To determine which cup the NaCl would most easily dissolve into, we need to consider the principles of solubility, specifically in relation to solutions that already contain a dissolved solute.
Cup A contains only pure water. There are no other solutes in the water that would affect the solubility of NaCl. The solubility of a solute in a pure solvent is primarily dependent on the solute's own solubility properties and the temperature of the solvent. Since NaCl is quite soluble in water, it would dissolve as expected until reaching its solubility limit at the given temperature.
Cup B, on the other hand, contains 100 grams of potassium nitrate (KNO3) that has already been dissolved. In addition to temperature and solute properties, the solubility of NaCl in this cup would be influenced by the presence of another solute, KNO3.
There are a few relevant effects to consider:
1. Common Ion Effect: Although NaCl and KNO3 don't share common ions, the presence of K+ and NO3- ions in solution could still impact the solubility of NaCl through changes in the ionic strength of the solution.
2. Ionic Strength: A higher ionic strength, due to the presence of K+ and NO3- ions, can reduce the solubility of other salts, such as NaCl, because of the reduced difference in ion concentration between the solid phase and the liquid phase, which drives the dissolution process.
3. Competing Interactions: In Cup B, water molecules are already interacting with K+ and NO3- ions, which could reduce the number of available water molecules to solvate Na+ and Cl- from NaCl. This competition can further inhibit the dissolution of NaCl.
These effects suggest that NaCl would be less soluble in Cup B compared to Cup A. Thus, NaCl would most easily dissolve into Cup A, which contains pure water with no additional solutes. This is because Cup A does not have the additional ionic interactions and competitive solvation that would occur in Cup B.
