Final answer:
Without the exact solubility curve for NaNO₃, we can't calculate the precise amount of precipitate. However, based on general solubility principles, we know some NaNO₃ would precipitate when the solution is cooled; thus, the precipitated amount is likely more than 0 grams.
Step-by-step explanation:
To determine how many grams of NaNO₃ settled out of the original solution when cooled from 60°C to 10°C, we need the solubility curve for NaNO₃. Unfortunately, the provided information does not give the exact solubilities at those temperatures, but we can use the principles exemplified with KNO₃ and NaCl in our explanation.
At a higher temperature, substances typically have higher solubility in water. When a solution is cooled, its solubility decreases, and the solute starts to precipitate. For KNO₃, at 60°C it had a solubility of 107 g per 100 g of water, but upon cooling to 0°C, only 14 g remained dissolved and the rest recrystallized. If we had the solubility curve for NaNO₃, we would subtract the amount soluble at 10°C from that at 60°C, and this difference would give us the weight of NaNO₃ that precipitated out.
Although we can't calculate the exact number without the solubility curve data for NaNO₃, we know from the provided examples that some amount of solute typically precipitates out when a solution is cooled significantly. Hence, the correct answer is likely not 0 grams, implying an answer like C. 20 grams or D. 90 grams might be more realistic depending on the exact solubility curve for NaNO₃.