Answer :
To determine which solution has the lowest freezing point, let's understand the concept of freezing point depression. Freezing point depression occurs when a solute is added to a solvent, reducing the temperature at which the solvent freezes.
The extent to which the freezing point is lowered depends on the number of particles into which the solute dissociates in the solution. This effect is quantified by the formula:
[tex]\[ \Delta T_f = i \cdot K_f \cdot m \][/tex]
where:
- \(\Delta T_f\) is the freezing point depression.
- \(i\) is the van't Hoff factor (number of particles the compound dissociates into).
- \(K_f\) is the freezing point depression constant of the solvent.
- \(m\) is the molality of the solution.
Let's evaluate each option:
A. Pure Water:
- Pure water freezes at 0°C.
- There are no solutes present to cause any freezing point depression.
B. 0.5 M Ionic \( NaCl \):
- Sodium chloride (\( NaCl \)) dissociates into 2 ions: \( Na^+ \) and \( Cl^- \).
- Thus, \( i = 2 \).
C. 0.5 M Ionic \( CaCl_2 \):
- Calcium chloride (\( CaCl_2 \)) dissociates into 3 ions: \( Ca^{2+} \) and 2 \( Cl^- \).
- Thus, \( i = 3 \).
D. 0.5 M Ionic \( AlCl_3 \):
- Aluminum chloride (\( AlCl_3 \)) dissociates into 4 ions: \( Al^{3+} \) and 3 \( Cl^- \).
- Thus, \( i = 4 \).
E. 0.5 M Molecular \( C_{12}H_{22}O_{11} \) (sucrose):
- Sucrose is a molecular compound and does not dissociate into ions in solution.
- Thus, \( i = 1 \).
To determine the lowest freezing point, look for the solution with the highest product of \( i \cdot m \) because it will exhibit the greatest freezing point depression.
Given the molality (\( m \)) is the same (0.5 M) for all solutions:
- Pure water (\( i = 0 \)) has no freezing point depression.
- \( NaCl \) (\( i = 2 \)) leads to an intermediate freezing point depression.
- \( CaCl_2 \) (\( i = 3 \)) results in a larger freezing point depression.
- \( AlCl_3 \) (\( i = 4 \)) results in the greatest freezing point depression.
- Sucrose (\( i = 1 \)) leads to the least freezing point depression among the solutions.
Therefore, the solution with the lowest freezing point is:
[tex]\[ \boxed{D. \, 0.5 \, M \, \text{ionic} \, AlCl_3} \][/tex]
The extent to which the freezing point is lowered depends on the number of particles into which the solute dissociates in the solution. This effect is quantified by the formula:
[tex]\[ \Delta T_f = i \cdot K_f \cdot m \][/tex]
where:
- \(\Delta T_f\) is the freezing point depression.
- \(i\) is the van't Hoff factor (number of particles the compound dissociates into).
- \(K_f\) is the freezing point depression constant of the solvent.
- \(m\) is the molality of the solution.
Let's evaluate each option:
A. Pure Water:
- Pure water freezes at 0°C.
- There are no solutes present to cause any freezing point depression.
B. 0.5 M Ionic \( NaCl \):
- Sodium chloride (\( NaCl \)) dissociates into 2 ions: \( Na^+ \) and \( Cl^- \).
- Thus, \( i = 2 \).
C. 0.5 M Ionic \( CaCl_2 \):
- Calcium chloride (\( CaCl_2 \)) dissociates into 3 ions: \( Ca^{2+} \) and 2 \( Cl^- \).
- Thus, \( i = 3 \).
D. 0.5 M Ionic \( AlCl_3 \):
- Aluminum chloride (\( AlCl_3 \)) dissociates into 4 ions: \( Al^{3+} \) and 3 \( Cl^- \).
- Thus, \( i = 4 \).
E. 0.5 M Molecular \( C_{12}H_{22}O_{11} \) (sucrose):
- Sucrose is a molecular compound and does not dissociate into ions in solution.
- Thus, \( i = 1 \).
To determine the lowest freezing point, look for the solution with the highest product of \( i \cdot m \) because it will exhibit the greatest freezing point depression.
Given the molality (\( m \)) is the same (0.5 M) for all solutions:
- Pure water (\( i = 0 \)) has no freezing point depression.
- \( NaCl \) (\( i = 2 \)) leads to an intermediate freezing point depression.
- \( CaCl_2 \) (\( i = 3 \)) results in a larger freezing point depression.
- \( AlCl_3 \) (\( i = 4 \)) results in the greatest freezing point depression.
- Sucrose (\( i = 1 \)) leads to the least freezing point depression among the solutions.
Therefore, the solution with the lowest freezing point is:
[tex]\[ \boxed{D. \, 0.5 \, M \, \text{ionic} \, AlCl_3} \][/tex]
