# Question 9d5e0

Jun 28, 2016

$- {1.9}^{\circ} \text{C}$

#### Explanation:

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

Once you know that, you can use the freezing point of the pure solvent to find the freezing point of the solution.

So, the freezing-point depression is calculated using the equation

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

${K}_{f} = {1.86}^{\circ} {\text{C kg mol}}^{- 1}$

Now, sodium chloride, $\text{NaCl}$, is a soluble ionic compound, which means that is dissociates completely in aqueous solution to form sodium cations and chloride anions

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

Notice that every mole of sodium chloride added to the solution produces $1$ mole of sodium cations and $1$ mole of chloride anions.

This means that the van't Hoff factor for sodium chloride, which essentially tells you how many moles of particles of solute are produced in solution per mole of solute dissolved, will be equal to $2$.

Plug in your values to find the freezing-point depression of the solution

DeltaT_f = 2 * 1.86""^@"C" color(red)(cancel(color(black)("kg"))) * color(red)(cancel(color(black)("mol"^(-1)))) * 0.50color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))

$\Delta {T}_{f} = {1.86}^{\circ} \text{C}$

The freezing-point depression is defined as

color(purple)(|bar(ul(color(white)(a/a)color(black)(DeltaT_f = T_f^@ - T_"f sol")color(white)(a/a)|)))" "

Here

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

Pure water freezes at ${0}^{\circ} \text{C}$, so your solution will freeze at

DeltaT_"f sol" = 0^@"C" - 1.86^@"C" = color(green)(|bar(ul(color(white)(a/a)color(black)(-1.9^@"C")color(white)(a/a)|)))#

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