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?

The boiling  point of a 8.5 m solution of Mg3(PO4)2 in water is 394.91 K.

Explanation:

The formula for molal boiling Point elevation is :

= elevation in boiling Point

= Boiling point constant( ebullioscopic constant)

m = molality of the solution

i = Van't Hoff Factor

Van't Hoff Factor = It takes into accounts,The abnormal values of Temperature change due to association and dissociation .

In solution Mg3(PO4)2 dissociates as follow :

Total ions after dissociation in solution :

= 3 ions of Mg + 2 ions of phosphate

Total ions = 5

i = Van't Hoff Factor = 5

m = 8.5 m

= 0.512 °C/m

Insert the values and calculate temperature change:

Boiling point of pure water = 100°C = 273.15 +100 = 373.15 K

= 373.15 K[/tex]

21.76 = T - 373.15

T = 373.15 + 21.76

T =394.91 K

a. hope i'm right

explanation: