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 : Calculation of molecular weight Provide

    Provide solution of this question. In an experiment, 1 g of a non-volatile solute was dissolved in 100 g of acetone (mol. mass = 58) at 298K. The vapour pressure of the solution was found to be 192.5 mm Hg. The molecular weight of the solute is (vapour pressure of ace

  • Q : Entropy on molecular basis. The

    The equation S = k in W relates entropy to W, a measure of the number of different molecular level arrangements of the system.In the preceding developments it was unnecessary to attempt to reach any "explana

  • Q : Problem on bubble point The following

    The following mixture of hydrocarbons is obtained as one stream in a petroleum refinery.

    Q : Calculate PH value for a acetic acid 1.

    1. A solution of 0.100 M acetic acid is prepared. a) What is its pH value? b) If 20% of the initial acetic acid is converted to the acetate form by titration with NaOH, what is the resultant pH?

  • 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

  • Q : Problem based on mole concept Choose

    Choose the right answer from following. An aqueous solution of glucose is 10% in strength. The volume in which mole of it is dissolved will be : (a) 18 litre (b) 9 litre (c) 0.9 litre (d) 1.8 litre

  • Q : Application of colligative properties

    Choose the right answer from following. Colligative properties are used for the determination of: (a) Molar Mass (b) Equivalent weight (c) Arrangement of molecules (d) Melting point and boiling point (d) Both (a) and (b)  

  • 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 : Difference among hcl gas and hcl acid

    What is the basic difference among hcl gas and hcl acid? Briefly state the difference?

  • Q : Formula of diesel Write a short note on

    Write a short note on the formula of diesel, petrol and also CNG?

©TutorsGlobe All rights reserved 2022-2023.