# Question #8956f

##### 1 Answer

#### Answer:

#### Explanation:

The trick here is to realize that you can use the volume of the *diluted solution* and the volume of the *stock solution* to determine the **dilution factor**.

As you know, in order to **dilute** a solution, you must **increase** its volume while keeping the *number of moles of solute * **constant**.

If you start from the molarity of the solution

#color(purple)(|bar(ul(color(white)(a/a)color(black)(c = n_"solute"/V_"solution" implies n_"solute" = c * V_"solution")color(white)(a/a)|)))#

you can use the fact that the number of moles of solute **remains unchanged** to write

#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

The **dilution factor**, which essentially tells you much **concentrated** the stock solution was compared with the *dilute solution*, will be equal to

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

In your case, you know that the volume of stock solution was **final volume** of the dilute solution is **do not** forget to convert the volume from *milliliters* to *liters* or vice versa!

#"D.F." = (2.50 color(red)(cancel(color(black)("L"))))/(250.0 * 10^(-3)color(red)(cancel(color(black)("L")))) = 10#

This tells you that the stock solution was **times** as concentrated as the dilute solution.

This means that the concentration of the dilute solution is

#"D.F." = c_1/c_2 implies c_2 = c_1/"D.F."#

which in your case is

#c_2 = "6.00 M"/10 = color(green)(|bar(ul(color(white)(a/a)"0.600 M"color(white)(a/a)|)))#

The answer is rounded to three **sig figs**.

So, by **increasing** the volume by a factor of **decreased** the concentration of the stock solution by a factor of **dilution**.