Question #792b1

1 Answer
Jul 26, 2017


#"0.83 L"#


Molarity is all about the number of moles of solute and the volume of the solution expressed in liters, so start by using the 8molar mass* of calcium chloride to calculate the number of moles present in your sample.

#212 color(red)(cancel(color(black)("g"))) * "1 mole CaCl"_2/(110.98color(red)(cancel(color(black)("g")))) = "1.91 moles CaCl"_2#

Now, you know that in order to have a #"2.3-M"# solution of calcium chloride, you need to have #2.3# moles of calcium chloride for every #"1 L"# of solution.

Since solutions are homogeneous mixtures, i.e. they have the same composition throughout, you can use the molarity of the solution as a conversion factor to find the volume of solution that would contain #1.91# moles of solute.

#1.91 color(red)(cancel(color(black)("moles CaCl"_2))) * overbrace("1 L solution"/(2.3color(red)(cancel(color(black)("moles CaCl"_2)))))^(color(blue)("= 2.3 M")) = color(darkgreen)(ul(color(black)("0.83 L solution")))#

The answer is rounded to two sig figs, the number of significant figures you have for the molarity of the solution.