An open tubular column with a diameter of 225 mu m and a stationary phase thickn

Question

An open tubular column with a diameter of 225 mu m and a stationary phase thickness on the inner wall of 0.66 mu m passes unretained solute through in 43 s. A particular solute has a retention time of 449 s. (a) What is the partition coefficient for this solute? (b) What fraction of time does this solute spend in the stationary phase? K = k V_m/V_s k = t_r - t_m/t_m where K = the partition coefficient k = the retention factor V_m = the volume of the mobile phase V_s = the volume of the stationary phase t = the retention time of the solute t_m = the retention time for an unretained solute The fraction of time the solute spends on the mobile phase is the time its retention is retarded divided by the total retention time.

Explanation / Answer

A)

k= 449-43 / 43 = 9.441

K = k* Vm / Vs = 9.441 * ( 225 / (225+2*0.66) )3

= 9.276

b)fraction of time spent in stationary phase = time spent in stationary phase/total time spent fraction of time spent in stationary phase = (449– 43s)/449 s = 0.904

90.4%

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