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

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 :

`\Delta T_(b) = iK_(b)m`

`\Delta T_(b)` = elevation in boiling Point

`K_(b)` = 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 :

`Mg_(3)(PO_(4))_(2)\rightarrow 3Mg^(2+) + 2 PO_(4)^(3-)`

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

`K_(b)` = 0.512 °C/m

Insert the values and calculate temperature change:

`\Delta T_(b) = iK_(b)m`

`\Delta T_(b) = 5* 0.512* 8.5`

`\Delta T_(b) = 21.76 K`

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

`\Delta T_(b) = T_(b) - T_(b)_(pure)`

`T_(b)_(pure)` = 373.15 K[/tex]

21.76 = T - 373.15

T = 373.15 + 21.76

T =394.91 K