Calculate the mass of ethylene glycol (C2H6O2) that must be added to 1.00 kg of ethanol (C2H5OH) to reduce its vapor pressure by 12.0 torr at 35∘C. The vapor pressure of pure ethanol at 35∘C is 1.00*102 torr.

Question

Chemistry

London Kirby

28 August, 21:07

Calculate the mass of ethylene glycol (C2H6O2) that must be added to 1.00 kg of ethanol (C2H5OH) to reduce its vapor pressure by 12.0 torr at 35∘C. The vapor pressure of pure ethanol at 35∘C is 1.00*102 torr.

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Answers (1)

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Rory Crane · Answer 1

183.7 g of ethylene glycol

Explanation:

This a case of colligative property, about vapour pressure lowering.

ΔPv = P°. Xst

ΔPv = 12 Torr

P° = Vapour pressure of pure solvent → 100 Torr

12 Torr = 100 Torr. Xst

12 Torr / 100 Torr = 0.12 → Xst

Xst is mole fraction for solute which means moles of solute / Total moles and, the total moles are the sum of moles of solvent + moles of solute.

We can make this equation:

Moles of solute / (Moles of solute + Moles of solvent) = 0.12

We don't know the moles of solute, but we do know the moles of solvent by the mass.

Mass / Molar mass = Mol

1000 g / 46 g/m = 21.7 moles

Moles of solute / Moles of solute + 21.7 moles = 0.12

Moles of solute = 0.12 (Moles of solute + 21.7 moles)

Moles of solute - 0.12 moles of solute = 2.60 moles

0.88 moles of solute = 2.60 moles

Moles of solute = 2.60 moles / 0.88 → 2.96 moles of solute

Mol. molar mass = Mass

2.96 m. 62g/m = 183.7 g