If 25.0 ml of a 6.00 m hcl solution is diluted to 2.00 l, what is the molarity of the new solution?

When you're diluting a solution, you're essentially keeping the number of moles of solute constant while changing the total volume of the solution.

Now, let's assume that you don't know the equation for dilution calculations.

In this case, you can use the molarity and volume of the concentrated solution to determine how many moles of hydrochloric acid you start with.

#c = n/V implies n = c * V#

#n_"HCl" = "18 M" * 190 * 10^(-3)"L" = "3.42 moles HCl"#

You then add water to get the total volume of the solution from #"190 mL"# to #"730 mL"#.

The number of moles of hydrochloric acid remains unchanged, which means that the molarity of the diluted solution will be

#c = "3.42 moles"/(730 * 10^(-3)"L") = color(green)("4.7 M")#

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