Water has a boiling point of 100.0°c and a kb of 0.512°c/m. what is the boiling
point of a 8.5 m solution of mg3(po4)2 in water?

 121.76 °C

Explanation:

Data Given:

Boiling point of water = 100.0°C

Kb (boiling point constant = 0.512°C/m

Concentration of the Mg₃(PO₄)₂ = 8.5 m

Solution:

Formula Used to find out boiling point

               ΔTb = m.Kb.i . . . . . . (1)

where

ΔTb = boiling point elevation

m = concentration of solution

kb = molal boiling point elevation constant

i = Van't Hoff factor

Van't Hoff factor : It is the number of ions of ionic compound in the solution produced by it when it dissolve in solvent.

Van't Hoff factor for Mg₃(PO₄)₂:

Assuming ideal Van't Hoff factor Look for ionization of Mg₃(PO₄)₂ in solution

        Mg₃(PO₄)₂ > 3 Mg²⁺ + 2 PO₄³⁻

From above equation it is clear that it give total 5 ions in solution form upon dissolving i.e. 3 Mg²⁺ ions and 2 PO₄³⁻ ions.

So,

Van't Hoff factor = 5

Now

As,

       ΔTb = boiling point of solution - boiling point of water

So,

we can write equation 1 as under

               Tb (Solution) -Tb (water) = m.Kb.i

As we have to find out boiling point so rearrange the above equation

              Tb (Solution)  = (m . Kb . i)+ Tb (water) . . . . . . . (2)

Put values in Equation 2

            Tb (Solution)  = (8.5 m x 0.512°C/m x 5) + 100.0°C

            Tb (Solution)  = 21.76 °C + 100.0°C

            Tb (Solution)  = 121.76°C

so the boiling point of 8.5m Mg₃(PO₄)₂ solution =  121.76 °C