Rolf prepares four solutions using different solutes as shown in the table below.
| Solution | Preparation | Solubility of the solutes |
|---|---|---|
| A | Rolf added 20 g of solute to 100 g of water and mixed the solution. | The solute in solution A has a solubility of [tex]$37 g / 100 g H _2 O$[/tex] at [tex]$20^{\circ} C$[/tex]. |
| B | Rolf added 10 g of solute to 100 g of water and mixed the solution. | The solute in solution B has a solubility of [tex]$10 g / 100 g H _2 O$[/tex] at [tex]$20^{\circ} C$[/tex]. |
| C | Rolf added 65 g of solute to 100 g of water, mixed the solution, and then heated the solution. | The solute in solution C has a solubility of [tex]$32 g / 100 g H _2 O$[/tex] at [tex]$20^{\circ} C$[/tex]. |
| D | Rolf added 7 g of solute to 100 g of water, mixed the solution, and then heated the solution. | The solute in solution D has a solubility of [tex]$4 g / 10 A H _2 O$[/tex] at [tex]$20^{\circ} C$[/tex]. |
Which solution is unsaturated?
A. Solution A
B. Solution B
C. Solution C
D. Solution D
Answer (2)
Compare the amount of solute added to 100g of water with the solubility of the solute in 100g of water for each solution.
Solution A: Since 20g < 37g, solution A is unsaturated.
Solution B: Since 10g = 10g, solution B is saturated.
Solution C: Since 65g > 32g, solution C is supersaturated.
Solution D: Since 7g > 4g, solution D is supersaturated. The solution that is unsaturated is S o l u t i o n A .
Explanation
Understanding Unsaturated Solutions We need to determine which of the four solutions (A, B, C, or D) is unsaturated. A solution is unsaturated if it contains less solute than the maximum amount that can be dissolved in the solvent at a given temperature. We will compare the amount of solute added to 100g of water with the solubility of the solute in 100g of water for each solution.
Analyzing Solution A Solution A: Rolf added 20 g of solute to 100 g of water. The solubility of the solute is 37 g / 100 g H2O. Since 20 g < 37 g, solution A is unsaturated.
Analyzing Solution B Solution B: Rolf added 10 g of solute to 100 g of water. The solubility of the solute is 10 g / 100 g H2O. Since 10 g = 10 g, solution B is saturated.
Analyzing Solution C Solution C: Rolf added 65 g of solute to 100 g of water. The solubility of the solute is 32 g / 100 g H2O. Since 65 g > 32 g, solution C is supersaturated.
Analyzing Solution D Solution D: Rolf added 7 g of solute to 100 g of water. The solubility of the solute is 4 g / 100 g H2O. Since 7 g > 4 g, solution D is supersaturated.
Conclusion Based on the analysis, solution A is unsaturated.
Examples
Understanding saturation is crucial in many real-world applications. For example, in cooking, knowing the solubility of sugar in water helps in making syrups or candies. In pharmaceuticals, it's important to know the solubility of a drug to ensure it dissolves properly in the body. In environmental science, understanding the solubility of pollutants helps in predicting their behavior in water systems.
To find the unsaturated solution among the four, we compared the amount of solute added to the solubility of each solute. Solution A is unsaturated since it contains less solute than can be dissolved at 20°C. Therefore, the answer is Solution A.
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