Question #13a6a

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 `"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.

`"NaNO"_ (3(aq)) -> "Na"_ ((aq))^(+) + "NO"_ (3(aq))^(-)`

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

`"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.

`"C"_ 12"H"_ 22"O"_ (11(s)) -> "C"_ 12 "H"_ 22 "O"_ (11(aq))`

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.

`"osmolarity solution" = "0.4 Osmol L"^(-1) + "0.3 Osmol L"^(-1)`

`color(darkgreen)(ul(color(black)("osmolarity solution" = "0.7 Osmol L"^(-1))))`

The answer is rounded to one significant figure.