the saturation point or the solubility of salt X in water is 27 g per 100 mL of solvent used. if a person adds 29 g of this salt to a 100 Ml of solvent what type of solution would form?
shouldnt the answer be supersaturated? since saturation point is 27g? the practice test says the answer is sauturated...
Saturated is the correct answer. The problem states that 29 g is ADDED to 100 mL of solvent--not that 29 g DISSOLVED. If it dissolved, yes, it would be supersaturated.
oh, ok. thank you for clearing that up i was confused.
In this case, let's first understand the concept of a saturated solution. A saturated solution is formed when the maximum amount of solute is dissolved in a given amount of solvent under a specific set of conditions, such as temperature.
The saturation point or solubility of salt X in water is given as 27 g per 100 mL of solvent. This means that under normal conditions, no more than 27 g of salt X can dissolve in 100 mL of water.
Now, let's consider the scenario where 29 g of salt X is added to 100 mL of solvent. Since the saturation point is 27 g, adding 29 g of salt X exceeds the maximum solubility limit. As a result, the excess amount of salt X will not be dissolved and will form a separate precipitate at the bottom of the container.
Based on this understanding, the correct answer should indeed be "supersaturated," as the amount of solute (29 g) exceeds the saturation point (27 g). It is possible that there was an error or confusion in the practice test answer.