Question

Which of the following $0.1M$ aqueous solutions will exert high osmotic pressure?
A. $A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}$
B. $N{{a}_{2}}S{{O}_{4}}$
C. $MgC{{l}_{2}}$
D. $KCl$

Answer 1

We know that osmotic pressure of an aqueous solution depends upon molar concentration, temperature and Van’t Hoff factor. As the molar concentration of each solution is considered constant i.e., $0.1M$ and the temperature is not specified. So, we need to compare the osmotic pressure of each given aqueous solution by the Van't Hoff factor.

Complete Step By Step Answer:
Osmotic pressure is defined as the minimum amount of pressure which must be applied to a solution to cease the flow of solvent molecules via a semipermeable membrane. It is one of the colligative properties which is independent of concentration of solute particles in the solution. Osmotic pressure for a given solution can be calculate as per following formula:
$\pi =iCRT$
Where, $\pi$ is the osmotic pressure, $i$ represents Van’t Hoff factor, C is the molar concentration of solution, R is the universal gas constant and T is the temperature.
Now, we need to compare the osmotic pressure of a given $0.1M$ aqueous solution with the help of its Van’t Hoff factor. The Van’t Hoff factor refers to the ratio of concentration of particles which are formed when a substance or molecule is dissolved to the concentration of the substance by mass. Basically, it describes the extent to which a substance dissociates in a solution i.e., the total number of ions that are present in one formula unit of the substance is its Van’t Hoff factor.
Each given compound dissociates in solution as follows:
Compound-A: $A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}$
$A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}\rightleftharpoons 2A{{l}^{3+}}+3SO_{4}^{2-}$
The total number of ions in the given compound $=5$
$\therefore i=5$
Compound-B: $N{{a}_{2}}S{{O}_{4}}$
$N{{a}_{2}}S{{O}_{4}}\rightleftharpoons 2N{{a}^{+}}+SO_{4}^{2-}$
The total number of ions in the given compound $=3$
$\therefore i=3$
Compound-C: $MgC{{l}_{2}}$
$MgC{{l}_{2}}\rightleftharpoons M{{g}^{2+}}+2C{{l}^{-}}$
The total number of ions in the given compound $=3$
$\therefore i=3$
Compound-D: $KCl$
$KCl\rightleftharpoons {{K}^{+}}+C{{l}^{-}}$
The total number of ions in the given compound $=2$
$\therefore i=2$
Since, osmotic pressure is directly proportional to Van’t Hoff factor that means higher the Van’t Hoff factor of a given solution, higher will be its osmotic pressure. So, among options, the Van’t Hoff factor is highest for aqueous solution of compound A i.e., $5$ .
Thus, we can conclude that $A{{l}_{2}}{{\left( S{{O}_{4}} \right)}_{3}}$ will exert highest osmotic pressure.
Hence, option (A) is correct.

Note:
It is important to note that instead of dissociation, association of substance may also take place while forming an aqueous solution. The major difference between association and dissociation is that the value of Van’t Hoff factor in case of association is less than one while in case of dissociation, it is greater than one.