--%>

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 : What are lattices and unit cells? The

    The repeating, atomic level structure of a crystal can be represented by a lattice and by the repeating unit of the lattice, the unit cell.It was apparent very early in the study of crystals that the shapes of crystals stem from an ordered array of smaller

  • Q : DNA Organic Explain DNA organic in

    Explain DNA organic in brief?

  • Q : What are emulsions?Describe its

    Emulsions are colloidal solutions in which disperse phase as well as dispersion medium is both liquids. Emulsions can be broadly classified into two types: (i) Oil in water (O/W type) emulsions: in this type of emulsions, oil acts disperse phase and water acts

  • Q : Vapour pressure over mercury Choose the

    Choose the right answer from following. At 300 K, when a solute is added to a solvent its vapour pressure over the mercury reduces from 50 mm to 45 mm. The value of mole fraction of solute will be: (a)0.005 (b)0.010 (c)0.100 (d)0.900

  • Q : Number of moles of potassium chloride

    Choose the right answer from following. The number of moles of KCL in 1000ml of 3 molar solution is: (a)1 (b)2 (c)3 (d)1.5

  • Q : Determining of normality of sodium

    Can someone please help me in getting through this problem. The normality of a solution of sodium hydroxide 100 ml of which includes 4 grams of NaOH is: (a) 0.1 (b) 40 (c) 1.0 (d) 0.4

  • Q : Avogadros hypothesis how avogadros

    how avogadros hypothesis used to deduce the atomicity of elementry gases ?

  • Q : Dipole moment Elaborate a dipole moment

    Elaborate a dipole moment?

  • Q : What is Spectroscopy? This is a very

    This is a very important aspect of Physical Chemistry in which knowledge of the size, shape, rigidity and electronic structure of molecules deduced from the experimental methods treated here goes hand in hand with the theoretical approaches of chemical reactions. Spec

  • Q : Problem on convection coefficient An

    An experiment to determine the convection coefficient associated with airflow over the surface of a thick stainless steel casting involves insertion of thermocouples in the casting at distances of 10 mm and 20 mm from the surface.  When the experiment was perform