Question #83c33

Mar 7, 2016

${\text{MgI}}_{\textrm{2 \left(a q\right]}}$

Explanation:

I'm going to go out on a limb here and assume that the problem wants you to determine which solution has the higher boiling point, since that would make sense in this context.

A solution's boiling point is a colligative property, which means that it depends on the concentration of the particles present in that solution and not on their nature.

In simple terms, a solution's boiling point will be affected by the number of molecules of solute it contains, not by the kind of molecules it contains.

More specifically, the higher the concentration of solute particles, the higher the boiling point.

In your case, you're dealing with two ionic compounds, potassium iodide, $\text{KI}$, and magnesium iodide, ${\text{MgI}}_{2}$, both soluble in water.

This means that in aqueous solution, these compounds dissociate completely to form cations and anions. The two solutions will contain

${\text{KI"_text((aq]) -> "K"_text((aq])^(+) + "I}}_{\textrm{\left(a q\right]}}^{-}$

and

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

So, what happens when you dissolve one mole of potassium iodide in water? The compound dissociates to produce one mole of potassium cations, ${\text{K}}^{+}$, and one mole of iodide anions, ${\text{I}}^{-}$.

This means that you have

$\text{1 mole KI " -> " 2 moles particles}$

What about when one mole of magnesium iodide dissolves in water? This time, the compound dissociates to produce one mole of magnesium cations, ${\text{Mg}}^{2 +}$, and two moles of iodide anions, $2 \times {\text{I}}^{-}$.

This means that you have

$\text{1 mole MgI"_2 -> "3 moles particles}$

So, which solution will have a higher boiling point?

Since a higher concentration of particles means a higher boiling point, the magnesium iodide solution will have a higher boiling point that the potassium iodide solution.

In other words, the magnesium iodide solution has a higher bolling-point elevation, which means that adding magnesium iodide to pure water will increase the boiling point of the solution more than in the case of potassium iodide.

Mathematically, boiling-point elevation is written like this

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

$i$ - the van't Hoff factor;
${K}_{b}$ - the ebullioscopic constant of the solvent;
$b$ - the molality of the solution.
$\Delta {T}_{b}$ - the poiling point elevation - defined as ${T}_{\text{b" - T_"b}}^{\circ}$

The van't Hoff factor tells you how many moles of particles you get per mole of dissolved solute. Since we've already determined this, you can say that the two solutions will have

$\text{1 mole KI " -> " 2 moles particles} \implies \textcolor{red}{i = 2}$

$\text{1 mole MgI"_2 -> color(white)(a)"3 moles particles} \implies \textcolor{red}{i = 3}$

So, as a conclusion, the solution with the higher concentration of solute particles will have the higher boiling point.