Calculate the vapor pressure at 25°C of a solution containing 55.3 g ethylene glycol (HOCH2CH2OH) and 285.2 g water. The vapor pressure of pure water at 25°C is 23.8 torr.

Vapor pressure of solution = 23.9 Torr

Explanation:

Let's apply the colligative poperty of vapor pressure to solve this:

ΔP = P°. Xm

ΔP = Vapor pressure of pure solvent - Vapor pressure of solution

We have solvent and solute mass, so let's find out the moles of each.

55.3 g / 62 g/mol = 0.89 moles

285.2 g / 18 g/mol = 15.84 moles

Let's determine the mole fraction of ethylene glycol.

Mole fraction = Moles of ethylene glyco / Total moles

0.89 moles / (0.89 + 15.84) = 0.053

25.3 Torr - Vapor pressure of solution = 25.3 Torr. 0.053

Vapor pressure of solution = 25.3 Torr. 0.053 - 25.3 Torr

Vapor pressure of solution = 23.9 Torr

P (25°C) = 22.523 torr

Explanation:

low volatile solute and diluted solution:

ΔP = P - P*a = - Xb. P*a

∴ a: water

∴ b: C2H6O2

∴ P*a (25°C) = 23.8 torr

Xb = (wb/Mb) / (wa/Ma + wb/Mb)

∴ wb = 55.3 g

∴ Mb = 62.07 g/mol

∴ wa = 285.2 g

∴ Ma = 18.015 g/mol

⇒ Xb = (55.3/62.07) / ((285.2/18.015) + (55.3/62.07))

⇒ Xb = 0.0536

⇒ P (25°C) = P*a - Xb. P*a

⇒ P (25°C) = 23.8 torr - ((0.0536) (23.8 torr))

⇒ P (25°C) = 22.523 torr