1) An open tubular GC column has an inner diameter of 300 micrometer and it is coated inside with a layer of stationary phase that is 2.0 micrometer in thickness. On this column, a derivatized form of adenosine has a retention time of 300 sec, and un-retained ethane has a sharp spike at 32 sec. Please calculate the partition coefficient for the derivatized adenosine.
Answer: 3.1*102
2) For the same separation as described in the last question, please calculate the number of theoretical plates for the separation of the derivatized adenosine if the column length is 50 m and the peak width for the derivatized adenosine is 2.5 sec
Answer: 2.3 x 105
3) . For the same separation as described in the last question, please calculate the plate height for the separation of the derivatized adenosine.
Answer: 2.2 x 10-4 M
I wonder how they got. Can you show me step by step?
1)
we know that
rentention factor is given by
k = ( tr- tm / tm)
so
k = ( 300-32) / 32
k = 8.375
now
b = r / 2d
where
r = column radius
d = thickness
given
r = 150 x 10-6 m
d = 2 x 10-6 m
so
b = 150 / 2 x 2
b = 37.5
now
partition coefficient = k x b = 8.375 x 37.5 = 314
so
the partition coefficient is 3.1 x 10^2
2)
we know that
N = 16 (tr/W)^2
where
N = number of theoretical plates
tr = retention time
W = peak width
given
tr = 300 sec
W = 2.5 sec
so
using given values
we get
N = 16 ( 300 / 2.5)^2
N = 230400
N = 2.304 x 10^5
so
number of theoretical plates are 2.304 x 10^5
3)
we know that
H= L / N
where
L = length of column
given
L = 50 m
so
H = 50 / 2.304 x 10^5
H = 2.17 x 10-4 m
so
the plate height is 2.2 x 10-4 m
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