D Benedict College m/v7/c test s inline.php?c-21960 If you hrve·TMstock of Ha an
ID: 269185 • Letter: D
Question
D Benedict College m/v7/c test s inline.php?c-21960 If you hrve·TMstock of Ha and you with to make 87mlof10nMNC, what would be the volume of stock NaCI you would use? o 8.7 m o 0.087 m 1.74 m How many grams of bovine serum albumin (8SA) do you need to add to 10 ml of buffer to make a 2.4% ESA splution? O0.240 milligram 240grams O 0.240 grams o 24 grams what volume oCa2.4% solution of ESA is required to prepare 1 ml of a 0.1% solution of BSA? o 0.417 m 64.17 m O 41.7 t 2 &SA; has a molecular weight of 66,000 solution (in ?")? /mole, what is the molarity of your 0.1% 151.2 uM 0.4 uMExplanation / Answer
Ans. #1. Using C1V1 (stock solution) = C2V2 (diluted solution)
Or, 1.0 M x V1 = 0.010 M x 87.0 mL ; [1 mM = 0.001 M]
Or, V1 = (0.010 M x 87.0 mL) / 1.0 M = 0.87 mL
Hence, required volume of stock NaCl = 0.87 mL
# So, correct option is- B.
#2. A 2.4% (w/v) solution gas 2.4 g of solute (BSA) per 100.0 mL of solution.
So,
Required mass of BSA = [BSA] x Required volume of solution
= (2.4 g/ 100 mL) x 10.0 mL
= 0.240 g = 240.0 mg
# So, Correct option is- A.
#3. Using C1V1 (stock solution) = C2V2 (diluted solution)
Or, 2.4% x V1 = 0.1% x 1.0 mL
Or, V1 = (0.1% x 1.0 mL) / 2.4 = 0.417 mL = 417 uL
# So, Correct option is- A.
#4. Given, [BSA] = 0.1 % = 0.1 g / 100 mL = (0.1 g / 66000 g mol-1) / 0.100 L
= 1.52 x 10-6 mol / 0.100 L
= 1.52 x 10-5 mol / L
= 1.52 x 10-5 M ; [1 M = 106 uM]
= 1.52 x 10-5 x 106 uM
= 15.2 uM
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