--%>

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 : Homework Silicon has three naturally

    Silicon has three naturally occurring isotopes. 28Si, mass = 27.976927; 29Si, mass = 28.976495; 30Si, mass = 29.973770 and 3.10% abundance. What is the abundance of 28Si?

  • Q : Molarity of cane sugar solution 171 g

    171 g of cane sugar (C12H22O11)  is dissolved in one litre of water. Find the molarity of the solution: (i) 2.0 M (ii) 1.0 M (iii) 0.5 M (iv) 0.25 M Choose the right answer from above.

  • Q : What is Henry law constant and its

    1. The units of Henry Law constant are same as those of pressure, i.e. torr or h bar. 2. Different gases have dissimilar values of Henry law constant. The values of KH for some gases in water are given in tabl

  • Q : Problem on making solution Select the

    Select the right answer of the question. The weight of H2C2O42H2O required to prepare 500ml of 0.2N solution is : (a) 126g (b) 12.6g (c) 63g (d) 6.3g

  • Q : Problems related to entropy change A)

    A) Two compartments each of 1 m3 capacity are joined by a valve and insulated from the surroundings and from one another. One compartment has saturated steam at 683.6 kPa and the other contains steam at the same temperature but at a pressure of 101.3 kPa. T

  • Q : Latent heat of vaporization Normal

    Normal butane (C4H10) is stored as a compressed liquid at 90°C and 1400 kPa. In order to use the butane in a low-pressure gas-phase process, it is throttled to 150 kPa and passed through a vaporizer. The butane emerges from the vaporizer as a

  • Q : Avogadro's hypothesis Law Principle

    Avogadro's hypothesis Law Principle- Berzelius, a chemist tried

  • Q : Problem on thermodynamic equilibrium In

    In the manufacture of sulphuric acid by the contact process, S02 is oxidized to SO3 over a vanadium catalyst: The reactor is adiabatic and operates at atmospheric pressure. The gases enter the reactor at 410&d

  • Q : Molality of Sulfuric acid Choose the

    Choose the right answer from following. The molality of 90% H2SO4 solution is: [density=1.8 gm/ml]  (a)1.8 (b) 48.4 (c) 9.18 (d) 94.6

  • Q : Coordination compounds discuss

    discuss practical uses of coordination compounds