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.

#"748 torr"#.

(what is the answer in #"mm Hg"#?)

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

#P_A = chi_(A(l)) P_A^"*"#

where:

• #P_A# is the vapor pressure of the solvent in the context of the solution.
• #"*"# indicates pure solvent.
• #chi_(A(l))# 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 #"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 #"CaNO"_3^(+)#.

We ignore the #"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 #"Ca"^(2+)# and #"NO"_3^(-)#, so:

#"Ca"("NO"_3)_2(aq) -> "Ca"^(2+)(aq) + 2"NO"_3^(-)(aq)#

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

#chi_("ions") = n_"ions"/(n_"ions" + n_(H_2O))#

#chi_"ions" + chi_(H_2O) = 1#

We would have three mols of ions, #n_"ions"#, for every one mol of calcium nitrate, #n_(Ca(NO_3)_2)#.

For #"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_"ions" = 3n_(Ca(NO_3)_2) = "0.2560 mols ions"#

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

#0.7n_"ions" = n_"ions"' = "0.1792 mols"#.

Given #"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_("ions")' = (n_"ions"')/(n_"ions"' + n_(H_2O))#

#= "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_2O) = 1 - chi_("ions")' = 0.9841#

Therefore, the solution vapor pressure is:

#color(blue)(P_A) = 0.9841 cdot "760 torr"#

#=# #color(blue)("748 torr")#