Question

Question #83b7d

Answer 1

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.