Question

Calculate the molecular weight of cellulose acetate if it's $0 - 2\% {\text{ }}\left( {wt.{\text{ }}/vol.} \right)$ solution in acetone (sp. gr. 0-8) shows an osmotic rise of 23.1 mm against pure acetone at ${27^o}C$

Answer 1

The osmotic pressure is the minimum pressure needed to avoid the inward passage of a solution's pure solvent over a semipermeable membrane. It can also be characterised as a measurement of a solution's proclivity to absorb a pure solvent through osmosis. The maximum osmotic pressure that could occur in a solution if it is isolated from its pure solvent by a semipermeable membrane is known as potential osmotic pressure.

Complete answer: The minimum pressure that must be applied to a solution to stop the passage of liquid molecules across a semipermeable membrane is known as osmotic pressure (osmosis). It is a colligative property that is influenced by the solute particle concentration in the solution. The following formula can be used to measure osmotic pressure: $\pi {\text{ }} = {\text{ }}iCRT$
A 0.2 $\%$ solution is one of which 0.2 gram of cellulose acetate are dissolved in 100 millilitres of water. 2.31 cm acetone osmotic pressure
${\text{P}} = 2.31 \times \dfrac{{0.80}}{{13.6}}\;{\text{cm of Hg}}$= 0.136g of Hg
${\text{P}} = \dfrac{{0.136}}{{76}}\;{\text{atm}}$
Assume M is the molecular weight of cellulose acetate.
${\mathbf{n}} = \dfrac{{0.2}}{{\text{M}}}$
V = 100 ml = 0.1 L
$R{\text{ }} = {\text{ }}0.0821{\text{ }}L{\text{ }}atm{\text{ }}mo{l^{ - 1}}{K^{ - 1}}$
T = 273 + 27 = 300 K
${\text{P}} = \dfrac{{{\text{nRT}}}}{{\text{V}}}$
$\dfrac{{0.136}}{{76}} = \dfrac{{\dfrac{{0.2}}{M}}}{{0.1}} \times 0.0821 \times 300$
M = 27500

Note:
The passage of solvent molecules across a semipermeable membrane from a region with low solute concentration to a region with high solute concentration is referred to as osmosis. The two sides of the semipermeable membrane eventually reach a state of equilibrium (equal solute concentration on both sides of the semipermeable membrane).