In terms of saturation, how would you classify a KBr solution with a concentration of 78 g/100 g of water at 47°C?
Here is a chart I found that lists a plot of solubility of KBr versus T. Perhaps you have the data and just didn't post it.
http://www.thestudentroom.co.uk/showthread.php?t=505138
If the read the graph right, the solubility of KBr at 47 C is abouat 40 g/100 mL; therefore, a solution with 78 g/100 mL would be supersaturated. If you have solubility data, use it to answer the question.
To determine the saturation of a KBr solution with a concentration of 78 g/100 g of water at 47°C, we need to compare the given concentration to the solubility of KBr at that temperature. Solubility refers to the maximum amount of solute (in this case, KBr) that can dissolve in a given amount of solvent (in this case, water) at a specific temperature.
To find the solubility of KBr at 47°C, you can refer to a solubility chart or database. These resources provide information on the solubility of various substances at different temperatures. Let's assume that we have determined that the solubility of KBr at 47°C is 70 g/100 g of water.
Since the given concentration of the KBr solution (78 g/100 g) exceeds the solubility of KBr at 47°C (70 g/100 g), we can conclude that the solution is supersaturated.
Supersaturation occurs when a solution holds more solute than it should be able to, given the solubility at a certain temperature. It is a metastable state, meaning it is not in equilibrium and has the potential to spontaneously precipitate the excess solute.
So, to classify the KBr solution with a concentration of 78 g/100 g of water at 47°C, we would classify it as a supersaturated solution.