Question

Calculate the osmotic pressure in pascals exerted by a solution prepared by dissolving 1.0 gram of polymer of molar mass 185000 in 450 ml of water at 37$^\circ $C.

Answer 1

Hint: The expression to find the osmotic pressure is given below:
$\pi = icRT$
Where $\pi $ is the osmotic pressure, $i$ is the Van’t Hoff index, $c$ is the concentration of the solution, $R$ is the gas constant, and $T$ is the temperature of the system.

Complete step by step answer:
Before heading directly towards solving the numerical part, we first need to understand what is an osmotic pressure. Osmotic pressure is one of the colligative properties of the solution. It is the minimum pressure required for the solution to avoid the inward flow of the solvent through a semipermeable membrane by the process of osmosis. Osmosis is the process in which the particles from the low concentration solution go towards the solution of high concentration through a semipermeable membrane which is generally made up of parchment paper, cellophane, etc.
The expression of osmotic pressure is given below:
$\pi = icRT$
Where $\pi $ is the osmotic pressure, $i$ is the van’t Hoff index, $c$ is the concentration of the solution, $R$ is the gas constant, and $T$ is the temperature of the system. Further we can write this equation as:
$\pi = \dfrac{n}{v}RT$
Where concentration (molarity) is written as number of moles $(n)$ divided by the volume of the solution $(v)$
In the question statement we are given:
Molar mass of polymer: \[M{{ }} = {{ }}185000{{ }}g\]
Volume: \[v{{ }} = {{ }}450{{ }}ml{{ }} = {{ }}0.45{{ }}Litres\]
Temperature: \[T = 37^\circ C\]
Given mass: \[m{{ }} = {{ }}1{{ }}gram\]
Firstly, we will find the number of moles which are not directly given in the question statement by using the formula:
\[N = \dfrac{m}{M}\]
Here, N is the number of moles present in the sample, m is the given mass, M is the molar mass of the polymer.
Thus, number of moles: $n = \dfrac{1}{{185000}}$
Now by using the formula for osmotic pressure:
\[\pi = \dfrac{1}{{185000}} \times \dfrac{1}{{0.45}} \times 8.314 \times {10^3} \times (37 + 273)K\]
\[\pi = 30.98Pa\]

Note: R here is the gas constant. It relates the energy scale to the temperature scale and its value is given by \[8.31446261815324{{ }}J \cdot {K^{ - 1}} \cdot mo{l^{ - 1}}\]. The artificial semipermeable membrane is made up of cellophane and parchment paper but the natural or biological semipermeable membrane is made up of lipid bilayer.