#"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
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
#color(blue)(ul(color(black)("osmolarity" = i xx "molarity")))#
Here
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
#"Na"_ 3"PO"_ (4(aq)) -> 3"Na"_ ((aq))^(+) + "PO"_( 4(aq))^(-)#
Now, notice that every mole of trisodium phosphate that dissociates in solution produces a total of
#"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
#color(darkgreen)(ul(color(black)("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)#