# What is the molality of NaCl if the freezing point of a NaCl solution is -5.58 C°?

Mar 18, 2016

${\text{1.50 mol kg}}^{- 1}$

#### Explanation:

A solution's molality will tell you how many moles of solute you get per kilogram of solvent.

In this case, the only information you have is that a given sodium chloride solution has a freezing point of $- {5.58}^{\circ} \text{C}$.

As you know, the freezing point of a solution is a colligative property, meaning that is depends exclusively on the concentration of solute particles.

Since sodium chloride is an electrolyte, it will dissociate completely in aqueous solution to form sodium cations and chloride anions

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

Notice that one mole of sodium chloride produces one mole of sodium cations and one mole of chloride anions in solution, i.e. two moles of particles.

This mans that sodium chloride has a van't Hoff factor, $i$, equal to $\textcolor{red}{2}$. The van't Hoff factor essentially tells you how many moles of particles you get per mole of solute dissolved in aqueous solution.

Now, you can compare the freezing point of the solution with that of pure water, which is equal to ${0}^{\circ} \text{C}$. The difference between the two freezing points will give you the freezing-point depression of the solution, $\Delta {T}_{f}$

color(blue)(|bar(ul(color(white)(a/a)DeltaT_f = T_f^@ - T_"f sol"color(white)(a/a)|)))" ", where

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

In your case, the freezing-point depression will be equal to

$\Delta {T}_{f} = {0}^{\circ} \text{C" - (-5.58^@"C") = 5.58^@"C}$

The freezing-point depression is also related to the molality of the solution

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \Delta {T}_{f} = i \cdot {K}_{f} \cdot b \textcolor{w h i t e}{\frac{a}{a}} |}}} \text{ }$, 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 cryoscopic constant of water is equal to $1.86 {\text{^@"C kg mol}}^{- 1}$

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

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 get

b = (5.58color(red)(cancel(color(black)(""^@"C"))))/(color(red)(2) * 1.86color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = color(green)(|bar(ul(color(white)(a/a)"1.50 mol kg"^(-1)color(white)(a/a)|)))

The answer is rounded to three sig figs.