What volume of 17.4 M HCl should be used to prepare 500 mL of a 6.00 M solution of HCl

Let's assume for a second that you're not familiar with the equation for dilution calculations.

What would your approach be here?

As you know, diluting a solution means keeping the number of moles of solute constant while increasing the total volume of the solution by adding more solvent.

Since molarity is defined as moles of solute per liters of solution, diluting a solution will result in a decrease in its concentration.

So, if the number of moles of solute must be constant in order for a dilution to be performed, you can use the molarity and volume of the target solution to figure out how many moles of solute must be present in the sample of stock solution you're diluting.

#color(blue)(c = n/V implies n = c * V)#

#n_(HCl) = "0.250 M" * "6.00 L" = "1.5 moles HCl"#

Now all you have to do is figure out what volume of #"6.0 M"# stock solution will contain #1.5# moles of hydrochloric acid

#color(blue)(c = n/V implies V = n/c)#

#V_"stock" = (1.5 color(red)(cancel(color(black)("moles"))))/(6.0 color(red)(cancel(color(black)("moles")))/"L") = "0.25 L"#

Expressed in milliliters, the answer will be

#V_"stock" = color(green)("250 mL") -># rounded to two sig figs

This is the principle behind the equation for dilution calculations, which looks like this

#color(blue)(c_1V_1 = c_2V_2)" "#, where

#c_1#, #V_1# - the molarity and volume of the concentrated solution
#c_2#, #V_2# - the molarity and volume of the dilute solution

Notice that you actually have

#overbrace(c_1V_1)^(stackrel( color(brown)("moles of solute"))(color(brown)("in stock solution"))) = overbrace(c_2V_2)^(stackrel(color(purple)("moles of solute"))(color(purple)("in target solution")))#

In your case, you'll once again get

#V_1 = c_2/c_1 * V_2#

#V_1 = (0.250 color(red)(cancel(color(black)("M"))))/(6.0color(red)(cancel(color(black)("M")))) * "6.00 L" = "0.25 L" = color(green)("250 L")#

To make the problem more interesting, let's assume that you don't know the formula for dilution calculations.

The idea with diluting a solution is that the number of moles of solute will remain constant after the initial solution is diluted. The only thing that changes in such cases is the volume of the solution.

This means that if you know how many moles of solute you have in the target solution, you also know how many moles of solute were present in the stock solution sample.

Use the molarity and volume of the target solution to determine how many moles of hydrochloric acid, #"HCl"#, you need in that solution

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

#n_"HCl" = "0.10 M" * 500 * 10^(-3)"L" = "0.050 moles HCl"#

Now the question is - what volume of stock solution would contain this many moles of hydrochloric acid?

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

#V_"stock" = (0.050color(red)(cancel(color(black)("moles"))))/(12color(red)(cancel(color(black)("moles")))/"L") = "0.0041667 L"#

I'll leave the answer rounded to two sig figs, despite the fact that you only gave one sig fig for the volume of the target solution