Question

Which solution will exert highest osmotic pressure?

• A. $1\mathrm{M}\mathrm{NaCl}$ solution
• B. $1\mathrm{M}$ glucose solution
• C. $1\mathrm{M}$ alum solution
• D. $1\mathrm{M}$ urea solution

Answer 1

The correct option is C

$1\mathrm{M}$ alum solution

The explanation for correct option

(C) $1\mathrm{M}$ alum solution

Osmotic pressure

• When a semipermeable membrane separates a solution from pure water, the pressure that must be supplied to the solution side to cease fluid flow is known as osmotic pressure.
• The formula to calculate osmotic pressure is as follows:

$\mathrm{\pi }=\mathrm{iMRT}$

• $\mathrm{\pi }$ is the osmotic pressure.
• $\mathrm{M}$ is the concentration.
• $\mathrm{R}$ is a gas constant.
• $\mathrm{i}$ is the Van't Hoff factor.
• $\mathrm{T}$ is the temperature.

Order of osmotic pressure

• Alum $\left(\mathrm{NaAl}{\left({\mathrm{SO}}_4\right)}_2·12{\mathrm{H}}_2\mathrm{O}\right)$ is a double salt.
• It will produce the most number of ions and thus generate the highest osmotic pressure.
• The reaction is as follows:

$\mathrm{NaAl}{\left({\mathrm{SO}}_4\right)}_2·12{\mathrm{H}}_2\mathrm{O}\left(\mathrm{aq}\right)\to {\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Al}}^{3+}\left(\mathrm{aq}\right)+2{\mathrm{SO}}_4^{2-}\left(\mathrm{aq}\right)+12{\mathrm{H}}_2\mathrm{O}\left(\mathrm{l}\right)$

• Hence for alum, $\mathrm{i}=4$
• For urea and glucose, $\mathrm{i}=1$
• The Sodium chloride $\left(\mathrm{HCl}\right)$ dissociates into sodium and chloride ions.
• The reaction is as follows:

$\mathrm{NaCl}\left(\mathrm{aq}\right)\to {\mathrm{Na}}^{+}\left(\mathrm{aq}\right)+{\mathrm{Cl}}^{-}\left(\mathrm{aq}\right)$

• For $\mathrm{NaCl}$, $\mathrm{i}=1$​
• Order of osmotic pressure will be:

$1\mathrm{M}$ alum solution > $1\mathrm{M}$ $\mathrm{NaCl}$ solution > $1\mathrm{M}$ urea $=$$1\mathrm{M}$ Glucose

Therefore, the correct option is the (C) $1\mathrm{M}$ alum solution.