# How many milliliters of a solution of 4.00M #KI# are needed to prepare 250.0 mL of 0.760M #KI#?

##### 1 Answer

#### Answer:

#### Explanation:

What you're essentially doing here is **diluting** a concentrated solution of potassium iodide, *known volume*.

The most important thing to remember about **dilutions** is that the *number of moles of solute* **must remain constant**, i.e. the *diluted solution* will have the same number of solute as the sample of *concentrated solution*.

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

#color(blue)(|bar(ul(color(white)(a/a)"molarity" = "moles of solute"/"liter of solution"color(white)(a/a)|)))#

This means that you can **decrease** the concentration of a solution by **increasing** its total volume **while** keeping the number of moles of solute constant.

To increase the total volume of the solution, you would **add solvent**, which in this case is *pure water*.

The problem wants you to figure out what volume of concentrated solution will contain **the same number of moles** of potassium iodide as

So, use the molarity and volume of the target solution to find the number of moles of solute it must contain

#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)|)))#

Plug in your values to find - **do not** forget to convert the volume from *milliliters* to *liters*!

#n_(KI) = "0.760 mol" color(red)(cancel(color(black)("L"^(-1)))) * 250.0 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.190 moles KI"#

Now use the molarity of the stock solution to find what volume would contain this many moles

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

Plug in your values to get

#V_"stock" = (0.190 color(red)(cancel(color(black)("moles"))))/(4.00color(red)(cancel(color(black)("mol"))) "L"^(-1)) = "0.0475 L" = color(green)(|bar(ul(color(white)(a/a)"47.5 mL"color(white)(a/a)|)))#

So, to prepare this solution, you would start with *enough water* until the total volume of the solution is equal to

This is the concept behind the formula for **dilution calculations**

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

Here

Plug in your values to get the exact same result

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

#V_1 = (0.760 color(red)(cancel(color(black)("M"))))/(4.00color(red)(cancel(color(black)("M")))) * "250.0 mL" = color(green)(|bar(ul(color(white)(a/a)"47.5 mL"color(white)(a/a)|)))#

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