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Answer :
Final answer:
The given solutions can be categorized as either saturated or unsaturated based on their solubilities at specific temperatures.
Explanation:
To determine if a solution is saturated or unsaturated, we need to compare the amount of solute dissolved in the solution with its solubility at a given temperature. If the amount dissolved is equal to or less than the solubility, the solution is saturated. If the amount dissolved is greater than the solubility, the solution is unsaturated. So, based on the given solutions:
- 2.00 g of PbCl2 at 80°C: saturated
- 100.2 g of NaNO3 at 60°C: saturated
- 128 g of KI at 0°C: saturated
- 75.5 g of LiCl at 20°C: unsaturated
- 8.22 g of Ba(OH)2 at 40°C: unsaturated
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Answer:
2.00 g of PbCl2 at 80°C is unsaturated
100.2 g of NaNO3 at 60°C is unsaturated
128 g of KI saturated
75.5 g of LiCl at 20°C is unsaturated
8.22 g of Ba(OH)2 at 40°C is saturated
Explanation:
To determine if a solution is (un)saturated, we should take a look at the solubility for the given temperature.
2.00 g of PbCl2 at 80°C
⇒ At 80 °C, the max solubility of PbCl2 is 2.54 grams / 100 mL
⇒ 2.00 < 2.54
⇒ This solution is unsaturated
100.2 g of NaNO3 at 60°C
⇒ At 60 °C, the max solubility of NaNO3 is 122 grams / 100 mL
⇒ 100.2 < 122
⇒ This solution is unsaturated
128 g of KI at 0°C
⇒ At 0 °C, the max solubility of KI is 128 grams / 100 mL
⇒ 128 = 128
⇒ This solution is saturated
75.5 g of LiCl at 20°C
⇒ At 20 °C, the max solubility of LiCl is 83.5 grams / 100 mL
⇒ 75.5<83.5
⇒ This solution is unsaturated
8.22 g of Ba(OH)2 at 40°C
⇒ At 40 °C, the max solubility of Ba(OH)2 is 8.22 grams / 100 mL
⇒ 8.22 = 8.22
⇒ This solution is saturated
