Ask Question
26 May, 14:01

Calculate the vapor pressure of water above a solution prepared by adding 24.5 g of lactose (C12H22O11) to 200.0 g of water at 338 K. (Vapor-pressure of water at 338 K 187.5 torr.)

+2
Answers (1)
  1. 26 May, 16:18
    0
    Vapor pressure of solution → 186.3 Torr

    Explanation:

    To solve this problem we must apply the colligative property of vapor pressure.

    ΔP = P°. Xm

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

    P° is vapor pressure of pure solvent → 187.5 Torr

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

    Let's determine the mole fraction,

    Moles of solute = Mass of solute / Molar mass

    24.5 g / 342 g/mol = 0.0716 moles

    Moles of solvent = Moles of solvent / Molar mass

    200 g / 18g/mol = 11.1 moles

    Total moles = 11.1 moles + 0.0716 moles → 11.1716 moles

    Mole fraction of solute = 0.0716 mol / 11.1716 mol = 6.40*10⁻³

    Let's apply the formula

    ΔP = P°. Xm

    Vapor P of pure solvent - Vapor P of solution = P°. 6.40*10⁻³

    Vapor P of pure solvent - 187.5 Torr. 6.40*10⁻³ = Vapor P of solution

    187.5 Torr - 187.5 Torr. 6.40*10⁻³ = Vapor P of solution

    Vapor pressure of solution → 186.3 Torr
Know the Answer?
Not Sure About the Answer?
Get an answer to your question ✅ “Calculate the vapor pressure of water above a solution prepared by adding 24.5 g of lactose (C12H22O11) to 200.0 g of water at 338 K. ...” in 📙 Chemistry if there is no answer or all answers are wrong, use a search bar and try to find the answer among similar questions.
Search for Other Answers