Which solution would have the lowest freezing point, 0.500 M NaCl or 0.500 M CaCl2?
To determine which solution would have the lowest freezing point, we need to compare the van't Hoff factor (i) of the two solutes, which corresponds to the number of particles produced when the solute dissolves in water.
The van't Hoff factor for NaCl is 2, which means that each NaCl molecule dissociates into two ions (Na+ and Cl-) when it dissolves in water.
The van't Hoff factor for CaCl2 is 3, which means that each CaCl2 molecule dissociates into three ions (Ca2+ and 2 Cl-) when it dissolves in water.
The colligative properties, such as freezing point depression, depend on the number of particles in the solution. Since CaCl2 dissociates into more ions than NaCl, it will have a higher number of particles in solution. As a result, a 0.500 M CaCl2 solution will have a lower freezing point than a 0.500 M NaCl solution.
To determine which solution would have the lowest freezing point between 0.500 M NaCl and 0.500 M CaCl2, we need to consider the concept of freezing point depression.
Freezing point depression is a colligative property that depends on the concentration of solute particles in a solution. The more solute particles present, the greater the freezing point depression.
Both NaCl and CaCl2 are ionic compounds that dissociate into ions when dissolved in water. NaCl dissociates into Na+ and Cl- ions, while CaCl2 dissociates into Ca2+ and 2Cl- ions.
For 0.500 M NaCl solution, it means there are 0.500 moles of NaCl in 1 liter of solution. Since NaCl dissociates into two ions (Na+ and Cl-), the total number of solute particles in the solution is 0.500 mol/L NaCl × 2 particles/mol = 1.00 mol/L.
For 0.500 M CaCl2 solution, it means there are 0.500 moles of CaCl2 in 1 liter of solution. Since CaCl2 dissociates into three ions (Ca2+ and 2Cl-), the total number of solute particles in the solution is 0.500 mol/L CaCl2 × 3 particles/mol = 1.50 mol/L.
Based on the calculation above, the 0.500 M CaCl2 solution has a higher concentration of solute particles compared to the 0.500 M NaCl solution. Therefore, the 0.500 M CaCl2 solution will have a greater freezing point depression and, consequently, a lower freezing point than the 0.500 M NaCl solution.