An artificial vesicle containing a 1 M glucose solution is composed of a phospholipid bilayer lacking any protein components oth

Question

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An artificial vesicle containing a 1 M glucose solution is composed of a phospholipid bilayer lacking any protein components oth

er than aquaporin channels. Assuming an ideal solution, what is the ratio of the osmotic pressure measured immediately after immersion of the vesicle in de-ionized water to the osmotic pressure measured immediately after immersion of an identical vesicle containing the original volume of 1 M glucose solution added to an equal volume of 1 M KCl solution in deionized water

Chemistry

2 answers:

Alchen [17] · Answer 1 · 2022-03-19T19:01:35+03:00

Answer:

A: 0.67

The situation described in the question is analogous to a semipermeable membrane. Water is able to pass through aquaporin channels present in the liposome, but large uncharged particles (glucose) and ions (K+ and Cl -) are impermeable and will remain trapped within the liposome. If assumed to be ideal, the osmotic pressure, π, exerted by the solution due to molarity differences across the membrane is defined as π = iMRT, where i is the van't Hoff factor, M is the molarity of the solution, R is the universal gas constant, and T is the absolute temperature of the solution. A change in osmotic pressure at constant temperature is due to changes in iM, a term that is equivalent to the concentration of dissolved particles produced by solute in solution. When compared to the original volume of 1 M glucose, the new combined solution has twice the volume and three times the number of dissolved particles (1 M KCl, a strongly electrolytic solution, produces 1 M concentrations of both K+ and Cl- in solution), or an increase in the concentration of dissolved particles by a factor of 1.5. This is equivalent to a combined molarity of dissolved particles of 1.5 M. The ratio of osmotic pressure is then [1 M dissolved glucose] / [1.5 M dissolved glucose + KCl] = 0.67

Explanation:

AlladinOne [14] · Answer 2 · 2022-03-19T20:46:35+03:00

Answer:

A: 0.67

The situation described in the question is analogous to a semipermeable membrane. Water is able to pass through aquaporin channels present in the liposome, but large uncharged particles (glucose) and ions (K+ and Cl -) are impermeable and will remain trapped within the liposome. If assumed to be ideal, the osmotic pressure, π, exerted by the solution due to molarity differences across the membrane is defined as π = iMRT, where i is the van't Hoff factor, M is the molarity of the solution, R is the universal gas constant, and T is the absolute temperature of the solution. A change in osmotic pressure at constant temperature is due to changes in iM, a term that is equivalent to the concentration of dissolved particles produced by solute in solution. When compared to the original volume of 1 M glucose, the new combined solution has twice the volume and three times the number of dissolved particles (1 M KCl, a strongly electrolytic solution, produces 1 M concentrations of both K+ and Cl- in solution), or an increase in the concentration of dissolved particles by a factor of 1.5. This is equivalent to a combined molarity of dissolved particles of 1.5 M. The ratio of osmotic pressure is then [1 M dissolved glucose] / [1.5 M dissolved glucose + KCl] = 0.67

Explanation:

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