which of the following solutes, dissolved in 1kg of water, would be expected to provide the fewest particles and to freeze at the highest temperature? (NH4)2SO4 , CaCl2, and/or NaCl.
To determine which solute would provide the fewest particles and freeze at the highest temperature, we need to consider the concepts of dissociation and colligative properties.
Dissociation is the process by which solutes break apart into ions when dissolved in a solvent. The number of particles a solute produces in a solution depends on its degree of dissociation. In this case, we have three solutes: (NH4)2SO4, CaCl2, and NaCl.
(NH4)2SO4 dissociates into three ions: 2NH4+ and SO42-.
CaCl2 dissociates into three ions: 1Ca2+ and 2Cl-.
NaCl dissociates into two ions: 1Na+ and 1Cl-.
Now, let's consider the colligative properties. Colligative properties depend on the number of particles present in a solution, regardless of their identity. One of these properties is freezing point depression, which means that a solution freezes at a lower temperature than the pure solvent.
According to Raoult's law, the extent of freezing point depression is directly proportional to the number of solute particles. Therefore, the solute with the fewest particles will produce the least freezing point depression and freeze at the highest temperature.
Among the given solutes, (NH4)2SO4 will provide the fewest particles because it dissociates into three ions, while CaCl2 dissociates into three ions and NaCl dissociates into two ions.
Therefore, NaCl would be expected to provide the fewest particles and freeze at the highest temperature among the three solutes.