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:

• A. saturated
• B. supersaturated
• C. unsaturated

Answer: (B)

Verified step by step guidance

1. 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.
2. 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.
3. 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: \( \text{moles} = \frac{10 \text{ g}}{122.55 \text{ g/mol}} \).
4. 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.
5. 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.