Ask Question
23 September, 10:25

What is the vapor pressure above a solution prepared by dissolving 0.500 mol of a non-volatile solute in 275 g of hexane (86.18 g/mol) at 49.6°C? P°hexane = 400.0 torr at 49.6°C.

+3
Answers (1)
  1. 23 September, 10:29
    0
    Vapor pressure of solution = 346 Torr

    Explanation:

    We must work with colligative property of vapor pressure, to solve this:

    ΔP = P°. Xm

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

    P° is Vapor pressure of pure solvent

    Xm is the mole fraction for solute (moles of solute / total moles)

    Let's find out Xm.

    We have the moles of solute, as data → 0.5 moles

    Moles of solvent → Mass of solvent / Molar mass of solvent

    275 g / 86.18 g/mol = 3.19 moles

    Total moles = Moles of solvent + Moles of solute

    Xm for solute = 0.5moles / (3.19moles + 0.5 moles) → 0.135

    400 Torr - Vapor pressure of solution = 400 Torr. 0.135

    Vapor pressure of solution = - (400 Torr. 0.135 - 400 Torr)

    Vapor pressure of solution = 346 Torr
Know the Answer?
Not Sure About the Answer?
Find an answer to your question 👍 “What is the vapor pressure above a solution prepared by dissolving 0.500 mol of a non-volatile solute in 275 g of hexane (86.18 g/mol) at ...” in 📗 Chemistry if the answers seem to be not correct or there’s no answer. Try a smart search to find answers to similar questions.
Search for Other Answers