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

Question

Henry Antrim · Accepted Answer

Dissolve 12 g of #"KI"# in enough water to make 500 mL of solution.

A solute's mass in one liter of solution is its molarity. #color(blue)(bar(ul(|color(white)(a/a)"Molarity" = "moles"/"litres"color(white)(a/a)|)))" "# Thus, you have to convert. #"millilitres of solution" → "litres of solution" → "moles of KI" → "grams of KI"# #"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 #"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 #"KI"#. If you need more precision, you will have to recalculate.

Kai Bowman · Answer

undefined To prepare 500 mL of a 0.15 M solution of potassium iodide (KI), you would need to dissolve 11.86 grams of KI in distilled water and then dilute it to a final volume of 500 mL.