# A solution is made by dissolving 25.0 g of magnesium chloride crystals in 1000 g of water. What will be the freezing point of the new solution assuming complete dissociation of the MgCl_2 salt?

## What will the boiling point of the new solution assuming complete dissociation of the $M g C {l}_{2}$ salt?

Jan 8, 2016

Here's what I got.

#### Explanation:

!! Long ANSWER !!

Magnesium chloride, ${\text{MgCl}}_{2}$, is a soluble ionic compound that dissociates in aqueous solution to form magnesium cations, ${\text{Mg}}^{2 +}$, and chloride anions, ${\text{Cl}}^{-}$

${\text{MgCl"_text(2(aq]) -> "Mg"_text((aq])^(2+) + 2"Cl}}_{\textrm{\left(a q\right]}}^{-}$

Now, if you assume that the salt dissociates completely, you can say that every mole of magnesium chloride will produce three moles of ions in solution

• one mole of magnesium cations
• two moles of chloride anions

As you know, the freezing point of a solution depends on how many particles of solute you have present, not on the nature of the solute $\to$ colligative property.

Mathematically, you can express the freezing-point depression of a solution by using the equation

$\textcolor{b l u e}{\Delta {T}_{f} = i \cdot {K}_{f} \cdot b} \text{ }$, 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

The van't Hoff factor tells you the ratio between the concentration of particles produces in solution when a substance is dissolved, and the concentration of said substance.

Since you know that every mole of magnesium chloride produces three moles of ions in solution, you can say that the van't Hoff factor will be equal to $3$.

In order to find the molality of the solution, you need to know how many moles of solute you have in that $\text{25.0-g}$ sample.

To do that, use the compound's molar mass

25.0 color(red)(cancel(color(black)("g"))) * "1 mole MgCl"_2/(95.21color(red)(cancel(color(black)("g")))) = "0.2626 moles MgCl"_2

Now, molality is defined as moles of solute per kilograms of solvent.

$\textcolor{b l u e}{b = {n}_{\text{solute"/m_"solvent}}}$

In your case, you will have

$b = \text{0.2626 moles"/(1000 * 10^(-3)"kg") = "0.2626 molal}$

This means that the freezing-point depression will be

$\Delta {T}_{f} = 3 \cdot {1.86}^{\circ} \text{C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.2626color(red)(cancel(color(black)("mol")))color(red)(cancel(color(black)("kg"^(-1)))) = 1.465^@"C}$

The freezing-point depression is defined as

color(blue)(DeltaT_f = T_f^@ - T_"f sol")" ", where

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

This means that you have

${T}_{\text{f sol}} = {T}_{f}^{\circ} - \Delta {T}_{f}$

${T}_{\text{f sol" = 0^@"C" - 1.465^@"C" = -1.465^@"C}}$

You should round this off to one sig fig, since that is how many sig figs you have for the mass of water, but I'll leave it rounded to two sig figs

T_"f sol" = color(green)(-1.5^@"C")

Now for the boiling point of this solution. The equation for boiling-point elevation looks like this

$\textcolor{b l u e}{\Delta {T}_{b} = i \cdot {K}_{b} \cdot b} \text{ }$, where

$\Delta {T}_{b}$ - the boiling-point elevation;
$i$ - the van't Hoff factor
${K}_{b}$ - the ebullioscopic constant of the solvent;
$b$ - the molality of the solution.

The ebullioscopic constant for water is equal to ${0.512}^{\circ} {\text{C kg mol}}^{- 1}$

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

Plug in your values to get

$\Delta {T}_{b} = 3 \cdot {0.512}^{\circ} \text{C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.2626 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1)))) = 0.403^@"C}$

The boiling-point elevation is defined as

color(blue)(DeltaT_b = T_"b sol" - T_b^@)" ", where

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

In your case, you have

${T}_{\text{b sol}} = \Delta {T}_{b} + {T}_{b}^{\circ}$

${T}_{\text{b sol" = 0.403^@"C" + 100^@"C" = 100.403^@"C}}$

I'll leave this answer as

T_"b 'sol" = color(green)(100.4^@"C")