Question #09585

Apr 12, 2015

The solution in which you dissolve the sodium chloride will have a lower freezing point.

This happens because freezing-point depression is a colligative property that depends on the concentration of particles in solution, not on what those particles are.

Mathematically, this is expressed as

$\Delta {T}_{f} = {K}_{f} \cdot {b}_{F} \cdot i$, where

$\Delta {T}_{f}$ - the freezing-point depression, defined as the freezing temperature of the pure solvent minus the freezing point of the solution;
${K}_{f}$ - the cryoscopic constant, which depends solely on the solvent;
${b}_{F}$ - the molality of the solution;
$i$ - the van't Hoff factor, which takes into account the number of particles produced by a compound when dissolved in solution.

The van't Hoff factor is the key to why the sodium chloride solution will have a lower freezing point.

Sodium chloride is a strong electrolyte, which means it dissociates completely in aqueous solution to give sodium cations and chloride anions

$N a C {l}_{\left(a q\right)} \to N {a}_{\left(A q\right)}^{+} + C {l}_{\left(a q\right)}^{-}$

Notice that 1 mole of $N a C l$ produces 1 mole of $N {a}^{+}$ and 1 mole of $C {l}^{-}$. This means that, regardless of how many moles of $N a C l$ you add to the solution, you'll have twice as many moles of ions present.

As a result, the van't Hoff factor for $N a C l$ will be 2.

This is not the case for ethanol. When placed in aqueous solution, a non-electrolyte does not dissociate to form ions. As a result, its van't Hoff factor will be equal to 1.

To compare the freezing points of the two solutions, just use

$\Delta {T}_{\text{f NaCl}} = {K}_{f} \cdot 0.5 \cdot 2 = {K}_{f}$

$\Delta {T}_{\text{f ethanol}} = {K}_{f} \cdot 0.75 \cdot 1 = 0.75 \cdot {K}_{f}$

or

${T}_{\text{f pure water" - T_"f NaCl" = K_f => T_"NaCl" = T_"f pure water}} - {K}_{f}$

${T}_{\text{f ethanol" = T_"f pure water}} - 0.75 \cdot {K}_{f}$

As you can see,

${T}_{\text{f NaCl" < T_"f ethanol}}$