# What is the molarity of a solution composed of 5.85 g of potassium iodide, #KI#, dissolved in enough water to make 0.125 L of solution?

##### 1 Answer

#### Explanation:

**Molarity** tells you how many **moles** of solute, which in your case is potassium iodide, **one liter** of solution.

The first thing to do here is use the **molar mass** of potassium iodide to figure out how many moles you have in that sample

#5.85 color(red)(cancel(color(black)("g"))) * "1 mole KI"/(166.0color(red)(cancel(color(black)("g")))) = "0.03524 moles KI"#

Now, you know that this many moles of potassium iodide are dissolved in *scale up* your current volume by a factor of

#(1 color(red)(cancel(color(black)("L"))))/(0.125color(red)(cancel(color(black)("L")))) = 8#

The volume goes up by a factor of **by the same factor**

#1 color(red)(cancel(color(black)("L solution"))) * "0.03524 moles KI"/(0.125color(red)(cancel(color(black)("L solution")))) = "0.282 moles KI"#

Since you have **moles** of potassium iodide in

#"molarity" = color(green)(|bar(ul(color(white)(a/a)color(black)("0.282 mol L"^(-1) = "0.282 M")color(white)(a/a)|)))#

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