Originally Posted by jmoore

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You start with 3.9% active ingredient, which is equal to 39 g / l (grams per litre) stock solution.

You need 350 micromoles in your final solution or 0.05 g / l ( based on the relative molecular mass you supplied of 144.something)

To reach that figure you divide your final mass of 0.05 grams ( as we are talking about 1 litre) by your starting concentration of 39 g / l which gives a figure of 1.28 ml / litre.
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