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23 August, 20:01

A 279.6 mL sample of an aqueous solution at 25°C contains 91.6 mg of an unknown nonelectrolyte compound. If the solution has an osmotic pressure of 8.44 torr, what is the molar mass (in g/mol) of the unknown compound?

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  1. 23 August, 20:15
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    The molar mass of the compound is 720.8 g/mol

    Explanation:

    Let's apply the colligative property of Osmotic pressure to solve this.

    Formula is π = M. R. T

    where π is pressure (atm)

    M is molarity (mol/L)

    R, Universal Constant Gases

    T, Absolute T° (T° in K = T° in C + 273)

    Let's replace the dа ta:

    8.44 Torr = M. 0.082 L. atm/mol. K. 298K

    As we have the pressure in Torr, we must convert to atm, to work properly.

    8.44 Torr. 1 atm / 760 Torr = 0.0111 atm

    0.0111 atm = M. 0.082 L. atm/mol. K. 298K

    0.0111 atm / (0.082 L. atm/mol. K. 298K) = M → 4.54*10⁻⁴ mol/L

    So molarity is the moles of solute (mass (g) / molar mass) / volume (L)

    Let's convert the volume to L → 279.6 mL. 1L / 1000 mL = 0.2796 L

    4.54*10⁻⁴ mol/L. 02796 L = 1.27*10⁻⁴ moles

    This moles are represented by the 91.6 mg, so let's convert the mass of solute from mg to g

    91.6 mg. 1 g / 1000 mg = 0.0916 g

    Molar mass → g/mol → 0.0916 g / 1.27*10⁻⁴ moles → 720.8 g/mol
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