# "Na"_3"PO"_4 dissolves in water to produce an electrolyte solution. What is the Osmolarity of a 2.0 * 10^(-3)"M Na"_3"PO"_4 solution? Thank you!

##### 1 Answer
Mar 16, 2017

$8.0 \cdot {10}^{- 3} {\text{osmol L}}^{- 1}$

#### Explanation:

The thing to keep in mind about osmolarity is that it takes into account the number of moles of particles of solute that are produced in a solution when a given number of moles of solute are dissolved to make said solution.

In other words, you can think about osmolarity as being a multiple of molarity

$\textcolor{b l u e}{\underline{\textcolor{b l a c k}{\text{osmolarity" = i xx "molarity}}}}$

Here $i$ represents the van't Hoff factor, which tells you the ratio that exists between the number of moles of particles of solute produced in solution and the number of moles of solute dissolved in solution.

In your case, trisodium phosphate is a strong electrolyte, which implies that it dissociates completely in aqueous solution to produce sodium cations and phosphate anions

${\text{Na"_ 3"PO"_ (4(aq)) -> 3"Na"_ ((aq))^(+) + "PO}}_{4 \left(a q\right)}^{-}$

Now, notice that every mole of trisodium phosphate that dissociates in solution produces a total of

$\text{3 moles Na"^(+) + "1 mole PO"_4^(-) = "4 moles ions}$

The number of moles of particles of solute produced in solution are actually called osmoles.

As a result, the van't Hoff factor will be equal to

i = "4 moles ions produced (osmoles)"/("1 mole Na"_3"PO"_4color(white)(.)"dissolved") = 4

Since you know that

["Na"_3"PO"_4] = 2.0 * 10^(-3)"M"

you can say that the solution will have an osmolarity equal to

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{osmolarity" = 4 xx 2.0 * 10^(-3)"M" = 8.0 * 10^(-3)color(white)(.)"osmol L}}^{- 1}}}}$

It's important to keep in mind that osmolarity is expressed in osmoles per liter because you have

(2.0 * 10^(-3)color(red)(cancel(color(black)("moles Na"_3"PO"_4))))/"1 L solution" * "4 osmoles"/(1color(red)(cancel(color(black)("mole Na"_3"PO"_4)))) = 8.0 * 10^(-3)color(white)(.)"osmol L"^(-1)