BCHM Module 4 Wkst

6. Explain how to make 50 mL of a 30% (w/v) solution of trichloroacetic acid in water. 7. You are given a protein solution with a concentration of 0.15 mg/mL a. You need 10 μ g for an experiment. What volume of the protein solution do you need? b. If the protein has a MW = 22,000 Da (1 Da ≅ 1 g/mol), express its initial concentration in M (mol/L) and μ M ( μ mol/L) c. If you need 20 μ moles for an experiment, what volume do you need? d. You want to make 1 mL of a 10 μ g/mL solution from the stock (0.15 mg/mL). What volume of the stock solution will you need to dilute to get this? e. You want to make 100 μ L of a 0.1 μ g/ μ L solution from the stock (0.15 mg/mL). What volume of the stock solution will you need to dilute to get this? 389 =0.3 g/mL 188mL SOmL x 0.3 g/m2= 15g trichloroacetic acid dissolve in water 10ng) ng = 0.01 mg 8 : 89mz = 0.067mL b.818uM 0.15 my 19 1800m2 1 mL 1888mg 12 =0.159/L ↑ x 1.06 um 0.19/demog = 6.818 x 18 - 6 M0Y : 6.818 x 18 - M 20 umoly, %umol = 2.0x10-s mol 6.818 x 18 - * M = 2.0 x 10-s mol -> 2.93 L L C,V, = zVz 2, = 0.15 mg/m2 -150 ng/mL (158ug/mL) v, = (10mg/m2) ((m)) V, = ? v, = (10nq(m)(1m)) C2 = 10 ng/mL (158ug/mL) V= = 1mL V, = 0.067 m C,V, = zVz 2, = 0.15 mg/m2 - 150ng(u)(158ug/m2)v, = (0.1ug(n)(100m2) V, = ? V, = (0.1ug(u)(100n)) C2 = 0.1ug/uL (158ug(x) V= = 108 eL V, = 0.067uL or 6.7 x 18-mL