--%>

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 : Some basic concepts of chemistry an

    an atom of an element is 10.1 times heavier than the mass of a carbon atom.What is its mass in amu?

  • Q : Which is polar HCl or HF Which one is

    Which one is polar HCl or HF?

  • Q : What is chemisorption or chemical

    When the forces of attraction existing between adsorbate particles and adsorbent almost of the same strength as chemical bonds, the adsorption is called chemical adsorption. This type of adsorption is also known as chemisorptions. Since forces of attraction existing b

  • Q : How much phosphorus is in superphosphate

    Superphosphate has the formulate: CaH4 (PO4)2 H2O calculate the percentage of Phosphorus in this chemical. Show your calculations

  • Q : Molecular crystals Among the below

    Among the below shown which crystal will be soft and have low melting point: (a) Covalent  (b) Ionic  (c) Metallic  (d) MolecularAnswer: (d) Molecular crystals are soft and have low melting point.

  • Q : Why aryl halides are less reactive?

    Aryl halides are much less reactive towards nucleophilic substitution reactions than haloalkanes. The less reactivity of aryl halides can be described

  • Q : Molarity of HCl solution 20 ml of HCL

    20 ml of HCL solution needs 19.85 ml of 0.01M NaOH solution for complete neutralization. Morality of the HCL solution is:  (i) 0.0099 (ii) 0.099 (iii) 0.99 (iv) 9.9 Choose the right answer from above.

  • Q : Problem related to molarity Provide

    Provide solution of this question. Increasing the temperature of an aqueous solution will cause: (a) Decrease in molality (b) Decrease in molarity (c) Decrease in mole fraction (d) Decrease in % w/w

  • Q : Mcq Give me answer of this question.

    Give me answer of this question. The normality of 10% (weight/volume) acetic acid is: (a)1 N (b)10 N (c)1.7 N (d) 0.83 N