Calculate the mass of ethylene glycol (c₂h₆o₂) that must be added to 1.00 kg of ethanol (c₂h₅oh) to reduce its vapor pressure by 11.0 torr at 35°c. the vapor pressure of pure ethanol at 35°c is 1.00×10² torr.

123.5 g of C₂H₅OH are required

Explanation:

Let's think the colligative property of vapour pressure to solve this.

ΔP = P°. Xm

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

P° = Vapor pressure of pure solvent

Xm = mole fraction of solute (moles of solute / (moles of solute + moles of solvent))

11 Torr = 100 Torr . Xm

11 Torr / 100 Torr = 0.11 → Xm

0.11 = moles of solute / moles of solute + moles of solvent

Let's determine the moles of solvent.

We have a mass of 1 kg, which is the same as 1000 g

Molar mass ethanol = 46 g/mol

1000g / 46g/mol = 21.7 moles

0.11 moles = moles of solute / moles of solute +21.7 moles

0.11 moles (moles of solute +21.7 moles) = moles of solute

0.11 moles of solute + 2.39 moles = moles of solute

2.39 moles = 1 - 0.11 moles of solute

2.39 moles = 0.89 moles of solute

2.39 / 0.89 = moles of solute → 2.68 moles

Let's convert the moles to mass

2.68 moles . 46 g/mol = 123.5 g

take a better picture and ill do it