The amount, in mL, of the concentrated acid required, would be 1.1875 mL
Dilution
From the dilution equation:
m1v1=m2v2 where m1 and m2 = molarity before and after dilution, and v1 and v2 = volume before and after dilution.
m2 = 0.285M, m1 = 12.0M v2 = 50.0 mL
v1 = m2v2/m1 = 0.285x50/12 = 1.1875 mL
Thus, 1.1875 mL of the acid would be taken and diluted with water up to the 50 mL mark.
More on dilution can be found here: https://brainly.com/question/13949222
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