Final answer:
Directly converting 123 g of K3P to liters is not possible without the density of K3P. The provided reference information does not relate to the conversion of K3P from grams to liters but pertains to different Chemistry concepts such as solution molarity and volume conversions.
Step-by-step explanation:
The question regarding 123 g of K3P is a conversion problem from grams to liters for a substance, which would typically require additional information such as the density of the substance or conditions under which the substance is measured. In the case of solid compounds like K3P (potassium phosphide), converting grams directly to liters isn't straightforward without knowing the substance's density.
However, the given reference information does not relate directly to K3P but rather to the substance potassium permanganate (KMnO4) in solution. The cited examples and the information about molarity, preparation of solutions, and conversions are helpful for understanding the concept of preparing solutions and converting units within the context of Chemistry.
It's worth noting that molarity and volume conversions are essential topics in Chemistry, which is helpful for high school and college students learning about solution preparation, and titration, and exploring solubility products (Ksp). An example used in the context of molarity is Example 13.6.4, where a chemist needs to prepare 3.00 L of a 0.250 M solution of potassium permanganate, which illustrates the process of determining the mass of a solute needed for a given volume and concentration of the solution. Similarly, understanding the conversion between liters and kiloliters is useful, as shown in the solution that explains the conversion factor of 1000L/1kL.