--%>

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 : BASIC CHARACTER OF AMINES IN GAS PHASE,

    IN GAS PHASE, BASICITIES OF THE AMINES IS JUST OPPOSITE TO BASICITY OF AMINES IN AQEUOUS PHASE .. EXPLAIN

  • Q : Number of moles present in water

    Provide solution of this question. How many moles of water are present in 180 of water: (a)1 mole (b)18 mole (c)10 mole (d)100 mole

  • Q : How alkyl group reactions takes place?

    Halogenations: ethers react with chlorine and bromine to give substitution products. The extent of halogenations depends upon the conditions of reacti

  • Q : Modes of concentration Which of the

    Which of the given modes of expressing concentration is fully independent of temperature: (1) Molarity (2) Molality (3) Formality (4) Normality Choose the right answer from above.

  • 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 : Means of molality Give me answer of

    Give me answer of this question. The number of moles of solute per kg of a solvent is called its: (a) Molarity (b) Normality (c) Molar fraction (d) Molality

  • 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

  • Q : Precipitation problem On passing H 2 S 

    On passing H2S  gas through a solution of Cu+ and Zn+2 ions, CuS is precipitated first because: (i) Solubility product of CuS is equal to the ionic product of ZnS (ii) Solubility product of CuS is equal to the solubility product o

  • Q : Mole fraction and Molality Select the

    Select the right answer of the following question.What does not change on changing temperature : (a) Mole fraction (b) Normality (c) Molality (d) None of these

  • Q : Rotational energy and entropy due to

    The entropy due to the rotational motion of the molecules of a gas can be calculated. Linear molecules: as was pointed out, any rotating molecule has a set of allowed rotational energies. For a linear molecule the