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14 August, 09:13

Calculate the vapor pressure at 25 °C of an aqueous solution that is 5.50% NaCl by mass. (Assume complete dissociation of the solute.)

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  1. 14 August, 09:24
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    Vapor pressure of solution = 25.1 Torr

    Explanation:

    Vapor pressure lowering is one of the colligative property. Its formula is:

    ΔP = P°. Xm. i

    ΔP = Vapor pressure pure solvent - Vapor pressure solution

    P° is Vapor pressure pure solvent

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

    i is the Van't Hoff factor, the number of ions particles dissolved.

    NaCl → Na⁺ + Cl⁻ (For this dissociation, i = 2)

    Let's determine the mole fraction of this solution.

    5.50% by mass is 5.50 g of solute in 100 g of solution.

    Total mass = 100 g → Mass of solute + Mass of solvent

    100 g - 5.5g = 94.5 g (mass of solvent)

    Moles of solute → Solute mass / molar mass

    5.5 g / 36.45 g/mol = 0.151mol

    Moles of solvent → Solvent mass / molar mass

    94.5 g / 18 g/mol = 5.25mol

    Total moles = Moles of solute + Moles of solvent

    0.151 mol + 5.25mol = 5.401 total moles

    Xm solute = 0.151 / 5.401 → 0.028

    Vapor pressure of water at 25°C → 23.8 Torr

    Let's replace the data

    Vapor pressure of solution - 23.8 Torr = 23.8 Torr. 0.028. 2

    Vapor pressure of solution = 23.8 Torr. 0.028. 2 + 23.8 Torr

    Vapor pressure of solution = 25.1 Torr
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