--%>

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 : Question on colligative property Choose

    Choose the right answer from following. Which of the following is a colligative property: (a) Osmotic pressure (b) Boiling point (c) Vapour pressure (d) Freezing point

  • Q : Problem on decomposition reaction

    Nitrogen tetroxide (melting point: -11.2°C, normal boiling point 21.15°C) decomposes into nitrogen dioxide according to the following reaction: N2O4(g) ↔ 2 NO2(g)<

  • Q : Problem on Molar solution Can someone

    Can someone please help me in getting through this problem. 2.0 molar solution is acquired, when 0.5 mole solute is dissolved in: (i) 250 ml solvent (ii) 250 g solvent (iii) 250 ml solution (iv) 1000 ml solvent

  • 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 : How haloalkanes are prepared from

    Alkyl halides can be prepared from alkanes through substitution and from alkenes through addition of halogen acids or through allylic substitution.    From alkanesWhen alkanes are treated with halogens, chlo

  • Q : Analytical chemistry 37% weight of HCl

    37% weight of HCl and density is 1.1g/ml. find molarity of HCl

  • Q : Forms a molecule to an organic molecule

    Briefly state what forms a molecule to an organic molecule?

  • Q : Group Cations Explain how dissolving

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid, establishes a buffer with a pH of approximately

  • Q : Examples of reversible reaction

    Describe some examples of a reversible reaction?

  • Q : Describe chemical properties of amines.

    Like ammonia, primary, secondary and tertiary amines have a single pair of electrons on N atom. Hence chemical behavior of amines is similar to ammonia. Amines are basic in nature, and in most of the reactions they act as nucleophiles.      1. Reaction wi