--%>

Sedimentation and Velocity

The first method begins with a well defined layer, or boundary, of solution near the center of rotation and tracks the movement of this layer to the outside of the cell as a function of time. Such a method is termed a sedimentary velocity experiment.


A particle of mass m at a distance x from center of rotation experiences a force given by 

ƒcentrif = m'xω2

Where w is the angular velocity in radiation per second m' is the distance effective mass of the solute particle, i.e. the actual mass corrected for the new effect of the solvent.

To express this buoyancy effect, we first recognize that v the specific volume of the solute, is the mass of 1 g of the solute. The volume of m g of solute is mv, and the mass of this volume of solvent is m of the solute is m - mvp = mj (1 - vp). We now can rewrite equation as:

Centrif = m (1 - vp) xω
2

Equating these two force expressions leads us to the constant drif velocity. A rearrangement of the equality:

M (1 - v) xω2 = 6∏r? dx/dt

Equating these two force expressions leads that collect the dynamic variables gives:

Dx/dt/xω2 = m (1 - vp)/6∏r?


The collection of dynamic terms on the left side of equation describes the results of sedimentation velocity experiments. This collection (dx/dt) xw2 can be looked on as the velocity with which the solute moves per unit centrifugal force. The sedimentation coefficient S is introduced as:

S = dx/dt/xω2

The experimentation results can therefore be tabulated as values of S. the value of S for many macromolecules is of the order of 10-13 has therefore been introduced, called a Svedberg, in honor of T. Svedberg, who did much of the early work with the ultracentrifuge.

Molar mass: s = dx/dt/xω2 = m )1 - vp)/ 6∏r?

Rearrangement and multiplication by Avogadro's number give:

M = Nm = 6∏r?NS/ 1- vp

Now the troublesome terms involving ? and r can be replaced by their effective values appear in the measurable values D of equation, to give the desired result:

M = RTS/ D (1 - vp)

Thus measurements of the substances of the sedimentation and diffusion coefficients and of the solvent and solute allow the deduction of the molar mass for a few macromolecules. The necessary data for such calculations for a few macromolecular materials are included.

A particular advantage of the sedimentation velocity technique is that a macromolecular solution containing two or more types of macromolecules is separated according to the molecular masses of the components. The type of sedimentation diagrams obtained for a system containing a number of macromolecular species.

Density gradient: better resolution can be obtained by allowing the sedimentation to occur in a density gradient solution, prepared, for example, by filling the centrifuge tube layer by layer with solutions of decreasing sucrose concentration. As the macromolecular substance or mixture of substances is centrifuged, it moves through a solvent with gradually increasing density. The result is more stable macromolecular zones and a better "spectrum" of the components. The technique is thus a modification of the sedimentation velocity method.

   Related Questions in Chemistry

  • Q : Advantages of doing your own chemistry

    What are the advantages of doing your own chemistry assignments? State your comment?

  • Q : Volumes of solution after concentration

    Hydrochloric acid solution A and B encompass concentration of 0.5N and 0.1N  corresspondingly. The volumes of solutions A and B needed to make 2liters of 0.2N of HCL are: (i) 0.5l of A + 1.5l of B (ii) 1.5l of A + 0.5 l of B  (iii) 1.0 l of A + 1.0l of B&nbs

  • Q : Explain the preparation of phenols. The

    The methods used for the preparation of phenols are given below:    From aryl sulphonic acids

  • Q : Molal elevation constant of water The

    The boiling point of 0.1 molal aqueous solution of urea is 100.18oC  at 1 atm. The molal elevation constant of water is: (a) 1.8    (b) 0.18   (c) 18    (d) 18.6Answer: (a) Kb

  • Q : Relative lowering in vapour pressure of

    Give me answer of this question. "Relative lowering in vapour pressure of solution containing non-volatile solute is directly proportional to mole fraction of solute". Above statement is: (a) Henry law (b) Dulong and Petit law (c) Raoult's law (d) Le-Chatelier's pri

  • Q : Hybridization Atomic orbitals can be

    Atomic orbitals can be combined, in a process called hybridization, to describe the bonding in polyatomic molecules. Descriptions of the bonding in CH4 can be used to illustrate the valence bond procedure. We must arrive a

  • Q : Molarity 20mol of hcl solution requires

    20mol of hcl solution requires 19.85ml of 0.01 M NAOH solution for complete neutralisation. the molarity of hcl solution

  • Q : Problem on Neutralization What weight

    What weight of hydrated oxalic acid should be added for complete neutralisation of 100 ml of 0.2N - NaOH solution? (a) 0.45 g  (b)0.90 g  (c) 1.08 g  (d) 1.26 g      Answer

  • Q : Concentration of Calcium carbonate Help

    Help me to go through this problem. 1000 gms aqueous solution of CaCO3 contains 10 gms of carbonate. Concentration of the solution is : (a)10 ppm (b)100 ppm (c)1000 ppm (d)10000 ppm

  • Q : Benzoic acid is weaker than paranitro

    Briefly state that Benzoic acid is weaker than paranitro benzoic acid?