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ludmilkaskok [199]
3 years ago
10

Water has a boiling point of 100.0°C and a Kb of 0.512°C/m. What is the boiling

Chemistry
1 answer:
Reil [10]3 years ago
8 0

Answer:

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

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

<em>i =</em> 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

<em>i =</em> 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\times 0.512\times 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

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