Ask Question
3 August, 13:26

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.

+4
Answers (1)
  1. 3 August, 13:53
    0
    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
Know the Answer?
Not Sure About the Answer?
Find an answer to your question ✅ “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 ...” in 📘 Chemistry if you're in doubt about the correctness of the answers or there's no answer, then try to use the smart search and find answers to the similar questions.
Search for Other Answers