--%>

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 : Effect on vapour pressure of dissolving

    Give me answer of this question. When a substance is dissolved in a solvent the vapour pressure of the solvent is decreased. This results in: (a) An increase in the b.p. of the solution (b) A decrease in the b.p. of the solvent (c) The solution having a higher fr

  • 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

  • Q : Inorganic Chemistry Inorganic

    Inorganic Chemistry:In the year 1869, Russian Chemist Dmitry Mendeleyev forms the periodic table of the element. Since Newlands did before him in the year 1863, Mendeleyev categorizes the el

  • Q : What are homogenous catalyst? Give few

    When a catalyst mixes homogeneously with the reactants and forms a single phase, the catalyst is said to be homogeneous and this type of catalysis is called homogeneous catalysis. Some more examples of homogeneous catalysis are:    SO2

  • Q : Problem on physical and thermodynamic

    The shells of marine organisms contain calcium carbonate CaCO3, largely in a crystalline form known as calcite. There is a second crystalline form of calcium carbonate known as aragonite. Physical and thermodynamic properties of calcite and aragonite at 298

  • Q : Organic and inorganic substances living

    living beings are made up of organic and inorganic substances.according to their complexity of their molecules how can ach of these substances be classified?

  • Q : Tetrahedral holes In zinc blende

    In zinc blende structure, zinc atom fill up:(a) All octahedral holes  (b) All tetrahedral holes  (c) Half number of octahedral holes  (d) Half number of tetrahedral holesAnswer: (d) In zinc blende (ZnS

  • Q : Neutralization of sodium hydroxide How

    How much of NaOH is needed to neutralise 1500 cm3 of 0.1N HCl (given = At. wt. of Na =23): (i) 4 g  (ii) 6 g (iii) 40 g  (iv) 60 g

  • Q : Explain oxygen and its preparation.

    Karl Scheele, the Swedish chemist, was

  • 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.