Question

Question #daedf

Answer 1

The osmolality of the solution is 304 mOsmol/kg.

Explanation:

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"^"-"`
`"1 mol"color(white)(mm)"1 mol"color(white)(ml)"2 mol"`
`"2 eq"color(white)(mmll)"2 eq"color(white)(mml)"2 eq"`

We see that `"2 mol Cl"^"-" = "2 eq Cl"^"-"`, but `"1 mol Ca"^"2+" = "2 eq Ca"^"2+"`.

For ions, the rule is, `"1 mol" = ncolor(white)(l) "eq"` or

`bb("1 eq" = 1/n color(white)(l)"mol")`,

where `n` 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.