Question cb01d

Jul 5, 2016

$\Delta {T}_{b} = {0.58}^{\circ} \text{C}$

Explanation:

Your tool of choice here will be the equation that helps you find the boiling-point elevation of the solution

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

Here

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

You already know the molality of the solution, which is given as

${\text{0.38 m" = "0.38 mol kg}}^{- 1}$

and the ebullioscopic constant of water, which is given as ${\text{0.51 K kg mol}}^{- 1}$.

You can rewrite the ebullioscopic constant of water in terms of degrees Celsius by using the fact that an increase of $\text{1 K}$ is equivalent to an increase of ${1}^{\circ} \text{C}$.

You will thus have

${K}_{b} = {\text{0.51"""^@"C kg mol}}^{- 1}$

Now, focus on find the van't Hoff factor. Calcium chloride, ${\text{CaCl}}_{2}$, is a soluble ionic compound that dissociates completely in aqueous solution to produce calcium cations, ${\text{Ca}}^{2 +}$, and chloride anions, ${\text{Cl}}^{-}$

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

Notice that every mole of calcium chloride that dissolves in solution produces $1$ mole of calcium cations and $\textcolor{red}{2}$ moles of chloride anions.

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

In this case, one mole of solute produces

$\text{1 mole Ca"^(2+) + color(red)(2)color(white)(a)"moles Cl"^(-) = "3 moles of ions}$

Therefore, the van't Hoff factor will be equal to $3$.

Plug in your values into the equation and solve for the boiling-point elevation of the solution

DeltaT_b = 3 * 0.51""^@"C" color(red)(cancel(color(black)("kg"))) color(red)(cancel(color(black)("mol"^(-1)))) * 0.38 color(red)(cancel(color(black)("mol"))) color(red)(cancel(color(black)("kg"^(-1))))#

$\Delta {T}_{b} = \textcolor{g r e e n}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{{0.58}^{\circ} \text{C}} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

The answer is rounded to two sig figs.

Pure water boils at a temperature of ${100}^{\circ} \text{C}$ at normal pressure, so your solution will boil at

${T}_{b} = {100}^{\circ} \text{C" + 0.58^@"C" = 100.58^@"C}$