Concentration questions about freezing point depression and osmolarity?

A) Will glucose or fructose have the greater freezing point depression? B) What are the osmolarities of $\text{0.10 M NaCl}$ and ${\text{0.080 M Na"_2"SO}}_{4}$?

Aug 15, 2017

A) Well, don't fructose and glucose have the same molar mass? Not that it matters...

$\Delta {T}_{f} = {T}_{f} - {T}_{f}^{\text{*}} = - i {K}_{f} m$

• ${T}_{f}$ is the freezing point. $\text{*}$ indicates pure solvent.
• $i$ is the van't Hoff factor, i.e. the effective number of solute particles in solution.
• $m$ is the molality, $\text{mols solute/kg solvent}$, but you are given the same concentration each time...

In principle, it doesn't matter what the identity of the compound even is...

That is the whole idea behind colligative properties. Under the general chemistry approximation, as long as it dissociates to produce the same number of particles in solution... $\Delta {T}_{f}$ is the same.

Hence, they have the same freezing point at the same concentration.

B) Osmolarity is analogous to molarity, and treats the dissociated particles on the same footing.

${\text{NaCl"(aq) -> "Na"^(+)(aq) + "Cl}}^{-} \left(a q\right)$

As ${\text{Na}}^{+}$ and ${\text{Cl}}^{-}$ go into solution, $\text{0.10 M}$ $\text{NaCl}$ is then...

$\underline{\text{0.20 Osm/L}}$.

${\text{Na"_2"SO"_4(aq) -> 2"Na"^(+)(aq) + "SO}}_{4}^{2 -} \left(a q\right)$

$\text{0.080 M}$ Sodium sulfate, ${\text{Na"_2"SO}}_{4}$, has three times the osmolarity as what is shown for its molarity, and it thus is...

$\underline{\text{0.240 Osm/L}}$