If 75 g of NaNO3 is dissolved in 100 g of H2O at 10°C, is the solution saturated, unsaturated or supersaturated?

To determine if the solution is saturated, unsaturated, or supersaturated, we need to compare the amount of solute dissolved to the maximum amount that can be dissolved at that temperature.

To do this, we can use the solubility table for NaNO3. According to the table, the solubility of NaNO3 in water at 10°C is approximately 88 g/100 g of water.

In this case, we have dissolved 75 g of NaNO3 in 100 g of water, which is less than the maximum solubility of 88 g/100 g.

Therefore, the solution is unsaturated.

To determine if the solution is saturated, unsaturated, or supersaturated, we need to compare the amount of solute (NaNO3) dissolved to the solubility of NaNO3 in water at the given temperature (10°C).

To find the solubility of NaNO3 in water at 10°C, you can consult a solubility table or a reference source. Let's assume that the solubility of NaNO3 at 10°C is 90 g/100 g of water.

Now we can compare the amount of NaNO3 dissolved to the solubility:

Amount of NaNO3 dissolved = 75 g
Solubility of NaNO3 at 10°C = 90 g/100 g

Comparing the amount of NaNO3 dissolved to the solubility, we can see that 75 g is less than 90 g. This means that the solution is unsaturated because it contains less NaNO3 than what could be dissolved at that temperature.

Therefore, the solution is unsaturated.

I don't know. Few people have the tables/graphs of solubility vs temperature memorized.

If you will post the solubility from you graphs/tables I can agree or disagree with you.