Answer:
The given parameter for the solubility of NaNO₃ in H₂O are;
The maximum solubility of NaNO₃ in H₂O at 10°C = 78 g
1. If we have 50 g of NaNO₃ in 100 g of water at 10°C, the solution is;
Undersaturated
2. If we have exactly 78 g of NaNO₃ in 100 g of water at 10°C the solution is;
A saturated solution
3. If will add 80 g of NaNO₃ in 100 g of water at 10°C,
The excess NaNO₃ which cannot be dissolved will be observed as solids in the mixture
4) The quantity of NaNO₃ that can be dissolved in H₂O at 40°C = 94 g
If the solution containing 94 g of NaNO₃ at 40°C is cooled to 10C about 16 g of NaNO₃ will precipitate out of the solution and exist as solids in the mixture
Explanation:
1. An undersaturated is a solution that holds smaller amount of solute that it can hold at a given temperature
2. A saturated solution holds as much solute as it can dissolve at a given temperature
3. When more solutes are added to a saturated solution, the excess solution will remain in the solid form in the mixture
4) Cooling a saturated solution below the saturation temperature will result in the precipitation or crystallization of the excess solutes in the solution.
