Map Consider a separation performed on a 30.0 mm long open tubular column with a

Question

Map Consider a separation performed on a 30.0 mm long open tubular column with a 0.50 mm diameter, ando a 2.0 m thick stationary phase. Compound A eluted at 12.56 min, compound B eluted at 13.29 min, and the unretained solvent eluted at 1.085 min What are the adjusted retention times, f, and retention factors, k, for compounds A and B? Number Number min Number Number min What is the relative retention, , for this separation? Number Scroll down to answer the rest of this question The width at half-height, w1/2, of the peak for compounds A and B are 0.137 min and 0.287 min respectively. Find the number of theoretical plates, N, and plate height, H, for each compound Number Number

Explanation / Answer

For the separation of compound A and B

Adjusted retention times,

t'r(A) = 12.56 - 1.085 = 11.475 min

t'r(B) = 13.29 - 1.085 = 12.205 min

Retention factors

k(A) = (12.56 - 1.085)/1.085 = 10.576

k(B) = (13.29 - 1.085)/1.085 = 12.29

Relative retention alpha,

= k(A)/k(B) = 10.576/12.29 = 0.86

Number of theoretical plates for N(A) = 5.54(11.475/0.137)^2 = 38867

Number of theoretical plates for N(B) = 5.54(12.205/0.287)^2 = 10021

Plate height,

H(A) = 30/38867 = 0.0008 mm

H(B) = 30/10021 = 0.003 mm

Resolution,

= sq.rt.(N(A).N(B) = sq.rt.(0.0008 x 0.003) = 0.00155

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