Final answer:
The solution will be saturated with KI at 20 degrees C and since 110g of KI exceeds the solubility in 200g of water, a precipitate of KI will form. Thus, the answer is (a) A precipitate will form.
Step-by-step explanation:
The solubility of KI (potassium iodide) indicates how much can be dissolved in a certain amount of water at a given temperature. At 20 degrees C, the solubility of KI is 50g per 100g of water. If we have 110 grams of KI being added to 200g of water, we need to find out if the solution will become saturated, unsaturated, or supersaturated. The question essentially asks if after dissolving as much KI as possible, there will be precipitate left over.
First, we should calculate the maximum amount of KI that can be dissolved in 200g of water based on its solubility at 20°C, which is 50g per 100g of water. By proportion:
- 100g of water dissolves 50g of KI.
- 200g of water would dissolve 50g * 2 = 100g of KI.
Since only 100g of KI can dissolve in 200g of water at 20 degrees C, and we have 110g of KI, the solution will be saturated and 10g of KI will remain undissolved, forming a precipitate. Therefore, the correct option in the final answer is that a precipitate will form, so the answer is (a) A precipitate will form.