An osmole (Osmol) is 1 mol of particles that contribute to the osmotic pressure of a solution.
The problem here is how to convert equivalents to osmoles.
The point to remember is that 1 eq of anything reacts with or produces 1 eq of anything else
For example,
"CaCl"_2 → "Ca"^"2+" + "2Cl"^"-"CaCl2→Ca2++2Cl-
"1 mol"color(white)(mm)"1 mol"color(white)(ml)"2 mol"1 molmm1 molml2 mol
"2 eq"color(white)(mmll)"2 eq"color(white)(mml)"2 eq"2 eqmmll2 eqmml2 eq
We see that "2 mol Cl"^"-" = "2 eq Cl"^"-"2 mol Cl-=2 eq Cl-, but "1 mol Ca"^"2+" = "2 eq Ca"^"2+" 1 mol Ca2+=2 eq Ca2+.
For ions, the rule is, "1 mol" = ncolor(white)(l) "eq"1 mol=nleq or
bb("1 eq" = 1/n color(white)(l)"mol")1 eq=1nlmol,
where nn is the absolute value of the charge on the ion.
Now, let's use this relationship to calculate the osmolarity of the solution.
Assume we have 1 L of solution.
We will set up a table for easy calculation.
bbul("Amt/meq"color(white)(mll)"Ion"color(white)(mll)ncolor(white)(mll)"Amt/mOsmol")
color(white)(mml)145color(white)(mmml)"Na"^"+"color(white)(mll)1color(white)(mmmmm)145
color(white)(mmmll)3color(white)(mmml)"K"^"+"color(white)(mm)1color(white)(mmmmmml)3
color(white)(mmmll)5color(white)(mmml)"Ca"^"2+"color(white)(m)2color(white)(mmmmmml)2.5
ul(color(white)(mmll)153color(white)(mmml)"Cl"^"-"color(white)(mm)1color(white)(mmmmm)153)
color(white)(mmmmmmmmmmml)"TOTAL = 303.5"
Note that there are 153 meq of cations and 153 meq of anions, as required.
"Osmolarity" = "303.5 mOsmol"/"1 L" = "304 mOsmol/L"
Since the solution is so dilute, its density is almost the same as that of water, and the mass of the water is approximately 1 kg.
∴ The concentration is also 304 mOsmol/kg.
