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26 March, 15:00

When 0.654 grams of a polymer were dissolved in 40.9 mL of benzene at 21.9 degrees C, the osmotic pressure was found to be 0.048 atm. Calculate the molar mass of the polymer.

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  1. 26 March, 15:18
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    Molar mass of the polymer is 8055.6 g/mol

    Explanation:

    Let's apply the colligative property of osmotic pressure:

    π = M. R. T where

    π is the osmotic pressure

    M is molarity (mol / L)

    R is the universal gases constant (0.082 L. atm/mol. K)

    T is Absolute T° (°C + 273)

    T° K = 294.9K

    Let's determine our unknown value in molarity

    Molarity is mol / L and mol is mass / molar mass

    Molarity is (mass / molar mass) / volume, that can be written as:

    (mass / molar mass). 1 / volume

    M = (0.654 g / molar mass). 1 / 0.0409 L

    Remember that volume must be in Litters. → 40.9 mL. 1 L/1000 mL = 0.0409 L

    0.048 atm = (0.654 g / mol mass).1 / 0.0409 L. 0.082 L. atm/mol. K. 294.9K

    0.048 atm / 0.082 L. atm/mol. K. 294.9K = (0.654 g / mol mass).1 / 0.0409 L

    1.98*10⁻³ L/mol = (0.654 g / mol mass).1 / 0.0409 L

    1.98*10⁻³ L/mol. 0.0409 L = 0.654 g / molar mass

    8.12*10⁻⁵ mol = 0.654 g / molar mass

    Molar mass = 0.654 g / 8.12*10⁻⁵ mol → 8055.6 g/mol
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