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
Jul 3, 2016

Answer:

#"0.282 mol L"^(-1)#

Explanation:

Molarity tells you how many moles of solute, which in your case is potassium iodide, #"KI"#, are dissolved in exactly 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 #"0.125 L"# of solution. Since you're interested in figuring out how many moles would be present in #"1 L"# of solution, all you have to do is 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 #8#, which can only mean that the number of moles of solute must go up 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 #0.282# moles of potassium iodide in #"1 L"# of solution, it follows that the molarity will be

#"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.