Answer :
To determine whether each solution will be saturated or unsaturated, we'll compare the amount of solute added to the solubility at 20°C as provided in the table. Here are the steps to determine the state of each solution:
1. Adding 16 g of NaNO3 to 25 g of H2O:
- First, convert the amount of NaNO3 added for 25 g of H2O to the amount that would be added to 100 g of H2O for ease of comparison with the given solubility.
- [tex]\[ \text{Concentration of NaNO3} = \left( \frac{16 \text{ g}}{25 \text{ g H2O}} \right) \times 100 = 64 \text{ g/100 g H2O} \][/tex]
- Compare 64 g/100 g H2O to the solubility of NaNO3 at 20°C, which is 88 g/100 g H2O.
- Since 64 g/100 g H2O is less than 88 g/100 g H2O, the NaNO3 solution is unsaturated.
2. Adding 471 g of malt sugar to 100 g of H2O:
- The added concentration is already per 100 g of H2O.
- Compare 471 g/100 g H2O to the solubility of malt sugar at 20°C, which is 204 g/100 g H2O.
- Since 471 g/100 g H2O is greater than 204 g/100 g H2O, the malt sugar solution is saturated.
3. Adding 29 g of KCl to 125 g of H2O:
- First, convert the amount of KCl added for 125 g of H2O to the amount that would be added to 100 g of H2O for ease of comparison with the given solubility.
- \
\text{Concentration of KCl} = \left( \frac{29 \text{ g}}{125 \text{ g H2O}} \right) \times 100 = 23.2 \text{ g/100 g H2O}
- Compare 23.2 g/100 g H2O to the solubility of KCl at 20°C, which is 29 g/100 g H2O.
- Since 23.2 g/100 g H2O is less than 29 g/100 g H2O, the KCl solution is unsaturated.
### Summary:
- Saturated solution:
- Adding 471 g of malt sugar to 100 g of H2O
- Unsaturated solution:
- Adding 16 g of NaNO3 to 25 g of H2O
- Adding 29 g of KCl to 125 g of H2O
1. Adding 16 g of NaNO3 to 25 g of H2O:
- First, convert the amount of NaNO3 added for 25 g of H2O to the amount that would be added to 100 g of H2O for ease of comparison with the given solubility.
- [tex]\[ \text{Concentration of NaNO3} = \left( \frac{16 \text{ g}}{25 \text{ g H2O}} \right) \times 100 = 64 \text{ g/100 g H2O} \][/tex]
- Compare 64 g/100 g H2O to the solubility of NaNO3 at 20°C, which is 88 g/100 g H2O.
- Since 64 g/100 g H2O is less than 88 g/100 g H2O, the NaNO3 solution is unsaturated.
2. Adding 471 g of malt sugar to 100 g of H2O:
- The added concentration is already per 100 g of H2O.
- Compare 471 g/100 g H2O to the solubility of malt sugar at 20°C, which is 204 g/100 g H2O.
- Since 471 g/100 g H2O is greater than 204 g/100 g H2O, the malt sugar solution is saturated.
3. Adding 29 g of KCl to 125 g of H2O:
- First, convert the amount of KCl added for 125 g of H2O to the amount that would be added to 100 g of H2O for ease of comparison with the given solubility.
- \
\text{Concentration of KCl} = \left( \frac{29 \text{ g}}{125 \text{ g H2O}} \right) \times 100 = 23.2 \text{ g/100 g H2O}
- Compare 23.2 g/100 g H2O to the solubility of KCl at 20°C, which is 29 g/100 g H2O.
- Since 23.2 g/100 g H2O is less than 29 g/100 g H2O, the KCl solution is unsaturated.
### Summary:
- Saturated solution:
- Adding 471 g of malt sugar to 100 g of H2O
- Unsaturated solution:
- Adding 16 g of NaNO3 to 25 g of H2O
- Adding 29 g of KCl to 125 g of H2O
