Answer:
104.352°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 . . . . . . (1)
where
ΔTb = boiling point of solution - boiling point of water
So,
we can write equation 1 as under
ΔTb = Tb (Solution) -Tb (water)
As we have to find out boiling point so rearrange the above equation
Tb (Solution) = m.Kb + Tb (water) . . . . . . . (2)
Put values in Equation 2
Tb (Solution) = (8.5 m x 0.512°C/m ) + 100.0°C
Tb (Solution) = 4.352 + 100.0°C
Tb (Solution) = 104.352°C
so the boiling point of Mg₃(PO₄)₂ 8.5 m solution = 104.352°C
