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