Question

$N{a_3}P{O_4}$ dissolves in water to produce an electrolyte solution. What is the osmolarity of a $2.0 \times {10^{ - 3}}M$ $N{a_3}P{O_4}$ solution?

Answer 1

Osmolarity means the number of osmoles of solute particles per unit volume of the solution. Osmolarity equals molarity times the van't Hoff $i$ factor. We need to Multiply the number of particles produced from dissolving the solution in water by the molarity to find the osmolarity.

Complete Step By Step Answer:
The concentration of osmotically active particles in solution, which may be quantitatively expressed in osmoles of solute per liter of solution.
In other words osmolarity is the multiple if molarity
Osmolarity = $i \times$ molarity
Here $i$ represents the van't Hoff factor, which tells you the ratio that exists between the number of moles of solute particles produced in solution and the number of moles of solute dissolved in solution.
Here, $N{a_3}P{O_4}$ is a strong electrolyte, which implies that it dissociates completely in aqueous solution to produce sodium cations and phosphate anions.
$N{a_3}P{O_4}(aq) \to 3N{a^ + }(aq) + P{O_4}^ - (aq)$
Now, every mole of trisodium phosphate that dissociates in solution produces a total of
$3$ Moles of $N{a^ + }$ $+$ 1 mole $P{O_4}^ - = 4$ 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 $N{a_3}P{O_4}$ .dissolved $= 4$
Since we know that,
$[N{a_3}P{O_4}] = 2.0 \times {10^{ - 3}}M$
You can say that the solution will have an osmolarity equal to:
Osmolarity $= 4 \times 2.0 \times {10^{ - 3}}M = 8.0 \times {10^{ - 3}}osmol{L^{ - 1}}$.

Note:
The difference between the term osmolarity and osmolality is that the term osmolarity refers to the number of particles of solute per liter of solution, whereas the term osmolality refers to the number of particles of solute per kilogram of solvent and it is important to keep in mind that osmolarity is expressed in osmoles per liter.
$\Rightarrow \dfrac{{2.0 \times {{10}^{ - 3}}molesN{a_3}P{O_4}}}{{1Lsolution}} \times \dfrac{{4osmoles}}{{1moleN{a_3}P{O_4}}} = 8.0 \times {10^{ - 3}}osmol{L^{ - 1}}$