
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
506.1k+ views
Hint: 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).
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).
Recently Updated Pages
Why are manures considered better than fertilizers class 11 biology CBSE

Find the coordinates of the midpoint of the line segment class 11 maths CBSE

Distinguish between static friction limiting friction class 11 physics CBSE

The Chairman of the constituent Assembly was A Jawaharlal class 11 social science CBSE

The first National Commission on Labour NCL submitted class 11 social science CBSE

Number of all subshell of n + l 7 is A 4 B 5 C 6 D class 11 chemistry CBSE

Trending doubts
Differentiate between an exothermic and an endothermic class 11 chemistry CBSE

10 examples of friction in our daily life

One Metric ton is equal to kg A 10000 B 1000 C 100 class 11 physics CBSE

Difference Between Prokaryotic Cells and Eukaryotic Cells

1 Quintal is equal to a 110 kg b 10 kg c 100kg d 1000 class 11 physics CBSE

State the laws of reflection of light

