Question

Calculate the osmotic pressure of \[5\% \;\] solution of glucose at \[{25^o}C\]. (Molecular weight of Glucose =\[180\;\] and \[R = 0.0821\;liter{\text{ }}atm\;{K^{ - 1}}mo{l^{ - 1}}\] )

Answer 1

Hint: Osmotic pressure is the equilibrium hydrostatic pressure of the column set up as a result of osmosis. Osmotic pressure can be calculated with the help of the following formula \[\pi {\text{ }} = {\text{ }}iCRT\]
It is denoted by \[\;\pi \].

Complete Step by step answer: Osmotic pressure is the minimum pressure that must be applied on the solution to prevent the entry of the solvent into the solution through the semipermeable membrane. It is the minimum pressure needed to apply on a solution to make its vapour pressure equal to vapour pressure of the solvent.
A s per formula - \[\pi {\text{ }} = {\text{ }}iCRT\]
\[\pi \] is the osmotic pressure,
\[\;i\] is the van’t Hoff factor
\[C\] is the molar concentration of the solute in the solution
\[\;R\] is the universal gas constant
\[\;T\] is the temperature
In van't Hoff theory describes that substances in dilute solution obey the ideal gas laws, resulting to the osmotic pressure formula \[\pi {\text{ }} = {\text{ }}\left( {n/V} \right)RT{\text{ }} = {\text{ }}\left[ {Ci} \right]RT\] where \[R\] is the gas constant,\[\;T\] the absolute temperature, and \[\left[ {Ci} \right]\] the molar concentration of solute i in dilute solution
\[\pi {\text{ }} = {\text{ }}\left( {\dfrac{n}{V}} \right)RT\]
\[n = {\text{ }}W/M\]
\[{\mathbf{\pi }}{\text{ }} = {\text{ }}\dfrac{{WRT}}{{MV}}\] $\therefore $ \[M\] = Molar mass of solute and \[W\] = Gram weight of solute
Now Putting the values in above equation -
Given = \[5\% \;\] solution of Glucose
\[5{\text{ }}gm\] of glucose in \[100{\text{ }}ml\] of solution
Now ,
The mass of solute \[w = {\text{ }}5g\]
Volume of solution \[V{\text{ }} = {\text{ }}100{\text{ }}ml{\text{ }} = {\text{ }}0.1{\text{ }}litre\]
Molar mass of solute \[M{\text{ }} = 180{\text{ }}gm/mol\]
Temperature = \[{25^o}C.{\text{ }} = {\text{ }}\left( {273{\text{ }} + 25} \right)\; = {\text{ }}298K\]
calculate the osmotic pressure
Osmotic pressure = \[{\mathbf{\pi }}{\text{ }} = {\text{ }}\dfrac{{WRT}}{{MV}}\]

\[\;\pi \] =\[\dfrac{5}{{180}} \times \dfrac{{0.0821}}{{0.1}} \times 298\]
\[\Rightarrow \;\pi \] = \[0.027{\text{ }} \times {\text{ }}244.65\] = \[6.605\]

Therefore the osmotic pressure is 6.605 Pa.

Note: Osmotic pressure shows a colligative property and is dependent on the concentration of solute particles in the solution. Colligative property depends on the no. of solute particles irrespective of their nature relative to the total no. of particles present in the solution.