--%>

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 : Ionic radius of chloride ion The edge

    The edge length of the unit cell of Nacl crystal lattice is 552 pm. If ionic radius of sodium ion is 95. What is the ionic radius of chloride ion:(a) 190 pm  (b) 368 pm  (c) 181 pm  (d) 276 pm     <

  • Q : Problem on Redlich-Kwong equation i)

    i) Welcome to Beaver Gas Co.! Your first task is to calculate the annual gross sales of our superpure-grade nitrogen and oxygen gases. a) The total gross sales of N2 is 30,000 units. Take the volume of the cylinder to be

  • Q : What is chemisorption or chemical

    When the forces of attraction existing between adsorbate particles and adsorbent almost of the same strength as chemical bonds, the adsorption is called chemical adsorption. This type of adsorption is also known as chemisorptions. Since forces of attraction existing b

  • Q : Normality of sulphuric acid Help me to

    Help me to go through this problem. Normality of sulphuric acid is: (a) 2N (b) 4N (c) N/2 (d) N/4

  • Q : Calculation of concentration of the

    Choose the right answer from following. 200ml of a solution contains 5.85 dissolved sodium chloride. The concentration of the solution will be(Na= 23: cl = 35.5 ) (a) 1 molar (b) 2 molar (c) 0.5 molar (d) 0.25 molar

  • Q : Water under pressure problem-henry law

    Can someone help me in going through this problem. The statement “When 0.003 moles of a gas are dissolved in 900 gm of water under a pressure of 1 atm, 0.006 moles will be dissolved under the pressure of 2 atm", signfies: (a)

  • Q : Chemical formula of detergent Describe

    Describe the chemical formula of detergent?

  • Q : Strength of dilute acid of Sulfuric acid

    Select the right answer of the question.10ml of conc.H2SO4 (18 molar) is diluted to 1 litre. The approximate strength of dilute acid could be: (a)0.18 N (b)0.09 N (c) 0.36 N (d)1800 N

  • Q : What is electrolysis? Explain with

    Passage of a current through a solution can produce an electrolysis reaction.Much additional information on the properties of the ions in an aqueous solution can be obtained from studies of the passage of a direct current (dc) through a cell containing a s

  • Q : How much phosphorus is in superphosphate

    Superphosphate has the formulate: CaH4 (PO4)2 H2O calculate the percentage of Phosphorus in this chemical. Show your calculations