# Question 83b7d

Jun 25, 2017

Here's what I got.

#### Explanation:

The idea here is that you need to figure out the mass of calcium chloride dihydrate, $\text{CaCl"_2 * 2"H"_2"O}$, that would deliver enough moles to a solution to get a concentration of $\text{2 mN}$ 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 $\text{1 L}$ of solution.

As you know, a solution's molarity tells you the number of moles of solute present in $\text{1 L}$ of solution. In this case, $\text{1 L}$ of a $\text{2-mM}$ 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)# $\text{moles}$

of anhydrous calcium chloride, ${\text{CaCl}}_{2}$.

Since you know that every mole of calcium chloride dihydrate contains

• one mole of anhydrous calcium chloride, $1 \times {\text{CaCl}}_{2}$
• two moles of water of hydration, $2 \times \text{H"_2"O}$

you can say that $2 \cdot {10}^{- 3}$ moles of anhydrous calcium chloride will be delivered to the solution by $2 \cdot {10}^{- 3}$ moles of calcium chloride dihydrate.

Moreover, every mole of anhydrous calcium chloride contains

• one mole of calcium cations, $1 \times {\text{Ca}}^{2 +}$
• two moles of chloride anions, $2 \times {\text{Cl}}^{-}$

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

$\left[{\text{Ca"^(2+)] = ["CaCl}}_{2}\right]$

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

${\text{110.98 g mol"^(-1) + 2 * "18.015 g mol"^(-) = "147.01 g mol}}^{- 1}$

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

$2 \cdot {10}^{- 3} \textcolor{red}{\cancel{\textcolor{b l a c k}{\text{moles CaCl"_2 * 2"H"_2"O"))) * "147.01 g"/(1color(red)(cancel(color(black)("mole CaCl"_2 * 2"H"_2"O")))) = color(darkgreen)(ul(color(black)("0.3 g}}}}$

of calcium chloride dihydrate in enough water to get the total volume of the solution to $\text{1 L}$.

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