A solution of KCI is saturated at 50°C. Use Table 9.1
(a) How many grams of solute are dissolved in 100 g of water?
(b) What is the total mass of the solution?
(c) What is the mass percent of this solution at 50°C?
(d) If the solution is heated to 100°C, how much more KCl can be dissolved in the solu- tion without adding more water?
(e) If the solution is saturated at 100ºC and then cooled to 30°C, how many grams of solute will precipitate out?
Table 9.1
Temperature Effect on Solubility of Four Salts in Water, g solute/100 g water
 |  |  |  |  |  |  |  |  |  |  | |
 | 27.6 | 31.0 | 34.0 | 37.0 | 40.0 | 42.6 | 45.5 | 48.3 | 51.1 | 54.0 | 55.6 |
 | 35.7 | 35.8 | 36.0 | 36.3 | 36.6 | 37.0 | 37.3 | 37.8 | 38.4 | 39.0 | 39.8 |
 | 53.5 | 59.5 | 65.2 | 70.6 | 75.5 | 80.2 | 85.5 | 90.0 | 95.0 | 99.2 | 104.0 |
 | 31.6 | 33.3 | 35.7 | 38.2 | 40.7 | 43.6 | 46.6 | 49.4 | 52.6 | 55.7 | 58.8 |
The correct answer and explanation is:
Based on Table 9.1, we will analyze the solubility of KCl in water at different temperatures and answer each question accordingly.
Given Solubility Data:
- At 50°C, the solubility of KCl is 42.6 g per 100 g of water.
- At 100°C, the solubility of KCl is 55.6 g per 100 g of water.
- At 30°C, the solubility of KCl is 37.0 g per 100 g of water.
(a) How many grams of solute are dissolved in 100 g of water?
From the table, at 50°C, the solubility of KCl is 42.6 g per 100 g of water. This means in 100 g of water, 42.6 g of KCl is dissolved.
(b) What is the total mass of the solution?
The solution consists of 100 g of water and 42.6 g of KCl. Total mass=100+42.6=142.6 g\text{Total mass} = 100 + 42.6 = 142.6 \text{ g}
(c) What is the mass percent of this solution at 50°C?
Mass percent of KCl is calculated as: mass of solutetotal mass of solution×100\frac{\text{mass of solute}}{\text{total mass of solution}} \times 100 42.6142.6×100=29.9%\frac{42.6}{142.6} \times 100 = 29.9\%
So, the mass percent is 29.9%.
(d) If the solution is heated to 100°C, how much more KCl can be dissolved without adding more water?
At 100°C, the solubility of KCl is 55.6 g per 100 g of water. Since we already have 42.6 g dissolved, the additional amount that can dissolve is: 55.6−42.6=13.0 g55.6 – 42.6 = 13.0 \text{ g}
So, 13.0 g more KCl can be dissolved.
(e) If the solution is saturated at 100°C and then cooled to 30°C, how many grams of solute will precipitate out?
At 100°C, the solution contains 55.6 g of KCl per 100 g of water. When cooled to 30°C, the solubility decreases to 37.0 g per 100 g of water. The amount of KCl that precipitates is: 55.6−37.0=18.6 g55.6 – 37.0 = 18.6 \text{ g}
So, 18.6 g of KCl will precipitate out.
Explanation:
Solubility is the maximum amount of solute that can dissolve in a given amount of solvent at a specific temperature. The solubility of KCl increases with temperature, meaning more KCl dissolves at higher temperatures. When a saturated solution is cooled, the solubility decreases, causing the excess solute to precipitate out. This is why when a 100°C saturated solution is cooled to 30°C, 18.6 g of KCl precipitates.
Now, I’ll generate an image to illustrate the solubility trend of KCl.
Here is a graph illustrating the solubility of KCl in water at different temperatures. The red points highlight key temperatures mentioned in the problem: 30°C, 50°C, and 100°C, where changes in solubility affect precipitation and dissolution of KCl. Let me know if you need any further clarification!
