What is the osmotic pressure of a solution made from 12.5 g of CaCl2 in enough water to make 500 mL, if CaCl2 is 78.5% dissociated at 30C?

says the answer is 14.37 I beg to differ. Shouldn't it be 12.5g/(111g)= .113/.5= .226M then (x-.226M)/.226M=.785 x=.402M

then 3*.402M*.08206*303= 30 atm (3 being the expected van't hoff's factor)

Question

What is the osmotic pressure of a solution made from 12.5 g of CaCl2 in enough water to make 500 mL, if CaCl2 is 78.5% dissociated at 30C?

says the answer is 14.37 I beg to differ. Shouldn't it be   12.5g/(111g)= .113/.5= .226M then (x-.226M)/.226M=.785 x=.402M

then 3*.402M*.08206*303= 30 atm (3 being the expected van't hoff's factor)

anonymous · Answer

I think you made a mistake where you calculated the molarity of the Ca+2 from the percent dissociated and the initial molarity of the CaCl2.  You've got 0.226 M of CaCl2, and you know it's only 78.5% dissociated -- yet you end up with a molarity of Ca+2 that is 0.402 M -- HIGHER than the initial molarity of CaCl2.  How's that possible?

I would have just multiplied 0.226 M by 0.785 to get 0.177 M for [Ca+2].  The rest of it then looks OK, but I do get an answer closer to 14 atm than 30 atm.

anonymous · Answer

Disassociation is the separation of ions. if I have a .226M solution of CaCl2 has .226 moles of CaCl2 with  78.5% disassociation meaning that only 78.5% of them separate into ions, giving you a larger number. .226M is before disassociation happened.

anonymous · Answer

Yeah, but that's what the van't Hoff factor is all about.  So you've multiplied for the creation of new ions twice.  You should only do that once.

anonymous · Answer

to hell with this problem.

anonymous · Answer

I wish there were more of these problems somewhere.