# When 3 grams of "NaCl" are added to 75.25 grams of water, what is the change in the water's freezing point?

Jul 29, 2016

${0.42}^{\circ} \text{C}$

#### Explanation:

The change in water's freezing point is simply the freezing-point depression that is produced by the dissolving of sodium chloride, $\text{NaCl}$, in $\text{75.25 g}$ of pure water.

As its name suggests, the freezing-point depression tells you by how many degrees the freezing point of the solution will decrease compared with that of the pure solvent.

The equation that allows you to calculate freezing-point depression looks like this

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

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

Water's cryoscopic constant is listed as

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

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

Now, sodium chloride is a strong electrolyte, which means that it dissociates completely in aqueous solution to form sodium cations, ${\text{Na}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$

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

Notice that every mole of sodium chloride that is dissolved in aqueous solution produces $2$ moles of ions.

This means that the van't Hoff factor, which tells you the ratio that exists between the number of moles of solute dissolved and the number of particles of solute produced in solution, will be equal to

$i = 2 \to$ one mole dissolved, two moles of ions produced

The next thing to do here is calculate the solution's molality, which is equal to the number of moles of solute present in $\text{1 kg}$ of solvent.

Use sodium chloride's molar mass to convert the grams of solute to moles

3 color(red)(cancel(color(black)("g"))) * "1 mole NaCl"/(58.443color(red)(cancel(color(black)("g")))) = "0.08555 moles NaCl"

Convert the mass of water from grams to kilograms

75.25 color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("kg")))) = "0.7525 kg"

The molality of the solution will be equal to

$b = {\text{0.08555 moles"/"0.7525 kg" = "0.1137 mol kg}}^{- 1}$

Plug in your values to calculate the value of $\Delta {T}_{f}$

DeltaT_f = 2 * 1.86^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.1137color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg")))

$\Delta {T}_{f} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{0.42}^{\circ} \text{C}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

I'll leave the answer rounded to two sig figs, but don't forget that you have one sig fig for the mass of sodium chloride.

So, this value tells you that the freezing point of the solution will be ${0.42}^{\circ} \text{C}$ lower than the freezing point of pure water.