# Question 75a08

Dec 9, 2015

${\text{0.0559 mol kg}}^{- 1}$

#### Explanation:

AN important part of this problem is the fact that you're dealing with a molecular compound. That means that it will dissolve in aqueous solution without dissociating into ions.

An important consequence of that is the fact that the van't Hoff factor, $i$, which gives you the ratio between the number of particles of solute you start with and the number of particles produced in solution by dissolving the solute, will be equal to $1$.

In simple words, one molecule dissolved, one molecule in solution.

Now, the equation that describes freezing-point depression looks like this

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

$\Delta {T}_{f}$ - the freezing-point depression;
$i$ - the van't Hoff factor, equal to $1$ for non-electrolytes;
${K}_{f}$ - the cryoscopic constant of the solvent;
$b$ - the molality of the solution.

The cryoscopic constant of water is equal to $1.86 {\text{^@"C kg mol}}^{- 1}$

http://www.vaxasoftware.com/doc_eduen/qui/tcriosebu.pdf

Now, the freezing-point depression is defined as

$\textcolor{b l u e}{\Delta {T}_{f} = {T}_{f}^{\circ} - {T}_{f}} \text{ }$, where

${T}_{f}^{\circ}$ - the freezing point of the pure solvent
${T}_{f}$ - the freezing point of the solution

In your case, the freezing-point depression will be

$\Delta {T}_{f} = {0}^{\circ} \text{C"- (-0.104^@"C") = 0.104^@"C}$

So, all you have to do from this point on is to plug your values into this equation and solve for $b$, the molality of the solution.

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

b = (0.104 color(red)(cancel(color(black)(""^@"C"))))/(1 * 1.86 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = color(green)("0.0559 mol kg"^(-1))#