Question

The degree of dissociation of `"Ca"("NO"_3)_2` in a dilute aqueous solution containing 14g of the salt per 200 g of water at `100^@"C"` is 70% .If the vapour pressure of water is 760 mmHg , what will be the vapour pressure of the solution ?

Answer 1

`"745 mmHg"`

Explanation:

The idea here is that you need to find the number of moles of particles of solute present in your solution.

Knowing this will allow you to find the mole fraction of the solvent, which will then get you the vapor pressure of the solution.

Notice that a formula unit of calcium nitrate, `"Ca"("NO"_3)_2`, contains

• one mole of calcium cations, `1 xx "Ca"^(2+)`
• two moles of nitrate anions, `2 xx "NO"_3^(-)`

This means that every mole of calcium nitrate that dissociates in aqueous solution will produce one mole of calcium cations and two moles of nitrate anions

`"Ca"("NO"_ 3)_ (2(aq)) -> "Ca"_ ((aq))^(2+) + 2"NO"_(3(aq))^(-)`

In other words, for every mole of calcium nitrate that you dissolve in solution, you get three moles of solute particles, i.e .of ions.

Now, use calcium nitrate's molar mass to determine how many moles you're adding to the solution

`14 color(red)(cancel(color(black)("g"))) * ("1 mole Ca"("NO"_3)_2)/(164.09color(red)(cancel(color(black)("g")))) = "0.0853 moles Ca"("NO"_3)_2`

At `100%` dissociation, this many moles of calcium nitrate would produce

`0.0853color(red)(cancel(color(black)("moles Ca"("NO"_3)_2))) * "3 moles ions"/(1color(red)(cancel(color(black)("mole Ca"("NO"_3)_2)))) = "0.256 moles ions"`

However, you know that the salt has a `70%` degree of dissociation in water at `100^@"C"`. This means that the actual number of moles of particles of solute present in solution will be

`0.256 color(red)(cancel(color(black)("moles ions"))) * overbrace("70 moles ions"/(100color(red)(cancel(color(black)("moles ions")))))^(color(purple)("= 70% dissociation")) = "0.179 moles ions"`

Use water's molar mass to determine how many moles of solvent you have present in the solution

`200color(red)(cancel(color(black)("g"))) * ("1 mole H"_2"O")/(18.015color(red)(cancel(color(black)("g")))) = "11.1 moles H"_2"O"`

Now, according to Raoult's Law, the vapor pressure of a solution that contains a non-volatile solute will depend on the mole fraction of the solvent, `chi_"solvent"`, and on the vapor pressure of the pure solvent, `P_"solvent"^@`

`color(blue)(|bar(ul(color(white)(a/a)P_"solution" = chi_"solvent" xx P_"solvent"^@color(white)(a/a)|)))`

The mole fraction of water will be equal to the number of moles of water divided by the total number of moles present in solution

`chi_"water" = (11.1 color(red)(cancel(color(black)("moles"))))/((11.1 + 0.179color(red)(cancel(color(black)("moles"))))) = 0.984`

The vapor pressure of the solution will thus be

`P_"solution" = 0.984 xx "760 mmHg" = color(green)(|bar(ul(color(white)(a/a)"745 mmHg"color(white)(a/a)|)))`

I'll leave the answer rounded to three sig figs, despite the fact that your values do not justify using this many sig figs.