A student was asked to separate two substances, A and B, on a 30 cm column. She

Question

A student was asked to separate two substances, A and B, on a 30 cm column. She obtained a chromatogram that gave retention times of 15.80 and 17.23 min for A and B, respectively. The base peak widths for A and B were 1.25 and 1.38 min, respectively. Please calculate the following:

A)What is the average number of theoretical plates for this column?

B)What is the plate height? (Cm/Plate)

C)What is the resolution of A and B?

D)What should be a length of the column to achieve resolution of 2? Hint: Resolution is directly proportional, but not equal, to the square root of column length.

Please Show Work

Explanation / Answer

a) NA = 16(t/W)^2 = 16* (15.80/ 1.25)^2 =2556.3

NB= 16* (t/W)^2 = 16*(17.23/1.38)^2 = 2494.2

so, the average number of theoritical plates is: (NA+NB)/2 = (2556.3+2494.2)/2 =2525.2

b) Plate Height = (column length / number of theoritical plates ) = (30 / 2525.2) cm/Plate = 0.01188 cm/Plate

c) Resolution = 2*(tB - tA )/(WA+WB) = (2*(15.80-17.23))/(1.25+1.38) = 1.08

d) Let the required length be x.

then (2/ sqrt(x) ) = ( 1.08 / sqrt (30) )

solving this equation, x = 102 cm

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