Question #6387d

1 Answer
Apr 6, 2016

Answer:

#"0.34 osmol L"^(-1)#

Explanation:

In order to find a solution's osmolarity, you need to know two things

  • how many osmoles of solute you have
  • the volume of the solution expressed in liters

Basically, a solution's osmolarity tells you how many osmoles of solute you get per liter of solution.

#color(blue)(|bar(ul(color(white)(a/a)"osmolarity" = "osmoles of solute"/"liters of solution"color(white)(a/a)|)))#

Now, an osmole represents a mole of solute particles that contributes to the solution's osmotic pressure, which is the pressure needed to prevent the flow of water across a semi-permeable membrane.

http://chemwiki.ucdavis.edu/Textbook_Maps/General_Chemistry_Textbook_Maps

Simply put, an osmole will tell you how many moles of particles you get per mole of solute dissolved in water.

Calcium chloride, #"CaCl"_2#, is a soluble ionic compound that dissociates completely in aqueous solution to form calcium cations, #"Ca"^(2+)#, and chloride anions, #"Cl"^(-)#

#"CaCl"_ (color(red)(2)(aq)) -> "Ca"_ ((aq))^(2+) + color(red)(2)"Cl"_((aq))^(-)#

Notice that one moles of calcium chloride produces three moles of ions in aqueous solution

  • one mole of calcium cations, #1 xx "Ca"^(2+)#
  • two moles of chloride anions, #color(red)(2) xx "Cl"^(-)#

All three moles of ions will contribute to the osmotic pressure of the solution, so you can say that you have #3# osmoles of solute particles for every #1# mole of solute.

Now, use calcium chloride's molar mass to calculate ho many moles you get in that #"12.5-g"# sample

#12.5 color(red)(cancel(color(black)("g"))) * "1 mole CaCl"_2/(111color(red)(cancel(color(black)("g")))) = "0.1126 moles CaCl"_2#

This means that the solution will contain

#0.1126 color(red)(cancel(color(black)("moles CaCl"_2))) * "3 osmoles of ions"/(1color(red)(cancel(color(black)("mole CaCl"_2)))) = "0.3378 osmoles"#

You can safely assume that the volume of water will be equivalent to the volume of the solution, which means that the osmolarity of the solution will be

#"osmolarity" = "0.3378 osmoles"/"1 L" = color(green)(|bar(ul(color(white)(a/a)"0.34 osmol L"^(-1)color(white)(a/a)|)))#

I'll leave the answer rounded to two sig figs, despite the fact that you only have one sig fig for the volume of water.