What volume of a 0.100 M HCl stock solution should be used to prepare 250.00 mL of 0.0250 M HCl

The most important thing to keep in mind when it comes to diluting solutions is that the number of moles of solute must remain constant at all times.

Simply put, the number of moles of solute present in the dilute solution must be equal to the number of moles of solute present in the concentrated sample.

This is the key to any dilution calculation. As you know, molarity is defined as

#color(blue)("molarity" = "moles of solute"/"liters of solution")#

In essence, when you dilute a solution, you decrease its concentration by increasing its volume, which in turn is done by adding more solvent to the solution.

#color(white)(a)#

So, you know the molarity and volume of the target solution, which means that you can determine how many moles of hydrochloric acid it must contain

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

#n_(HCl) = "0.040 mol" color(red)(cancel(color(black)("L"^(-1)))) * 1.00color(red)(cancel(color(black)("L"))) = "0.040 moles HCl"#

Now your task is to determine what volume of the concentrated solution would contain this many moles of hydrochloric acid.

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

Plug in your values to get

#V = (0.040color(red)(cancel(color(black)("moles"))))/(0.25color(red)(cancel(color(black)("mol")))"L"^(-1)) = "0.16 L"#

Expressed in milliliters and rounded to two sig figs, the answer will be

#V = color(green)("160 mL")#

This is exactly what the formula for dilution calculations allows you to do

#color(blue)(overbrace(c_1 xx V_1)^(color(black)("moles of solute in concentrated solution")) = overbrace(c_2 xx V_2)^(color(black)("moles of solute in diluted solution"))#

Here you have

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

Once again you will have

#V_1 = c_2/c_1 * V_2#

#V_1 = (0.040color(red)(cancel(color(black)("M"))))/(0.025color(red)(cancel(color(black)("M")))) * "1.00 L" = color(green)("160 mL")#

So, you would prepare this solution by adding enough water to your #"160-mL"# sample of #"0.25-M"# hydrochloric acid solution to make the total volume of the resulting solution equal to #"1.00 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