# Question #6e951

##### 1 Answer

Here's what I got.

#### Explanation:

**!! LONG ANSWER !!**

The idea here is that you can **decrease** the concentration of a solution by **increasing** its volume while keeping the *number of moles of solute* **constant**.

This is the underlying principle of a **dilution**. In order to *dilute* a solution, you must make sure that you keep the number of moles of solute **constant** and proceed to increase the total volume of the solution.

Since doing this would leave you with the **same number of moles of solute** in a bigger volume, the target solution will be *less concentrated* than the starting solution, i.e. more *dilute*.

Now, you're dealing with two solutions labeled

As you know, a solution's **molarity** tells you how many moles of solute you get **per liter of solution**.

Use the *molarities* and *volumes* of the two solutions to determine how many moles of solute you get

#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 will have

#"For A: " n_A = "1.00 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(100 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters")) = "0.10 moles A"#

#"For B: " n_B = "2.00 mol" color(red)(cancel(color(black)("L"^(-1)))) * overbrace(200 * 10^(-3)color(red)(cancel(color(black)("L"))))^(color(blue)("volume in liters")) = "0.40 moles B"#

Now, you mix these two solutions and add an extra **total volume** of the resulting solution will be

#V_"total" = V_A + V_B + V_"water"#

#V_"total" = "100 mL" + "200 mL" + "100 mL" = "400 mL"#

Now all you have to do is use the known numbers of moles of solute to determine the new concentrations of

#"For A: " c_A = "0.10 moles"/(400 * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)"0.25 mol L"^(-1)color(white)(a/a)|)))#

#"For B: " c_B = "0.40 moles"/(400 * 10^(-3)"L") = color(green)(|bar(ul(color(white)(a/a)"1.0 mol L"^(-1)color(white)(a/a)|)))#

I'll leave the answers rounded to two **sig figs**.

**ALTERNATIVE APPROACH**

Alternatively, you can get the same result a bit quicker by using the **dilution factor**, which essentially tells you how much **concentrated** the initial solution was compared with the *dilute solution*.

If you keep in mind the idea that the number of moles of solute **remains constant** in a dilution, you can use the formula for molarity to write

You can express this by using molarities and volumes

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

Here

Rearrange this equation to get

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

This represents your *dilution factor*

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

Let's take solution **final volume** is

#"For A: " "D.F." = (400 color(red)(cancel(color(black)("mL"))))/(100color(red)(cancel(color(black)("mL")))) = 4#

Therefore, the initial solution was **times as concentrated** as the final solution. This means that the new concentration of

#"D.F." = c_1/c_2 implies c_2 = c_1/"D.F." = "1.00 mol L"^(-1)/4 = color(green)(|bar(ul(color(white)(a/a)"0.25 mol L"^(-1)color(white)(a/a)|)))#

Now look at solution **final volume** of

#"D.F." = (400color(red)(cancel(color(black)("mL"))))/(200color(red)(cancel(color(black)("mL")))) = 2#

Its final concentration will thus be

#c_2 = c_1/"D.F." = "2.00 mol L"^(-1)/2 = color(green)(|bar(ul(color(white)(a/a)"1.0 mol L"^(-1)color(white)(a/a)|)))#