# Question a59e1

Sep 15, 2016

The concentration was 3.3 nmol per gram of fresh weight.

#### Explanation:

Let's work backwards from your final result.

You took 1 mL of the cooled supernatant, and it had a concentration of 0,66 nmol/mL.

That 1 mL came from 4 mL of solution, so the solution must have contained

4 color(red)(cancel(color(black)("mL"))) × "0,66 nmol"/(1 color(red)(cancel(color(black)("mL")))) = "2,6 nmol"# of compound.

The 2 mL of original supernatant must have contained 2,6 nmol of compound before dilution.

That was a portion of the original 5 mL of extract, so the extract must have contained

$\text{2,6 nmol" × (5 color(red)(cancel(color(black)("mL"))))/(2 color(red)(cancel(color(black)("mL")))) = "6,6 nmol}$ of compound.

The mass of your original sample was 2 g.

The concentration in your sample must have been

$\text{6,6 nmol"/"2 g" = "3,3 nmol/g}$ of fresh weight.