Question

Calculate the osmotic pressure by a solution prepared by dissolving 1.0g of polymer of molar mass 185,000 in450 ml of water at 37°c

Answer 1

Solution :-

Given :-

Weight of polymer = 1g

Mw of polymer = 185,000g/mol

Volume of solution = 450ml

Temperature = 37°C

To Find :-

Osmotic pressure of the solution

Concept :-

▪ The net spontaneous flow of the solvent molecules from the solvent to the solution or from a less concentrated solution to a more concentrated solution through a semipermeable membrane is called osmosis.

▪ The minimum excess pressure that has to be applied on the solution to prevent the entry of the solvent into the solution through membrane is called the osmotic pressure.

Calculation :-

$\implies\bf\:No.\:of\:moles\:of\:polymer=\dfrac{1}{185000}\\ \\ \implies\sf\:\pi=CRT=\dfrac{n}{V}RT\\ \\ \implies\sf\:\pi=\dfrac{1}{185000}\times \dfrac{8.314\times 10^3\times 310}{0.45}\\ \\ \dag\:{\blue{\bf{R=8.314\times\:10^3\:Pa\:LK^{-1}mol^{-1}}}}\\ \\ \implies\underline{\boxed{\bf{\purple{\pi=30.95\:Pa}}}}$