Which has the lowest freezing point?

Question

A.
pure water
B.
0.5 M ionic NaCl
C.
0.5 M ionic CaCl2
D.
0.5 M ionic AlCl3
E.
0.5 M molecular C12H22O11

Answer 1

To determine which solution has the lowest freezing point, we can use the concept of freezing point depression, which is given by the formula:

$$\mathrm{\Delta }{T}_f=i\cdot K_f\cdot m$$

where:

• $\mathrm{\Delta }{T}_f$ is the change in freezing point (how much lower the freezing point is),
• $i$ is the van 't Hoff factor (the number of particles the solute breaks into),
• $K_f$ is the freezing point depression constant (which is specific to the solvent, and we assume it's the same for all since they are in water),
• $m$ is the molality of the solution.

We will analyze each option based on $i$:

1. Pure water:

   • Freezing point: 0 °C
   • $i=1$ (1 particle)

2. 0.5 M ionic NaCl:

   • NaCl dissociates into Na${}^{+}$ and Cl${}^{-}$, so:
   • $i=2$
   • Effective concentration = $0.5\times 2=1.0$

3. 0.5 M ionic CaCl2:

   • CaCl2 dissociates into Ca${}^{2+}$ and 2 Cl${}^{-}$, so:
   • $i=3$
   • Effective concentration = $0.5\times 3=1.5$

4. 0.5 M ionic AlCl3:

   • AlCl3 dissociates into Al${}^{3+}$ and 3 Cl${}^{-}$, so:
   • $i=4$
   • Effective concentration = $0.5\times 4=2.0$

5. 0.5 M molecular C12H22O11 (sucrose):

   • Sucrose does not dissociate, so:
   • $i=1$
   • Effective concentration = $0.5\times 1=0.5$

Now we summarize the effective concentrations:

• Pure water: 0
• NaCl: 1.0
• CaCl2: 1.5
• AlCl3: 2.0
• C12H22O11: 0.5

The solution with the highest effective concentration (and thus the lowest freezing point) is 0.5 M ionic AlCl3.

Thus, D. 0.5 M ionic AlCl3 has the lowest freezing point.