CHM 5414 Thermodynamics

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Quiz #5 Solutions

26 November 1996

  1. If a protein is soluble to the extent of 100 g/L in aqueous solution, and an osmotic pressure experiment on that protein can read accurately only to 1 mm of H2O, what is the upper limit on the molecular weight which can be determined at 25°C?

    [ rho (H2O) = 1 g/cc; rho (Hg) = 13.6 g/cc ]

We want to use the osmotic pressure formula, MW=wRT/V pi , where the pressure pi is that of 1 mm of water, the smallest measurable difference between solution and the surrounding water, and w/V is 100 g/L.

Since water weighs only 1/13.6 the weight of the equivalent column of mercury,

pi = 1 mm H2O × (1 mm Hg/13.6 mm H2O) × (1 atm/760 mm Hg) = 9.67×10-5 atm

Hence,

MW = 100 g × 0.08206 L atm/mol°K × 298°K / ( 1 L × 9.67×10-5 atm ) = 2.53×107 g/mol

which is equisitely sensitive! Since protein molecular weights are typically two orders of magnitude smaller than this, we could measure them with fair accuracy even if their solubility was only 1 g/L; phenomenal.

  • Estimate the solubility (molality*) of anthracene, C14H10, in naphthalene, C10H8, at 100.0°C. (Their solution is ideal.)

    Molecule TM, °C Delta H°fus, kJ/mol TB, °C
    C10H8 80.2 19.06 217.9
    C14H10 215.0 28.83 339.9

    * moles solute per kg solvent

    • In a solubility problem, the dissolving component is treated as the solute regardless of its mole fraction, so abbreviating anthracene by phi phi phi ,

    ln(X phi phi phi ) = - [ Delta Hfus( phi phi phi ) / R ]×[ 1/T - 1/TM( phi phi phi ) ]

    ln(X phi phi phi ) = - [28.83×103 J/mol / 8.314 J/mol°K ]×[ 1/(273.15+100.0) - 1/(273.15+215.0) ]

    ln(X phi phi phi ) = - 2.189
    or
    X phi phi phi = e-2.189 = 0.112 hence X phi phi = 0.888

    Since MW( phi phi ) = 10×12 + 8×1 = 128 g/mol, 1 kg of phi phi contains 1000/128 = 7.81 mol phi phi , and the extra anthracene adds (0.112/0.888)×7.81 = 0.985 mol phi phi phi to the mixture.

    So m( phi phi phi ) = 0.985 molal.

    (If you believed the quiz's footnote typo, you'd have used MW( phi phi phi ) = 14×12 + 10×1 = 178 g/mol to quote the solubility as 178 g/mol × 0.985 mol = 175 g phi phi phi per kg phi phi . Also correct, but not molality of course.)
    Last modified 1 December 1996