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iris [78.8K]
4 years ago
11

Se prepara una solución de KCl disolviendo 25.0 g de KCl en 250.0 g de agua a 25 ° C. ¿Cuál es la presión de vapor de la solució

n si la presión de vapor del agua a 25 ° C es 23.76 mm Hg?
Chemistry
1 answer:
Nesterboy [21]4 years ago
3 0

Answer:

Vapor pressure of solution is 24.85 mmHg

Explanation:

Colligative property of vapor pressure lowering to solve this:

ΔP = P° . Xm . i

Xm = molar fraction (moles of solute / total moles)

i = Van't Hoff factor ( amount of ions dissolved in solution)

P° = Vapor pressure of pure solvent

ΔP = Vapor Pressure of solution - Vapor pressure of pure solvent (P°)

KCl dissociates like this:

KCl → K⁺  + Cl⁻

So i = 2

Let's calculate the Xm

Moles of KCl = Mass KCl / Molar mass

Moles of H₂O = Mass H₂O / Molar mass H₂O

Moles KCl + Moles H₂O = Total moles

25 g/74.55 g/m = 0.335 moles KCl

250 g/18 g/m = 13.89 moles H₂O

13.88 + 0.335 = 14.22 Total moles

Xm = 0.335 / 14.22 = 0.023

Let's replace in the formula

Vapor pressure of solution - 23.76 mmHg = 23.76 mmHg .  0.023  .2

Vapor pressure of solution = (23.76 mmHg .  0.023  .2) + 23.76 mmHg

Vapor pressure of solution = 1.09 mmHg + 23.76 mmHg

Vapor pressure of solution = 24.85 mmHg

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