--%>

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 : Describe Point Groups. For any

    For any symmetric object there is a set of symmetry operations that, together, constitute a mathematical group, called a point group.It is clear from the examples that most molecules have several elements of symmetry. The H2O

  • Q : Anti-aromatic and the non-aromatic

    What is main difference among anti-aromatic and the non-aromatic compounds?

  • Q : Extensive property Choose the right

    Choose the right answer from following. Which one of the following is an extensive property: (a) Molar volume (b) Molarity (c) Number of moles (d) Mole fraction

  • Q : Calculating value of molar solution

    Choose the right answer from following. An X molal solution of a compound in benzene has mole fraction of solute equal to 0.2. The value of X is: (a)14 (b) 3.2 (c) 4 (d) 2

  • Q : Problem on colligative properties

    Choose the right answer from following. The magnitude of colligative properties in all colloidal dispersions is : (a) Lowerthan solution (b)Higher than solution(c) Both (d) None

  • Q : Molar mass Select the right answer of

    Select the right answer of the question. Which is heaviest: (a)25 gm of mercury (b)2 moles of water (c)2 moles of carbon dioxide (d)4 gm atoms of oxygen

  • Q : Osmotic Pressure The O.P. (Osmotic

    The O.P. (Osmotic Pressure) of equimolar solution of Urea, BaCl2 and AlCl3, will be in the order:(a) AlCl3 > BaCl2 > Urea  (b) BaCl2 > AlCl3 > Urea  (c) Urea > BaCl2<

  • Q : Molarity of Nacl solution When 5.85 g

    When 5.85 g of NaCl (having molecular weight 58.5) is dissolved in water and the solution is prepared to 0.5 litres, the molarity of the solution is: (i) 0.2 (ii) 0.4 (iii) 1.0 (iv) 0.1

  • Q : What are the chemical properties of

    Haloalkanes are extremely reactive category of aliphatic compounds. Their reactivity is due to the presence of polar carbon-halogen bond in their mole

  • Q : Explain Ionic Bond with examples. The

    The bonding in ionic molecules can be described with a coulombic attractive term.For some diatomic molecules we take quite a different approach from that used in preceding sections to describe the bonding. Ionic bonds are interpreted in terms of the coulom