The solubility of sodium chloride is about 0.36 g per gram of water at 25 degrees Celsius. g (2 dec places) of NaCl will dissolve in 96 g of water at 25 degrees Celsius. If there were 43.92 g of NaCl dissolved in 122 g of water, the solution would be (unsaturated/saturated/supersaturated).

In this case, we understand that solubility accounts for the maximum amount of a solute to completely dissolve into a solvent, in such a way, sodium chloride's solubility at 25 °C is 0.36 g of NaCl per gram of water, thus, if a relationship is greater than this value, the solution will be supersaturated, if it is equal saturated and if is less it will be unsaturated. In such a way, for 43.92 g of sodium chloride and 122 g of water, the relationship result:

[tex]\frac{43.92gNaCl}{122gH_2O}=0.36gNaCl/gH_2O[/tex]

Thus, such solution would be saturated as the relationship equals the solubility.

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