If 6.4 grams of KOH crystals are palced within 1.5 L of water(distilled), and after 2 days several crystals are still visible, what is the condition of the system?Is it saturated,closed,unsaturated,open or supersaturated?
Thank you
To determine the condition of the system, we need to understand the concept of saturation.
A saturated solution is one where the solvent (in this case, water) has dissolved as much solute (in this case, KOH crystals) as it possibly can at a given temperature. In other words, the solvent has reached its maximum capacity to dissolve the solute.
An unsaturated solution is one where the solvent has not dissolved as much solute as it is capable of dissolving. There is still room for more solute to be dissolved.
A supersaturated solution is one where the solvent contains more solute than it can normally dissolve at a given temperature. This occurs when a saturated solution is heated and then slowly cooled down, allowing more solute to dissolve and remain in the solution at the lower temperature.
Given that several crystals are still visible after 2 days, it indicates that the solution is unsaturated. If the solution were saturated, all the KOH crystals would have dissolved, leaving no crystals visible. Therefore, we can conclude that the condition of the system is unsaturated.