Colligative properties, such as boiling point elevation,

depend on the number of dissolved particles in

solution. For nonelectrolytes, no dissociation occurs,

and so you can use the number of moles of solute to

calculate both molality and molarity. In contrast,

electrolytes dissociate, and therefore the molality and

molarity must be calculated based on the number of

moles of dissociated particles or ions.

There are two ions per formula unit of NaCl.

Therefore, we would expect the freezing$-$point

depression $T_f$ of a NaCl solution to be twice that

of a sugar solution of the same concentration.

However, it has been experimentally determined that

in the typical unsaturated solution $T_f$ for the salt

solution is only $1\mathrm{.}9$ times that of the sugar solution.

This indicates that not all ion pairs in the NaCl

solution are dissociated. The number $1\mathrm{.}9$ is called the

van't Hoff factor $($symbolized by $i)$ and can be thought

of as the number of dissociated particles per NaCl

formula unit. This factor changes based on the

concentration of the solution, and each salt will have a

series of experimentally determined values.

How to express the van't Hoff factor

Here are three different methods of expressing the van't Hoff factor:

$i=\frac{\text{a}\text{p}\text{p}\text{a}\text{r}\text{e}\text{n}\text{t}\text{m}\text{o}\text{l}\text{e}\text{s}of\text{p}\text{a}\text{r}\text{t}\text{i}\text{c}\text{l}\text{e}\text{s}in\text{s}\text{o}\text{l}\text{u}\text{t}\text{i}\text{o}\text{n}}{\text{m}\text{o}\text{l}\text{e}\text{s}of\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}}$

$i=\frac{\text{a}\text{p}\text{p}\text{a}\text{r}\text{e}\text{n}\text{t}\text{p}\text{a}\text{r}\text{t}\text{i}\text{c}\text{l}\text{e}\text{c}\text{o}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{t}\text{i}\text{o}\text{n}}{\text{s}\text{o}\text{l}\text{u}\text{t}\text{e}\text{c}\text{o}\text{n}\text{c}\text{e}\text{n}\text{t}\text{r}\text{a}\text{t}\text{i}\text{o}\text{n}}$

$i=\frac{\text{o}\text{b}\text{s}\text{e}\text{r}\text{v}\text{e}\text{d}\text{c}\text{o}\text{l}\text{l}\text{i}\text{g}\text{a}\text{t}\text{i}\text{v}\text{e}\text{p}\text{r}\text{o}\text{p}\text{e}\text{r}\text{t}\text{y}}{\text{c}\text{o}\text{l}\text{l}\text{i}\text{g}\text{a}\text{t}\text{i}\text{v}\text{e}\text{p}\text{r}\text{o}\text{p}\text{e}\text{r}\text{t}\text{y}\text{a}\text{s}\text{s}\text{u}\text{m}\text{i}\text{n}\text{g}no\text{d}\text{i}\text{s}\text{s}\text{o}\text{c}\text{i}\text{a}\text{t}\text{i}\text{o}\text{n}}$

Part A

At what temperature would a $2\mathrm{.}00$mNaCl solution freeze, given that the van't Hoff factor for NaCl is $1\mathrm{.}9?K_f$ for water is

$1\mathrm{.}86\frac{C}{m}.$

Express your answer with the appropriate units.

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Part B

A $0\mathrm{.}050M$ solution of $AlC{l}_3$ had an observed osmotic pressure of $3\mathrm{.}85$atm at $20C.$ Calculate the van't Hoff factor i for $AlC{l}_3.$

Express your answer numerically.

View Available Hint$($s$)$