# Question 6387d

Apr 6, 2016

${\text{0.34 osmol L}}^{- 1}$

#### Explanation:

In order to find a solution's osmolarity, you need to know two things

• how many osmoles of solute you have
• the volume of the solution expressed in liters

Basically, a solution's osmolarity tells you how many osmoles of solute you get per liter of solution.

$\textcolor{b l u e}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \text{osmolarity" = "osmoles of solute"/"liters of solution} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

Now, an osmole represents a mole of solute particles that contributes to the solution's osmotic pressure, which is the pressure needed to prevent the flow of water across a semi-permeable membrane. Simply put, an osmole will tell you how many moles of particles you get per mole of solute dissolved in water.

Calcium chloride, ${\text{CaCl}}_{2}$, is a soluble ionic compound that dissociates completely in aqueous solution to form calcium cations, ${\text{Ca}}^{2 +}$, and chloride anions, ${\text{Cl}}^{-}$

${\text{CaCl"_ (color(red)(2)(aq)) -> "Ca"_ ((aq))^(2+) + color(red)(2)"Cl}}_{\left(a q\right)}^{-}$

Notice that one moles of calcium chloride produces three moles of ions in aqueous solution

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

All three moles of ions will contribute to the osmotic pressure of the solution, so you can say that you have $3$ osmoles of solute particles for every $1$ mole of solute.

Now, use calcium chloride's molar mass to calculate ho many moles you get in that $\text{12.5-g}$ sample

12.5 color(red)(cancel(color(black)("g"))) * "1 mole CaCl"_2/(111color(red)(cancel(color(black)("g")))) = "0.1126 moles CaCl"_2

This means that the solution will contain

0.1126 color(red)(cancel(color(black)("moles CaCl"_2))) * "3 osmoles of ions"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = "0.3378 osmoles"

You can safely assume that the volume of water will be equivalent to the volume of the solution, which means that the osmolarity of the solution will be

"osmolarity" = "0.3378 osmoles"/"1 L" = color(green)(|bar(ul(color(white)(a/a)"0.34 osmol L"^(-1)color(white)(a/a)|)))#

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig fig for the volume of water.