From the following:
1) pure water
2) 0.01 m solution of table sugar (C12H22O11)
3) 0.01 m solution of NaCl
4) 0.01 m solution of CaCl2
Choose the one with the largest van't Hoff factor and hence the lowest freezing point.
You want the one which dissolves in water producing the most particles. Since the solutions are the same molality, then just count the particles produced when a mole of the solute is dissolved. Sugar produces 1, NaCl produces 2, etc etc.
To determine the van't Hoff factor (i), we need to first understand what it represents. The van't Hoff factor is a measure of the extent to which a compound dissociates or breaks apart into ions in a solution. It is denoted by the symbol "i". The higher the van't Hoff factor, the greater the degree of dissociation.
Now, let's analyze each solution and determine their respective van't Hoff factors:
1) Pure Water:
Pure water does not dissociate into ions; it remains as H2O molecules. Therefore, the van't Hoff factor for pure water is 1.
2) 0.01 M Solution of Table Sugar (C12H22O11):
Table sugar, or sucrose (C12H22O11), does not dissociate into ions in aqueous solutions. Therefore, the van't Hoff factor for table sugar is also 1.
3) 0.01 M Solution of NaCl:
Sodium chloride (NaCl) readily dissociates into sodium ions (Na+) and chloride ions (Cl-) in aqueous solutions. Each NaCl molecule dissociates into two ions, thus the van't Hoff factor for NaCl is 2.
4) 0.01 M Solution of CaCl2:
Calcium chloride (CaCl2) dissociates into three ions: one calcium ion (Ca2+) and two chloride ions (2Cl-). So, each CaCl2 molecule dissociates into three ions, giving a van't Hoff factor of 3.
Based on the analysis above, the solution with the largest van't Hoff factor is the 0.01 M solution of CaCl2. Therefore, it also has the lowest freezing point among the given options, as a higher van't Hoff factor leads to a lower freezing point according to the colligative properties of solutions.