml volumetric flask.*
4. Dilute to mark and mix.
*Of course they assumed we cannot afford volumetric flasks. So you would use the crude, imprecise but cheap instrument, the graduated cylinder
3. (JUNE 1999: 416): You are given 200 mL of a solution that has a concentration of 30 g/100 mL and are asked to dilute it to obtain a concentration of 10 g/100 mL.
What will be the volume of the diluted solution?
A. 66.6 mL
B. 400 mL
C. 600 mL
D. 800 mL
4. (JUNE 2000: 436): Using 300 mL of a 2.0 mol/L solution of CuSO4 (copper sulfate), a student must prepare a 0.50 mol/L solution of copper sulfate.
What volume of H2O must be added to prepare the 0.50 mol/L solution?
Answer:
C1V1 = C2V2
2(0.300 L) = 0.50 V2. (Notice they say "using 300 mL". Don't confuse this problem with the type that reads:" there is 10 L available." The fact that so much is available doesn't imply you have to use it all. Only a certain amount has to be pipetted.
V2 = 1.2 L
But V2 = final volume. This means that you will be adding
0.9 L = 900 mL (1.2 - 0.3) to obtain a total of 1.2 L.
Notes:
Concentration Units
Concentration in 416 is measured strictly in g/L. This means you divide the mass of the solute (minor component of solution) by the liters of solution.
In 436, we express concentration in mol/L. This means we have to first convert mass into moles by dividing by molar mass, and then we divide by liters of solution.
Preparing Solutions
(Don’t confuse the two situations outlined in the table below; they depend on starting materials!)|
Starting Material |
Preliminary Calculation |
Procedure |
|
Solute and water have to be turned into a solution of known concentration |
Mass = CV, if C is in g/L (for 436 : moles = CV, if C is in mol/L) |
|
|
An already prepared solution has to be diluted to create a less concentrated solution |
C1V1 = C2V2 C1= concentration of original solution V1 = volume actually used from original C2 = final concentration of the newly prepared solution V2 = volume of the new solution (it is total of the original volume and the volume of water added) |
|