# Question 514b9

Aug 22, 2015

$\text{Osmolarity" = "0.297 Osmol/L}$

#### Explanation:

Before doing anything else, try to get a clear understanding of what it is you're dealing with here.

The first important thing to notice is that you were given the normality of the sodium and potassium cations. For both these cations, the normality will be equal to the molarity.

That happens because both cations have a valence of $1$, which implies that their equivalent weight will be equal to their molecular weight.

Another thing to look out for is the fact that you don't have the same units gioven for all the species present in the serum. Sodium and potassium are given per liter of serum, but glucose and BUN are given per deciliter - keep that in mind.

So, start by calculating the osmolarity of each species. For the sodium and potassium cations keep in mind that each of them can pair up with a negative ion to give 2 Osmoles of particles in the serum.

The osmolarity of the sodium cations will be

$\text{Na"^(+) = 140color(red)(cancel(color(black)("mmol")))/"L" * "1 mol"/(10""^(3)color(red)(cancel(color(black)("mmol")))) * "2 Osmol"/(1color(red)(cancel(color(black)("mol")))) = "0.280 Osmol/L}$

The osmolarity of the potassium cations will be

$\text{K"^(+) = 4.0color(red)(cancel(color(black)("mmol")))/"L" * "1 mol"/(10""^(3)color(red)(cancel(color(black)("mmol")))) * "2 Osmol"/(1color(red)(cancel(color(black)("mol")))) = "0.008 Osmol/L}$

To get the osmolarities of glucose and BUN, use their molar masses. For glucose you'd get

$\text{glucose" = 95color(red)(cancel(color(black)("mg")))/color(red)(cancel(color(black)("dL"))) * (10color(red)(cancel(color(black)("dL"))))/"L" * "1 g"/(10""^(3)color(red)(cancel(color(black)("mg")))) * (1color(red)(cancel(color(black)("mole"))))/(180.16color(red)(cancel(color(black)("g")))) * "1 Osmol"/(1color(red)(cancel(color(black)("mol")))) = "0.00527 Osmol/L}$

Do the same for BUN to get

$\text{BUN" = 10color(red)(cancel(color(black)("mg")))/color(red)(cancel(color(black)("dL"))) * (10color(red)(cancel(color(black)("dL"))))/"L" * "1 g"/(10""^(3)color(red)(cancel(color(black)("mg")))) * (1color(red)(cancel(color(black)("mole"))))/(28.01color(red)(cancel(color(black)("g")))) * "1 Osmol"/(1color(red)(cancel(color(black)("mol")))) = "0.00357 Osmol/L}$

Add all these values to get the total osmolarity of the serum

$\text{osm. serum} = 0.280 + 0.008 + 0.00527 + 0.00375$

"osm. serum" = color(green)("0.297 Osmol/L")#