An open tubular column with a diameter of 195 um and a stationary phase thicknes

Question

An open tubular column with a diameter of 195 um and a stationary phase thickness on the inner wall of 0.54 m passes unretained solute through in 41 s. A particular solute has a retention time of 493 s. What is the partition coefficient for this solute? What fraction of time does this solute spend in the stationary phase? K= t_s/t_total= You must first calculate the retention factor, k, from the retention time of the solute and then use ratio of the mobile phase volume to the stationary phase volume to determine the partition coefficient. 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_r = the retention time of the solute

Explanation / Answer

Remember the geometry of a cylinder. Its volume is Base area x hight.

Area is 3.14x r2.. For the stationary phase the volume is the inner space between 2 cylinders.

But you have to calculate directly Vm/Vs

Vm= (195/2 – 0.54)2 x3.14 x column length

Vs = [(195/2)2 - (195/2 – 0.54)2] x3.14 x column length

Vm/Vs= (195/2 – 0.54)2 / [(195/2)2 - (195/2 – 0.54)2]

            = 96.962 / (97.502 – 96.962)

            = 9401/105 = 89.5

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