# How can I prepare 500 mL of a 0.15 M solution of potassium iodide?

Aug 3, 2014

Dissolve 12 g of $\text{KI}$ in enough water to make 500 mL of solution.

#### Explanation:

Molarity is the mass of solute in 1 L of solution:

color(blue)(bar(ul(|color(white)(a/a)"Molarity" = "moles"/"litres"color(white)(a/a)|)))" "

So, you must convert

$\text{millilitres of solution" → "litres of solution" → "moles of KI" → "grams of KI}$

$\text{Mass of KI" = 500 color(red)(cancel(color(black)("mL soln"))) × (1color(red)(cancel(color(black)("L soln"))))/(1000color(red)(cancel(color(black)("mL soln")))) × (0.15color(red)(cancel(color(black)("mol KI"))))/(1color(red)(cancel(color(black)("L soln")))) × "166.0 g KI"/(1color(red)(cancel(color(black)("mol KI")))) = "12 g KI}$

So, you would place 12 g of $\text{KI}$ in a 500 mL volumetric flask and add enough water to dissolve the solid. Then you would add enough more water to reach the 500 mL mark.

Note: The answer can have only two significant figures, because that is all you gave for the molarity of the $\text{KI}$. If you need more precision, you will have to recalculate.