Answer:
121.76 °C
Step-by-step 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