Question

Question1: Multiple Choice: circle the best answer (3 points) 150 mL of an aqueous analyte solution is extracted five times with 50 mL of ether (an organic solvent). The partition constant K for the analyte is 1.99. If the starting solution is 1.00 M in analyte concentration, what is the concentration of analyte remaining in the aqueous layer after the 5th extraction? a. (i) 0.07 (ii) 0.12 (iii) 0.19 (iv) 0.22 (v) 0.26 b. Reverse-phase chromatography has what type of stationary phase? (i) (ii) (ii) (iv) (v) No stationary phase (whatever the column is made of) Size-exclusion gel Ion-exchange gel Non-polar Polar c. In order to triple the resolution in a liquid chromatography experiment, the number of theoretical plates, N, must be increased by what factor? (i) 0.333 (1/3) (ii) 1 ii) 1.732 (V3) (iv) 3 (v) 9

Answer 1

(a) Voume of Jhe oqeous xoluton 150 mL Volume of the orsoni phase or Initial concentnation of the anaJyte Now e Kno fo nnumbe of exteaction the frtion of onalyte pemainin oc K phion co-esficient =1.99 and n=5 7.8. -0,048 1.79 x 50 + 150 Hence, afHen 5+h extrac concentoation o the analyte wil be Pemanin in the aqueu phase The concentation in the ogeous phase .

(h) In Revere phe chometugraphy th 24adionary phabe 1,3 Non-Polar, (c) We Know resolution is propor tional to the 2gare root of the number of theoretical plate R2 So the Heonetical plate must be in creabed b\ 9