Question: You have 2 5 0 m L stock solution containing 0 . 6 2 g boric acid ) = ( 6 2 g m Mol.

You have 250mL stock solution containing 0.62g boric acid )=(62gmMol. A millilitre of this solution is used to prepare 1L nutrient medium. What is the molarity and % boric acid content in the final solution?
How many litres of nutrient medium could you prepare if you use the entire amount of stock solution in the previous question?
You have 250mL stock solution containing 0.62g boric acid )=(62gmMol. A

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