# What is the final volume when 2.50 mL of a 11.0 M #HCl# solution is diluted to 0.100 M #HCl# solution?

##### 1 Answer

#### Explanation:

The thing to remember when diluting a solution is that you can determine the **dilution factor** by using the *concentrations or the volumes* of the stock and target solutions.

The underlying principle of a **dilution** is that the concentration of the solution is **decreased** by **increasing** its volume while keeping the number of moles of solute **constant**.

You can express this by using molarities and volumes

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

Here

Now, you can find a solution's *dilution factor* be rearranging the above equation to get

#c_1V_1 = c_2V_2 implies c_1/c_2 = V_2/V_1#

This will be the solution's dilution factor

#color(blue)(|bar(ul(color(white)(a/a)"D.F." = c_1/c_2 = V_2/V_1color(white)(a/a)|)))#

In essence, the dilution factor tells you how much higher the concentration of the stock solution was compared with that of the *diluted solution*.

In your case, you know that the stock solution had a concentration of

This means that your solution was diluted by a factor of

#"D.F." = (11.0 color(red)(cancel(color(black)("M"))))/(0.100color(red)(cancel(color(black)("M")))) = 110#

If the concentration of the stock solution was **times higher** than that of the diluted solution, it follows that the *volume* of the diluted solution must be **times bigger** than that of the stock solution

#color(purple)(|bar(ul(color(white)(a/a)color(black)("D.F." = V_2/V_1 implies V_2 = "D.F." xx V_1)color(white)(a/a)|)))#

In your case, you'll have

#V_2 = 110 * "2.50 mL" = color(green)(|bar(ul(color(white)(a/a)"275 mL"color(white)(a/a)|)))#

So, to prepare this solution, you'd start with