Posted by Lindsay on Wednesday, December 7, 2011 at 1:39am.
We need the molarity so we will need to be working in moles and therefore need the molar mass of the steroid.
I make this
C. 12 x 20 = 240
H. 29 x 1 = 29
F. 19 x 1 = 19
O. 3 x 16 = 48
total = 336 i.e. 336 g mol^-1
If 10.0 mg is dissolved taken then this is
0.0100 g /336 g mol^-1
= 2.98 x 10^-5 mol
If this is dissolved in 500 ml which is 0.500 L then the concentration is
2.98 x 10^-5 mol / 0.500 L
= 5.95 x 10^-5 M
If 100 microlitre (0.100 ml) is diluted to 100 ml then this is a 1000x dilution (100 ml / 0.100 ml = 1000) so the final solution will be 1000 x more dilute. so
5.95 x 10^-5 M /1000
= 5.95 x 10^-8 M
(please check the maths, and you will need to decide on the final sig figs, which will be 3 sig figs I think)
Thank you! That was a giant help, I understand now!
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