How Do You Calculate the Sedimentation Coefficient of Escherichia coli?

[quietviolence]

Homework Statement

Escherichia coli data:
molecular weight = 74000g/mol
specific volume = 0.737cm³/g
temperature = 293 K

In a dilute aqueous buffer (0.02M potassium phosphate) at 293.75 K
speed of rotation = 56050 rpm

gave the following results:

Time (min)...x_1/2
0.....5.9
20...6
40...6.1
60...6.2
80...6.3
100.....6.4
120.... 6.5
140.....6.6

Homework Equations

This is the equation given to us:

M=(RTs)/(D(1-($$\rho$$ _f *V _o))

The Attempt at a Solution

I used an R value of 8.31m²kgmol^-1s^-2K^-1

I rearranged the equation to equal s

However, I got stuck at $$\rho$$_f*V_o. Using the $$\rho$$ and V given in the question results in a division by 0.

Could someone explain to me how I find $$\rho$$ _f and V _o? I don't need anyone to solve the question, I would like to do that on my own.

 

[quietviolence]

I'm sorry for the terrible equations.
The rho is supposed to be rho with a subscript f, and V is V subscript o

 

[kerimek]

I just need help understanding what \rho f and V o represent and how to find them in this specific context.

The sedimentation coefficient (s) is a measure of how fast a particle or molecule settles in a centrifuge, and is influenced by the size and shape of the particle, as well as the density and viscosity of the surrounding medium. In this context, \rho f represents the density of the buffer solution and V o represents the partial specific volume of the molecule (in this case, E. coli).

To find \rho f, you can look up the density of 0.02M potassium phosphate buffer at 293.75 K. To find V o, you can use the specific volume given in the question and the molecular weight of E. coli to calculate the partial specific volume using the formula V o = V/m, where V is the specific volume and m is the molecular weight.

Once you have both \rho f and V o, you can plug them into the equation and solve for s. Keep in mind that s is a measure of sedimentation rate, so it will have units of seconds or minutes.