# Question 4b1a8

Feb 9, 2017

${\text{63 g CaCl}}_{2}$

#### Explanation:

Your tool of choice here will be the equation

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\Delta T = i \cdot {K}_{f} \cdot b}}}$

Here

• $\Delta T$ is the freezing-point depression
• $i$ is the van't Hoff factor
• ${K}_{f}$ is the cryoscopic constant of the solvent, which in your case is water
• $b$ is the molality of the solution

Now, water has a cryoscopic constant equal to

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

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

Calcium chloride, ${\text{CaCl}}_{2}$, is a soluble ionic compound that dissociates completely in aqueous solution to form calcium cations and chloride anions

${\text{CaCl"_ (2(aq)) -> "Ca"_ ((aq))^(2+) + 2"Cl}}_{\left(a q\right)}^{-}$

Notice that every mole of calcium chloride that dissolves in water produces $3$ moles of ions

• one mole of calcium cations, $1 \times {\text{Ca}}^{2 +}$
• two moles of chloride anions, $2 \times {\text{Cl}}^{-}$

The van't Hoff factor tells you the ratio that exists between the number of moles of solute dissolved in solution and the number of moles of particles of solute produced in solution.

In this case, $1$ mole of calcium chloride produces $3$ moles of ions, so you have

$i = 3 \to$ one mole of calcium chloride dissolved, three moles of ions produced

Now, water has a normal freezing point of ${0}^{\circ} \text{C}$, which means that a solution that freezes at $- {4.2}^{\circ} \text{C}$ will have a freezing-point depression of

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\Delta T = {T}_{\text{f pure water" - T_"f solution}}}}}$

DeltaT = 0^@"C" - (-4.2^@"C")

$\Delta T = {4.2}^{\circ} \text{C}$

Rearrange the above equation and solve for $b$, the molality of the solution

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

Plug in your values to find

b = (4.2 color(red)(cancel(color(black)(""^@"C"))))/(3 * 1.86 color(red)(cancel(color(black)(""^@"C"))) "kg mol"^(-1)) = "0.753 mol kg"^(-1)

The molality of the solution is calculated by taking the number of moles of solute present in $\text{1 kg}$ of solvent.

In this case, your solution contains $0.753$ moles of calcium chloride for every $\text{1 kg}$ of water. Your solution contains

750. color(red)(cancel(color(black)("g"))) * "1 kg"/(10^3color(red)(cancel(color(black)("g")))) = "0.750 kg"

of water, which means that it also contains

0.750 color(red)(cancel(color(black)("kg water"))) * "0.753 moles CaCl"_2/(1color(red)(cancel(color(black)("kg water")))) = "0.565 moles CaCl"_2#

To convert this to grams, use the molar mass of calcium chloride

$0.565 \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CaCl"_2))) * "110.98 g"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = color(darkgreen)(ul(color(black)("63 g}}}}$

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