Ultra high-performance liquid chromatography (UHPLC) uses sub 2.0 m diameter par

Question

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 × 3.0-mm-diameter phenyl UHPLC column, a compound has a retention time of 0.719 min and a width at half height of 2.76 s. Calculate the number of theoretical plates, N, and the plate height, H, for this column under the given conditions.

3/15/2017 11:55 PM A 84.2/100 3/3/2017 11:06 AM Gradebook Print i Calculator Periodic Table Question 15 of 19 Map Sapling Learning 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 1 x3.0-mm-diameter phenyl UHPLC column, a compound has a retention time of 0.719 min and a width at half height of 2.76 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 1.87 pm in diameter, how many particles placed side-by- side equal one theoretical plate? Number particles/ plate A O Previous Check Answer Next Ext Hint

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 = 0.719 min = 0.719 x 60 = 43.14 s

N = 5.45 (43.14 /2.76)^2 = 1331.5

N = 1332 plates

the height equivalent to a theoretical plate

H = L/N

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

H = 150/1332

H = 0.1126 mm = 112.6 micrometers

H = 113 micrometers

For:

particle size --> 1.87 micrometer

112.6 micrometers --> 113 /1.87 = 60.22

number of particles = 60

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