Question #83b7d

1 Answer
Jun 25, 2017

Answer:

Here's what I got.

Explanation:

The idea here is that you need to figure out the mass of calcium chloride dihydrate, #"CaCl"_2 * 2"H"_2"O"#, that would deliver enough moles to a solution to get a concentration of #"2 mN"# of calcium cations.

Since you didn't provide a volume for your solution, let's make the calculations easier by assuming that we're working with #"1 L"# of solution.

As you know, a solution's molarity tells you the number of moles of solute present in #"1 L"# of solution. In this case, #"1 L"# of a #"2-mM"# solution of calcium chloride will contain

#2 color(red)(cancel(color(black)("mmoles"))) * "1 mole"/(10^3color(red)(cancel(color(black)("mmoles")))) = 2 * 10^(-3)# #"moles"#

of anhydrous calcium chloride, #"CaCl"_2#.

Since you know that every mole of calcium chloride dihydrate contains

  • one mole of anhydrous calcium chloride, #1 xx "CaCl"_2#
  • two moles of water of hydration, #2 xx "H"_2"O"#

you can say that #2 * 10^(-3)# moles of anhydrous calcium chloride will be delivered to the solution by #2 * 10^(-3)# moles of calcium chloride dihydrate.

Moreover, every mole of anhydrous calcium chloride contains

  • one mole of calcium cations, #1 xx "Ca"^(2+)#
  • two moles of chloride anions, #2 xx "Cl"^(-)#

This means that the molarity of the calcium cations will match that of the anhydrous calcium chloride.

#["Ca"^(2+)] = ["CaCl"_2]#

Now, in order to find the molar mass of the hydrate, use the molar mass of anhydrous calcium chloride and the molar mass of water.

You will have

#"110.98 g mol"^(-1) + 2 * "18.015 g mol"^(-) = "147.01 g mol"^(-1)#

This means that in order to make your solution, you must dissolve

#2 * 10^(-3) color(red)(cancel(color(black)("moles CaCl"_2 * 2"H"_2"O"))) * "147.01 g"/(1color(red)(cancel(color(black)("mole CaCl"_2 * 2"H"_2"O")))) = color(darkgreen)(ul(color(black)("0.3 g")))#

of calcium chloride dihydrate in enough water to get the total volume of the solution to #"1 L"#.

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