Question

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?​

Answer 1

104.352°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 . . . . . . (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