# How do you calculate the osmolarity of a 6.0 * 10^-2 M NaCl solution?

Oct 31, 2015

$\text{0.12 osmolar}$

#### Explanation:

As you know, osmolarity tells you how many particles of a solute you get per liter of solution.

Osmolarity is expressed in number of osmoles, or solute particles, per liters of solution.

When dealing with soluble ionic compounds, you have to keep track of how many ions you get per formula unit. Sodium chloride, $\text{NaCl}$, will dissociate completely in aqueous solution to produce sodium cations, ${\text{Na}}^{+}$, and chloride anions, ${\text{Cl}}^{-}$

${\text{NaCl"_text((aq]) -> "Na"_text((aq])^(+) + "Cl}}_{\textrm{\left(a q\right]}}^{-}$

As you can see, each mole of sodium chloride will produce two osmoles in solution.

This means that you will have

6.0 * 10^(-2)color(red)(cancel(color(black)("moles")))/"L" * "2 Osmoles"/(1color(red)(cancel(color(black)("mole")))) = 12 * 10^(-2)"Osmoles/L"