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?

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

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