How many milliliters of 12.0 M #HCl# (aq) are needed to prepare 215.0 mL of 1.00 M #HCl# (aq)?
Use the relationship molarity = no. of moles / no. of litres.
Then rearrange for no. of moles = no. of litres x molarity
and no. of litres no. of moles / molarity.
215 ml of 1 M HCl (aq) contains (0.215 x 1) = 0.215 moles of HCl.
The no. of ml of 12 M HCl (aq) that contains 0.215 mol of HCl is (0.215 /12) = 0.0179 litre = 17.9 ml.
Therefore you would use 17.9 ml of 12 M HCl,and then dilute up to total volume 215 ml with water.