# Question #4859e

Sep 26, 2015

$2.709$

#### Explanation:

Even without doing any calculations, you could say that the answer should be $3$, since sodium sulfate dissociates in aqueous solution to give

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

This means that, in theory, the van't Hoff factor, which gives you the number of particles produced when one molecule or formula unit, depending if you have a covalent or an ionic solute, of the solute dissolves in the solvent, will be equal to $3$.

So, the equation that gives you the freeezing-point depression is

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

$\Delta {T}_{f}$ - the freezing-point depression;
$i$ - the van't Hoff factor;
${K}_{f}$ - the cryoscopic constant of the solvent;
$b$ - the molality of the solution.

The freezing-point depression is defined as

$\Delta {T}_{f} = {T}_{\text{f"^0 - T_"f sol}}$

In your case, you know that the freezing point of the solution is equal to $- 4.218 \text{^@"C}$. This means that $\Delta {T}_{f}$ is

$\Delta {T}_{f} = {0}^{\circ} \text{C" - (-4.218""^@"C") = 4.218""^@"C}$

You know that the Cryoscopic constant of water is equal to

${K}_{f} = 1.853 {\text{^@"C kg mol}}^{- 1}$

Rearrange the equation and solve for $i$ to get

$i = \frac{\Delta {T}_{f}}{{K}_{f} \cdot b}$

$i = \left(4.218 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{^@"C"))))/(1.853color(red)(cancel(color(black)(""^@"C"))) color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-)))) * 0.8402color(red)(cancel(color(black)("mol")))/color(red)(cancel(color(black)("kg}}}}\right) = \textcolor{g r e e n}{2.709}$

The van't Hoff factor turned out to be smaller than the predicted value.

Since the solute does not produce as many particles per formula unit as it would produce if completely dissolved, you can conclude that not all the ions are completely separated from each other.

In other words, some sodium cations will actually bind to the sulfate anions and exist as ${\text{Na"^(+)"SO}}_{4}^{2 -}$ solvation cells, so instead of getting three ions for every sodium sulfate formula unit, you get something like

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

and

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