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1A. Ultra high-performance liquid chromatography (UHPLC) uses sub 2.0 ?m diamete

ID: 1062124 • Letter: 1

Question

1A. Ultra high-performance liquid chromatography (UHPLC) uses sub 2.0 ?m diameter particles and pressures up to 100 MPa to achieve increased resolution or decrease the run time while maintaining adequate resolution. When run on a 150.0-mm-long × 2.1-mm-diameter phenyl UHPLC column, a compound has a retention time of 0.927 min and a width at half height of 3.23 s. Calculate the number of theoretical plates, N, and the plate height, H, for this column under the given conditions.

1B. If this particular column contains particles that are 1.86 ?m in diameter, how many particles placed side-by-side equal one theoretical plate?

Ultra high-performance liquid chromatography (UHPLC) uses sub 2.0 um diameter particles and pressures up to 100 MPa to achieve increased resolution or decrease the run time while maintaining adequate resolution. When run on a 150.0-mm-long x2.1-mm-diameter phenyl UHPLC column, a compound has a retention time of 0.927 min and a width at half height of 3.23 s. Calculate the number of theoretical plates, N, and the plate height, H, for this column under the given conditions. Number plates Number Um/ plate f this particular column contains particles that are 1.86 pm in diameter, how many particles placed side-by- side equal one theoretical plate? Number particles/ plate

Explanation / Answer

The formula for calculating number of theoretical plates is

N = 5.45 (tr/W)2

where tr: retention time, and W: peak width at half height

t = 0.927 min = 0.927*60 = 55.62 s

N = 5.45 (55.62 /3.23)^2 = 1616.04

N = 1617 plates

the height equivalent to a theoretical plate

H = L/N

L = length in mm and N is the number of theoretical plates

H = 150/1617

H = 0.0927 mm = 92.7 micrometers

For:

particle size --> 1.86 micrometer

92.7 micrometers --> 92.7/1.86 = 49,83 particles = 49 particles for sure

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