# To 225 mL of a 0.80M solution of #KI#, a student adds enough water to make 1.0 L of a more dilute #KI# solution. What is the molarity of the new solution?

##### 1 Answer

#### Explanation:

Right from the start, you know that the molarity of solution **decreased** upon the addition of water, which is what *diluting* a solution implies.

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

So, you can use the molarity and volume of the initial solution to find the *number of moles* of potassium iodide,

#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)|)))#

Remember, molarity uses **liters of solution**, so don't forget to convert the volume from *milliliters* to *liters*

#n_(KI) = "0.80 mol" color(red)(cancel(color(black)("L"^(-1)))) * 225 * 10^(-3)color(red)(cancel(color(black)("L"))) = "0.180 moles KI"#

This is **exactly** how many moles of solute must be present in the *diluted* solution, so you can say that

#c_"diluted" = n_(KI)/V_"solution"#

#c_"diluted" = "0.180 moles"/"1.0 L" = color(green)(|bar(ul(color(white)(a/a)"0.18 mol L"^(-1)color(white)(a/a)|)))#

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

Here

Plug in your values to get

#c_2 = V_/1V_2 * c_1#

#c_2 = (225 * 10^(-3)color(red)(cancel(color(black)("L"))))/(1.0color(red)(cancel(color(black)("L")))) * "0.80 mol L"^(-1) = color(green)(|bar(ul(color(white)(a/a)"0.18 mol L"^(-1)color(white)(a/a)|)))#

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