Thursday, June 23, 2011

C.5 Questions #1-3, p. 62

1) Suppose you dissolved 40g potassium chloride (KCl) in 100g water at 50°C. You then let the solution cool to room temperature, about 25°C.
a. What changed would you see in the beaker as the solution cooled? See Figure 1.40.
b. Draw models of what the contents in the beaker would look like at the particulate level at 50°C, 40°C, and 25°C.


a. As the solution cooled, it would become a supersaturated solution, and therefore, if the beaker was undisturbed and no more KCl was added, nothing would happen to the solution. However, since the solution would be unstable, if one crystal of KCl was added or the beaker was knocked, the extra solutes in the solution would become precipitate, and the solution would rebalance itself into a saturated solution.
b. See drawing

2) An unsaturated solution will become more concentrated if you add more solute. Decreasing the total volume of water in the solution (such as by evaporation) also increases the solution's concentration. Consider a solution made by dissolving 20g KCl in 100g water at 40°C.
a. Draw a model of this solution.
b. Suppose that while the solution was kept at 40°C, one-fourth of the water evaporated.
i. Draw a model of this final solution and describe how it differs from your model of the original solution.
ii. How much water must evaporate at this temperature to create a saturated solution?


a and b i: See drawings.
b ii: 50 grams of water would have to evaporate at this temperature to create a saturated solution.

3) A solution may be diluted (made less concentrated) by adding water.
a. Draw a model of solution containing 10.0 KCl in 100g water at 25°C.
b. Suppose you diluted this solution by adding another 100g water with stirring. Draw a model of this 25°C new solution.
c. Compare your models in Questions 3a and 3b. What key feature is different in the two models? Why?


a and b: See drawings.
c. In 3a, the concentration of the solution expressed as a percent KCl by mass is 10%. Since the amount of water is double in 3b, 10g/200g X 100% = 5%. The percent concentration of 3a is 10%, where as the percent concentration of 3b is 5%.

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