At 10C, 100g of H2O can dissolve only 78g of NaNO3 1.If we have 50g of NaNO3, in 100g of water at 10C the solution is?

2.If we have exactly 78g of NaNO3 in 100g of water at 10C the solution is?

3.What will happen if we add 80g of NaNO3 in 100g of water at 10C?

4) 40C we can dissolve 94g of NaNO3 in 100g of water. What will happen if the solution is cooled to 10C?

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.

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the answer is c

tell me if wrong