Question

The following solutions are prepared by dissolving the requisite amount of solute in water to obtain the desired concentrations. Rank the solutions according to their respective osmotic pressures in decreasing order assuming the complete dissociation of ionic compounds.
Rank from highest to lowest osomotic pressure.
a) 1 M CaCl2
b) 2 M CH3OH
c) 1 M C6H12O6
d) 1 M LiCl

Answer 1

Answer:

CaCl₂ > CH₃OH = LiCl > C₆H₁₂O₆

Explanation:

The osmotic pressure of a compound is calculated using the following expression:

π = MRT (1)

This expression is used when the substance is nonelectrolyte. If the solution is electrolyte solution then we need to count the van't hoff factor into the expression so:

π = MRTi  (2)

Now, we have 4 solutions here, only two of them are electrolyte solution, this means that these solutions can be dissociated into separate ions. These solutions are LiCl and CaCl₂. It can be shown in the following reactions:

LiCl -------> Li⁺ + Cl⁻     2 ions   (i = 2)

CaCl₂ ---------> Ca²⁺ + 2Cl⁻      3 ions (i = 3)

The methanol (CH₃OH) and glucose (C₆H₁₂O₆) are non electrolyte solutions, therefore they are not dissociated. So, let's use expression (1) for methanol and glucose, and expression (2) for the salts:

CaCl₂:   π = 1 * 3 * RT = 3RT

CH₃OH:   π = 2 * RT = 2RT

C₆H₁₂O₆:  π = 1 * RT = 1RT

LiCl: π = 1 * 2 * RT = 2RT

Finally with these results we can conclude that the decreasing order of these solutions according to their osmotic pressures are:

CaCl₂ > CH₃OH = LiCl > C₆H₁₂O₆