Question

The stock solution is 100 μM the protocol calls for 10 mL solution of 80, 60, 40 and 20 μM. Compute the table.Target Concentration(μM) Volume of stock(mL) Volume of water(mL)20406080

Answer 1

Answer:

See the answer below

Explanation:

Using the dilution equation:

m1v1 = m2v2

m1 =100 μM, v1 = ?, v2 = 10 mL

In order to prepare 80 μM, then m2 = 80 μM

v1 = $\frac{80*10}{100}$ = 8 mL

In order to prepare 60 μM

v1 = $\frac{60*10}{100}$ = 6 mL

In order to prepare 40 μM

v1 = $\frac{40*10}{100}$ = 4 mL

In order to prepare 20 μM

v1 = $\frac{20*10}{100}$ = 2 mL

Since the final solution has to be up to the 10 mL mark, the complete table would be:

Target Concentration(μM)       Volume of stock(mL)        Volume of water(mL)

              20                                           2                                            8

              40                                           4                                             6

              60                                           6                                             4

              80                                           8                                             2