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

Answer:

The time required to elute the two species is 53.3727 min

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:

$Re_{s} =\frac{2(t_{B}-t_{A})}{W_{A}+W_{B} } =\frac{2*(17.63-16.63)}{1.11+1.21} =0.8621$

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

$\frac{t_{2}}{t_{1}} =(\frac{Re_{B} }{Re_{s} } )^{2}$

Here

ReB = 1.5

$t_{2} =t_{1} *(\frac{Re_{B} }{Re_{s} } )^{2} =17.63*(\frac{1.5}{0.8621} )^{2} =53.3727min$