Table of contents

1. Intro to General Chemistry3h 48m
2. Atoms & Elements4h 16m
3. Chemical Reactions4h 10m
4. BONUS: Lab Techniques and Procedures1h 38m
5. BONUS: Mathematical Operations and Functions48m
6. Chemical Quantities & Aqueous Reactions3h 53m
7. Gases3h 49m
8. Thermochemistry2h 37m
9. Quantum Mechanics2h 58m
10. Periodic Properties of the Elements3h 9m
11. Bonding & Molecular Structure3h 29m
12. Molecular Shapes & Valence Bond Theory1h 57m
13. Liquids, Solids & Intermolecular Forces2h 23m
14. Solutions3h 1m
15. Chemical Kinetics2h 53m
16. Chemical Equilibrium2h 29m
17. Acid and Base Equilibrium5h 1m
18. Aqueous Equilibrium4h 47m
19. Chemical Thermodynamics1h 50m
20. Electrochemistry2h 42m
21. Nuclear Chemistry2h 36m
22. Organic Chemistry5h 7m
23. Chemistry of the Nonmetals2h 39m
24. Transition Metals and Coordination Compounds3h 16m

14. Solutions

Types of Aqueous Solutions

General Chemistry

14. Solutions

Types of Aqueous Solutions

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Multiple Choice

The solubility of KClO₃ in water at 30ºC is 10 g per 100 mL of water. A 0.95 M solution of KClO₃ in water at 30ºC is:

saturated

supersaturated

unsaturated

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Verified step by step guidance

First, understand the concept of solubility. Solubility is the maximum amount of solute that can dissolve in a solvent at a given temperature to form a saturated solution. Here, the solubility of KClO3 at 30ºC is given as 10 g per 100 mL of water.

Next, convert the solubility from grams to moles to compare it with the molarity of the solution. Use the molar mass of KClO3, which is approximately 122.55 g/mol, to perform this conversion.

Calculate the number of moles of KClO3 that corresponds to the solubility limit. Use the formula: \( \text{moles} = \frac{\text{grams}}{\text{molar mass}} \). Substitute the values: \( \text{moles} = \frac{10 \text{ g}}{122.55 \text{ g/mol}} \).

Determine the molarity of the saturated solution. Since the solubility is given per 100 mL, convert this to liters to find molarity: \( \text{Molarity} = \frac{\text{moles}}{\text{volume in liters}} \). Use the volume conversion: 100 mL = 0.1 L.

Compare the calculated molarity of the saturated solution with the given molarity of the solution (0.95 M). If the calculated molarity is less than 0.95 M, the solution is supersaturated. If it is equal, the solution is saturated. If it is more, the solution is unsaturated.