Question

I have been provided with a stock solution of 17β-estradiol at 60 mg / L (solution A).
From this must be prepared 100 ml of solution B at a concentration of 60 g / L.
Question 1:
How many ul of solution A to be transferred?
Question 2:
What is the molarity of 17β-estradiol in the stock solutions A and B?

Question 3:
From the stock solution A is desired to prepare 10 ml with a concentration of 6 ng / ml.
How much should be taken from solution A? Can you do that? If not - then what?

Answer 1

in your problem you have some typo mistake I assume that you want to made a 60 ug/L solution B. Otherwise you can not made a more concentrated "dilution" from the A solution. when you calculate dilutions you can use the following formula\[V _{1}C _{1}=V _{2}C _{2}\] where V1 is the initial volume, C1 the initial concentration V2 the final volume and C2 the final concentration. The concentrations and volumes has to be in the same units. You have to convert the ug to mg
\[60 \mu g\frac{ 1 mg }{ 1000 \mu g }=0.06 mg\]
\[V _{1}=\frac{ V _{2}C _{2} }{ C _{1} }\]
\[V _{1}=\frac{ 100ml 0.060 mg/L }{ 60 mg/L }=0.100 ml \frac{ 1000 \mu L }{ 1mL}= 100 \mu L\]

Question 2: you need to know the molecular weigh of the 17bestradiol
Formula C18H24O2
Mol. mass 272.38 g/mol
Soluc A
\[60 \frac{ \mu g }{ L } \times\frac{ 1 g }{ 10^{6} \mu g } \times\frac{ 1 mol }{ 272.38 g } = 2.2 \times 10^{-7}M\]
Soluc B
\[2.2 \times 10^{-7} M \times \frac{ 0.1 mL }{ 100 mL}= 2.2 \times 10^{-10}\]

Answer 2

\[60\frac{ mg }{ L }\times \frac{ 10^{6}ng }{ 1 mg }\times \frac{ 1 L }{ 1000mL }= 60000 ng/mL\]

\[V _{1}= \frac{ 10mL \times 6 ng/mL }{ 60000 ng/mL}\]

V1 = 0.001 mL = 1 uL

if you have a nice automatic pipet you can do it otherwise you have to do a 1:10 or 1:100 dilution and then take from this new dilution 10 or 100 uL