Your strategy here will be to use the molarity and volume of the diluted solution to determine how many moles of solute, which in your case is hydrochloric acid, #"HCl"#, it must contain.. Once you know this value, you an sue the molarity of the stock solution as a conversion factor to see how many milliliters would contain this many moles. The underlying principle of a dilution is the fact that the number of moles of solute must remain constant. Basically, a dilution decreases the concentration of a solution by increasing its volume. As you know, a solution's molarity tells you how many moles of solute you get per liter of solution. The diluted solution has a molarity of #"3.00 mol L"^(-1)#, which means that every liter of this solution will contain #3.00# moles of hydrochloric acid. In your case, the #"500.-mL"# sample will contain
Now use the molarity of the stock solution to determine how many milliliters would contain #1.5# moles of #"HCl"#. Since a concentration of #"12.0 mol L"^(-1)# means that you get #12.0# moles of hydrochloric acid per liter of solution, you an say that you have
The answer is rounded to three sig figs. ALTERNATIVE APPROACH You can get the same result by using the equation for dilution calculations, which looks like this
Here you have #c_1#, #V_1# - the molarity and volume of the concentrated solution Rearrange to solve for #V_1# and plug in your values to find
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