Question

Substances A and B have retention times of 16.63 and 17.63 min, respectively, on a 30 cm column. An unretained species passes through the column in 1.30 min. The peak widths (at base) for A and B are 1.11 and 1.21 min, respectively. Calculate the time in minutes required to elute the two species with resolution Rs

Answer 1

The time required to elute the two species is 53.3727 min

Step-by-step explanation:

Given data:

tA = retention time of A=16.63 min

tB=retention time of B=17.63 min

WA=peak of A=1.11 min

WB=peak of B=1.21 min

The mathematical expression for the resolution is:

`Re_(s) =(2(t_(B)-t_(A)))/(W_(A)+W_(B) ) =(2*(17.63-16.63))/(1.11+1.21) =0.8621`

The mathematical expression for the time to elute the two species is:

`(t_(2))/(t_(1)) =((Re_(B) )/(Re_(s) ) )^(2)`

Here

ReB = 1.5

`t_(2) =t_(1) *((Re_(B) )/(Re_(s) ) )^(2) =17.63*((1.5)/(0.8621) )^(2) =53.3727min`