# If "14 g" of calcium nitrate is dissolved in "200 g" of water, and it experiences a 70% dissociation, what is the solution's vapor pressure if the solvent began at the normal boiling point of water? Hint: the normal boiling point occurs at 760 torr.

Jun 7, 2017

$\text{748 torr}$.

(what is the answer in $\text{mm Hg}$?)

Vapor pressure of the solution relative to the solvent is given by Raoult's law for ideal solutions:

${P}_{A} = {\chi}_{A \left(l\right)} {P}_{A}^{\text{*}}$

where:

• ${P}_{A}$ is the vapor pressure of the solvent in the context of the solution.
• $\text{*}$ indicates pure solvent.
• ${\chi}_{A \left(l\right)}$ is the mol fraction of solvent in the solution phase.

When nonvolatile solute is added to a solvent, it decreases its vapor pressure. Thus, the answer will be less than $\text{760 torr}$.

In this case, we have a strong electrolyte. Normally, we assume 100% dissociation of strong electrolytes, but in this case there is ion pairing to form ${\text{CaNO}}_{3}^{+}$.

We ignore the ${\text{CaNO}}_{3}^{+}$ species in solution for simplicity (which is supported by the fact the solution is dilute so that the ions have a hard time finding each other)

The salt dissociates into ${\text{Ca}}^{2 +}$ and ${\text{NO}}_{3}^{-}$, so:

${\text{Ca"("NO"_3)_2(aq) -> "Ca"^(2+)(aq) + 2"NO}}_{3}^{-} \left(a q\right)$

To calculate the new vapor pressure, we'll need the mol fraction $\chi$ of the ions or of the water:

${\chi}_{\text{ions") = n_"ions"/(n_"ions} + {n}_{{H}_{2} O}}$

${\chi}_{\text{ions}} + {\chi}_{{H}_{2} O} = 1$

We would have three mols of ions, ${n}_{\text{ions}}$, for every one mol of calcium nitrate, ${n}_{C a {\left(N {O}_{3}\right)}_{2}}$.

For $\text{14 g}$ of the salt...

n_(Ca(NO_3)_2) = 14 cancel("g Ca"("NO"_3)_2) xx ("1 mol Ca"("NO"_3)_2)/(164.088 cancel("g Ca"("NO"_3)_2))

$=$ "0.08532 mols Ca"("NO"_3)_2

Again, three times the mols of ions result from 100% dissociation, so:

${n}_{\text{ions" = 3n_(Ca(NO_3)_2) = "0.2560 mols ions}}$

With 70% dissociation, we simply have 70% of the mols, or

$0.7 {n}_{\text{ions" = n_"ions"' = "0.1792 mols}}$.

Given $\text{200 g}$ of water, we have:

200 cancel("g H"_2"O") xx ("1 mol H"_2"O")/(18.015 cancel("g H"_2"O")) = "11.10 mols H"_2"O"

This gives us a mol fraction of:

${\chi}_{\text{ions")' = (n_"ions"')/(n_"ions} ' + {n}_{{H}_{2} O}}$

= "0.1792 mols"/("0.1792 mols" + "11.10 mols")

$= 0.01588$

or since all mol fractions for a solution when added together equal $1$,

${\chi}_{{H}_{2} O} = 1 - {\chi}_{\text{ions}} ' = 0.9841$

Therefore, the solution vapor pressure is:

$\textcolor{b l u e}{{P}_{A}} = 0.9841 \cdot \text{760 torr}$

$=$ $\textcolor{b l u e}{\text{748 torr}}$