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    HPLC Column

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    An HPLC column of length 15cm and ID of 4.6mm is tightly packed with monodisperse ODS particles of diameter 3:m surface area 300m/g^2 and density 1.2g/mL. Dead volume is 1.2mL. Assuming that the particles are tightly packed and that the void volume is essentially due to the pore volumes of the particles, calculate:

    1. the # of ODS particles in the column
    2. the weight in grams of the ODS needed to obtain a
    tight packing
    3. the mass ingrams of each ODS particle
    4. the pore volume of each ODS particle

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    https://brainmass.com/chemistry/experimental-design-and-methods-in-chemistry/hplc-column-17528

    Solution Preview

    First, we need to calculate the volume of the column. The volume of a cylinder is given as pi*r^2*h = 3.141 * (2.3)^2 * 150 = 2492.38 square mm or 24.92 mL. Since we are to assume in this question that the dead volume is made up completely ...

    Solution Summary

    This solution is provided in 220 words and provides equations to calculate volume and mass to solve the problem.

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