# The freezing point of an aqueous solution that contains a nonelectrolyte is -4.0^@"C". What is the molal concentration of the solution? Kf = 1.86^@"C/m"

Jul 10, 2016

${\text{2.2 mol kg}}^{- 1}$

#### Explanation:

The first thing to do here is use the freezing point of the solution to calculate the freezing-point depression, $\Delta {T}_{f}$.

As you know, the freezing-point depression is a measure of how low the freezing point of a solution is compared with the freezing point of the pure solvent.

You're dealing with an aqueous solution, so right from the start you know that the solvent is water. Pure water freezes at ${0}^{\circ} \text{C}$ under normal pressure.

Your solution freezes at $- {4.0}^{\circ} \text{C}$. This means that the freezing-point depression account for a ${4.0}^{\circ} \text{C}$ decrease in the freezing temperature of the solution compared with that of pure water

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{T}_{\text{f sol" = T_"f pure solvent}} - \Delta {T}_{f}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

$- {4.0}^{\circ} \text{C" = - DeltaT_f implies DeltaT_f = 4.0^@"C}$

Now, the freezing-point depression is calculated sing the equation

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \Delta {T}_{f} = i \cdot {K}_{f} \cdot b \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Here

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

Your solute is a non-electrolyte, which means that its molecules do not dissociate in aqueous solution. This implies that the van't Hoff factor will be equal to $i = 1$.

Rearrange the above equation to solve for $b$

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

Plug in your values to find

$b = \left(4.0 \textcolor{red}{\cancel{\textcolor{b l a c k}{{\text{^@"C"))))/(1 * 1.86 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = color(green)(|bar(ul(color(white)(a/a)color(black)("2.2 mol kg}}^{- 1}} \textcolor{w h i t e}{\frac{a}{a}} |}}\right)$

The answer is rounded to two sig figs, the number of sig figs you have for the freezing point of the solution.