Part complete how much water would you add to 175 ml of 6.00 m hydrochloric acid to prepare a 2.00 m solution of this acid?

Molarity (M) = moles of solute (mol) / Volume of the solution (L)

Molarity of the given solution = 6.00 M
Volume taken to prepare 2.00 M = 175 mL
moles in 175 mL = 6.00 M x 0.175 L
                             = 1.05 mol
Hence, moles of HCl in 2.00 M solution = 1.05 mol
By applying formula,
2.00 M = 1.05 mol / Volume
Volume = 0.525 L

Hence, volume needed to prepare the solution = 0.525 L
Volume of  6.00 M solution + volume of water   = 0.525 L

Hence, the volume of water needed to add = 0.525 L - 0.175 mL
                                                                         = 0.350 L = 350 mL

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The correct answer is "atomic weight".