--%>

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 : 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 : Molarity 20mol of hcl solution requires

    20mol of hcl solution requires 19.85ml of 0.01 M NAOH solution for complete neutralisation. the molarity of hcl solution

  • Q : Q what is the basicity of primary

    what is the basicity of primary secondary and tertiary amines in chlorobenzene

  • Q : Calculating Formulae Superphosphate has

    Superphosphate has the formula CaH4(PO4)2 H2O, what is the calculation to get the percentage of Phosphorus, I need to show the calculation. I know it is 30.9737622 u in weight and 2 atoms of the formula, but not sure how to work the calculation backwards.

  • Q : Question related to colligative

    The colligative properties of a solution depend on: (a) Nature of solute particles present in it (b) Nature of solvent used (c) Number of solute particles present in it (d) Number of moles of solvent only

  • 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 : Molecular Structure type The ionic

    The ionic radii of Rb+ and I- respectively are 1.46 Å and 2.16Å. The very most probable type of structure exhibited by it is: (a) CsCl type  (b) ZnS type  (c) Nacl type  (d) CaF2 type

    Q : Problem on partial pressure i) Show

    i) Show that the equilibrium constant Kp for the reaction CaCo3(s) ↔ CaO(s) +CO2(g)is about unity (i.e. = 1.0) at 895 °C.ii) If two grams of calcium carbonate are pl

  • Q : Question based on normality Provide

    Provide solution of this question. A 5 molar solution of H2SO4 is diluted from 1 litre to 10 litres. What is the normality of the solution : (a) 0.25 N (b) 1 N (c) 2 N (d) 7 N

  • Q : Solution and colligative properties

    what is molarity of a solution of hcl which contains 49% by weight of solute and whose specific gravity is 1.41