--%>

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 : Molarity of solution Help me to go

    Help me to go through this problem. When 7.1gm Na2SO4 (molecular mass 142) dissolves in 100ml H2O , the molarity of the solution is: (a) 2.0 M (b) 1.0 M (c) 0.5 M (d) 0.05 M

  • Q : Pressure Phase Diagrams The occurrence

    The occurrence of different phases of a one component system can be shown on a pressure temperature. The phases present in a one line system at various temperatures can be conveniently presented on a P- versus-T diagram. An example is pro

  • Q : Problem on molality Select the right

    Select the right answer of the question. Calculate the molality of 1 litre solution of 93% H2SO4 (weight/volume). The density of the solution is 1.84 g /ml : (a) 10.43 (b) 20.36 (c) 12.05 (d) 14.05

  • Q : Distribution law Help me to go through

    Help me to go through this problem. The distribution law is applied for the distribution of basic acid between : (a) Water and ethyl alcohol (b) Water and amyl alcohol (c) Water and sulphuric acid (d) Water and liquor ammonia

  • Q : Solutions The normality of 10 lit.

    The normality of 10 lit. volume hydrogen peroxide is: (a) 0.176 (b) 3.52 (c) 1.78 (d) 0.88 (e)17.8

  • Q : Problem on vapor-liquid equilibrium Two

    Two tanks which contain water are connected to each other through a valve. The initial conditions are as shown (at equilibrium): 683_tank question.jpg

  • Q : Electrochemistry ( electrolysis of

    1. Define Faraday's first law of electrolysis 2. define Faraday's second law of electrolysis

  • Q : Define thermal energy The thermal part

    The thermal part of the internal energy and the enthalpy of an ideal gas can be given a molecular level explanation. All the earlier development of internal energy and enthalpy has been "thermodynamic". We have made no use o

  • 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 : What are isotonic and hypotonic

    The two solutions which are having equivalent osmotic pressure are called isotonic solutions. The isotonic solutions at the same temperature also have same molar concentration. If we have solutions having different osmotic pressures then the solution having different