# Question #83b7d

##### 1 Answer

Here's what I got.

#### Explanation:

The idea here is that you need to figure out the mass of calcium chloride dihydrate,

Since you didn't provide a *volume* for your solution, let's make the calculations easier by assuming that we're working with

As you know, a solution's **molarity** tells you the number of moles of solute present in

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

of anhydrous calcium chloride,

Since you know that **every mole** of calcium chloride dihydrate contains

,one moleof anhydrous calcium chloride#1 xx "CaCl"_2# ,two molesof water of hydration#2 xx "H"_2"O"#

you can say that **moles** of anhydrous calcium chloride will be delivered to the solution by **moles** of calcium chloride dihydrate.

Moreover, **every mole** of anhydrous calcium chloride contains

,one moleof calcium cations#1 xx "Ca"^(2+)# ,two molesof 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

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