Which of the following aqueous solution have the lowest freezing point?

Question

Accepted Answer

To determine which aqueous solution has the lowest freezing point, we need to consider the concept of freezing point depression, which is a colligative property. Colligative properties depend on the number of solute particles in a solution, not on the identity of the solute. The freezing point of a solution is lowered in proportion to the number of solute particles dissolved in the solvent.

1. **Option 4: 0.1 m $BaCl_{2}$**  
   - $BaCl_{2}$ dissociates into three ions in solution: one $Ba^{2+}$ ion and two $Cl^{-}$ ions. Therefore, a 0.1 m solution of $BaCl_{2}$ will produce 0.3 moles of ions per kilogram of water (0.1 moles of $Ba^{2+}$ and 0.2 moles of $Cl^{-}$). This results in the highest number of solute particles among the given options, leading to the greatest freezing point depression. Thus, the solution with $BaCl_{2}$ will have the lowest freezing point.

2. **Option 1: 0.1 m Glucose**  
   - Glucose is a non-electrolyte and does not dissociate into ions in solution. Therefore, a 0.1 m solution of glucose will produce only 0.1 moles of solute particles per kilogram of water. This results in a smaller freezing point depression compared to solutions that dissociate into more particles.

3. **Option 2: 0.1 m NaCl**  
   - NaCl is an electrolyte and dissociates into two ions: one $Na^{+}$ ion and one $Cl^{-}$ ion. A 0.1 m solution of NaCl will produce 0.2 moles of ions per kilogram of water. While this results in a greater freezing point depression than glucose, it is still less than that of $BaCl_{2}$, which produces more ions.

4. **Option 3: 0.1 m Sucrose**  
   - Like glucose, sucrose is a non-electrolyte and does not dissociate into ions. A 0.1 m solution of sucrose will also produce only 0.1 moles of solute particles per kilogram of water, resulting in a similar freezing point depression to glucose, which is less than that of ionic compounds like NaCl and $BaCl_{2}$.

In conclusion, the solution with 0.1 m $BaCl_{2}$ has the lowest freezing point due to the highest number of solute particles resulting from its dissociation into three ions. This leads to the greatest freezing point depression, making Option 4 the correct answer.