1. How would you make 1 L of a solution that is 0.5 M NaCl and 0.75 M 2. If you
ID: 196355 • Letter: 1
Question
1. How would you make 1 L of a solution that is 0.5 M NaCl and 0.75 M 2. If you added 59 g of NaCl and 118 g of succinate to 500 mL of H20, 3. If you add 5 L of a protein solution that has a concentration of 2.0 succinate? (MW of NaCl = 58.4 g/mol, Mw of succinate-118.1) what would be the resulting composition (in M) of that solution? mg/mL to a test tube and then add 95 L of H2O what is the final concentration (g/L) of BSA in the tube? 4. If you add 7.5 L of a protein solution that has a concentration of 2.0 mg/mL to a test tube and then add 100 L of H2O what is the final concentration (g/L) of BSA in the tube?Explanation / Answer
1. NaCl: M = W/MW x 1000/1000
0.5 = W/58.4
0.5 x 58.4 = W
W = 29.2 gm
Succinate : M = W/MW x 1000/1000
0.75 = W/118.1
0.75 x 118.1 = W
= 88.57 gm
2. NaCl: M = 59/58.4 x 1000/500
= Approximately 2M
Succinate : Approximately 2M
3. M1V1 = M2V2 (1mg/ml = 1000ppm)
2000 ppm x 5ul = M2 x 100ul
M2 = 2000 x 5 / 100
M2 = 10000/100
M2 = 100 ppm
M2= 100 ug/ml
M2 = 0.1mg/ml
M2 = 0.1g/ltr
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