# Question 13a6a

Jul 10, 2017

${\text{0.7 Osmol L}}^{- 1}$

#### Explanation:

As you know, the osmolarity of a solution tells you the number of moles of particles of solute, i.e. the number of osmoles, present for every $\text{1 L}$ of solution.

So in order to find the osmolarity of this solution, you must determine exactly how many moles of particles of solute are present in the solution.

Now, sodium nitrate is a strong electrolyte, which implies that it dissociates completely when dissolved in water to produce sodium cations and nitrate anions.

${\text{NaNO"_ (3(aq)) -> "Na"_ ((aq))^(+) + "NO}}_{3 \left(a q\right)}^{-}$

Notice that every $1$ mole of sodium nitrate that dissolves in water produces $2$ moles of particles of solute.

$\text{1 mole Na"^(+) + "1 mole NO"_3^(-) = "2 moles ions}$

This means that $1$ mole of sodium nitrate will be equivalent to $2$ osmoles,

On the other hand, sucrose is a non-electrolyte, which implies that it does not dissociate when dissolved in water.

${\text{C"_ 12"H"_ 22"O"_ (11(s)) -> "C"_ 12 "H"_ 22 "O}}_{11 \left(a q\right)}$

Since every $1$ mole of sucrose that dissolves in water produces $1$ mole of particles of solute, you can say that $1$ mole of sucrose is equivalent to $1$ osmole.

The osmolarity of the two solutes will thus be

0.2 color(white)(.)color(red)(cancel(color(black)("moles NaNO"_3)))/"L" * "2 Osmoles"/(1color(red)(cancel(color(black)("mole NaNO"_3)))) = "0.4 Osmol L"^(-1)

and

0.3 color(white)(.)color(red)(cancel(color(black)("moles sucrose")))/"L" * "1 Osmol"/(1color(red)(cancel(color(black)("mole sucrose")))) = "0.3 Osmol L"^(-1)#

Finally, to find the osmolarity of the solution, simply add the osmolarities of the two solutes.

${\text{osmolarity solution" = "0.4 Osmol L"^(-1) + "0.3 Osmol L}}^{- 1}$

$\textcolor{\mathrm{da} r k g r e e n}{\underline{\textcolor{b l a c k}{{\text{osmolarity solution" = "0.7 Osmol L}}^{- 1}}}}$

The answer is rounded to one significant figure.