--%>

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 : Precipitation Addition of conc. HCl to

    Addition of conc. HCl to saturated Bacl2 solution precipitates Bacl2 ; because of the following reason : (a) It follows from Le Chatelier's principle (b) Of common-ion effect (c) Ionic product (Ba++)(cl) remains constant in a saturated sol

  • Q : Question associated to vapour pressure

    Choose the right answer from following. The vapour pressure lowering caused by the addition of 100 g of sucrose(molecular mass = 342) to 1000 g of water if the vapour pressure of pure water at 25degree C is 23.8 mm Hg: (a)1.25 mm Hg (b) 0.125 mm Hg (c) 1.15 mm H

  • Q : Explain the molecular mass with respect

    During the formation of polymers, different macromolecules have different degree of polymerisation i.e. they have varied chain lengths. Thus, the molecular masses of the individual macromolecules in a particular sample of the polymer are different. Hence, an average value of the molecular mass is

  • Q : Biodegradable polymers what are the

    what are the examples of biodegradable polymers

  • Q : What are different mechanisms for

    Nucleophilic substitution reactions in halides containing  - X bond may take place through either of the two different mechanisms,S<

  • Q : Coordination compounds discuss the

    discuss the practical uses of coordination compounds, give reactions involves and explain whats happening in the process

  • Q : Strength of the Hydrochloric acid

    Provide solution of this question. 1.0 gm of pure calcium carbonate was found to need 50 ml of dilute HCL for complete reaction. The strength of the HCL solution is specified by : (a) 4 N (b) 2 N (c) 0.4 N (d) 0.2 N

  • Q : Problem on vapour pressure Choose the

    Choose the right answer from following. If P and P are the vapour pressure of a solvent and its solution respectively N1 and N2 and are the mole fractions of the solvent and solute respectively, then correct relation is: (a) P= PoN1 (b) P= Po N2 (c)P0= N2 (d)

  • Q : What is Elevation in boiling point? The

    The boiling of a liquid may be defused by the temperature at which its vapour pressure which is equal to atmospheric pressure. The effect of addition in a non-volatile solute on the boiling point shown and its solution containing non-volatile solute with tempe

  • Q : Molar concentration Choose the right

    Choose the right answer from following. Molar concentration (M) of any solution : a) No. of moles of solute/Volume of solution in litre (b) No. of gram equivalent of solute / volume of solution in litre (c) No. of moles os solute/ Mass of solvent in kg  (