1
$\begingroup$

I’d like to ask a question and clear my misunderstanding.

Which of the following 0.01 M solutions have the highest osmotic pressure?
-$\ce{NaCl}$
-$\ce{C12H22O11}$
-$\ce{BaCl2}$
-$\ce{CO(NH2)2}$
-$\ce{[Cr(NH3)4Cl2]Cl}$

I know that the equation for osmotic pressure is
$$\Pi = cRTi$$ I have eliminated sucrose, sodium chloride, and urea from my answer, because sucrose and urea are non-electrolyte. And sodium chloride has smaller van ’t Hoff factor than barium chloride. I ended up in barium chloride and the complex compound. I’m not really sure with the complex compound since from what I learnt, a complex compound is easier to dissolved in water (my teacher in solubility chapter taught me that $\ce{AgI + NH3 -> [Ag(NH3)2]I}$ have a high solubility because it is a complex compound). However I’m not sure if it has a high van ’t Hoff factor as well.

$\endgroup$
3
$\begingroup$

You are correct in thinking that urea and sucrose would have van't hoff factor (i) = 1, since they are non-electrolytes and don't undergo disassociation(or ion pairing).

What you are left with are the following: $\ce{NaCl}$ , $\ce{BaCl_2}$ and $\ce{[Cr(NH_3)_4Cl_2]Cl}$.

Let us begin with the the two chloride salts: $\ce{NaCl}$ , $\ce{BaCl_2}$. we assume 100% disassociation for the two (strong electrolytes) and we have i = 2 and i = 3 respectively for both.

Now, the complex would indeed disassociate but not completely (the extent would be determined $K_f$ for the complex). Thus 1 < i < 2.

Based on this, I believe you can order them on the basis of increasing osmotic pressure.

$\endgroup$
  • $\begingroup$ Thank you for your answer and your time! I learned some today! $\endgroup$ – Prajogo Atmaja May 31 '15 at 16:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.