0.10 M glucose and 0.10 M K C l are separated by a semi-permeable membrane and the direction of net flow of water has to be predicted. Concept introduction: The passage of solvent molecule occurs across the semi-permeable membrane due to the difference in osmotic pressure of solution and pure solvent. Osmotic pressure is directly proportional to the concentration of the solute. That is, whenever there is a concentration gradient setup between the semi-permeable membrane, the solvent will move from its higher concentration to the lower.

Question

Want to see the full answer?

Answer 1

Question

Chapter 6, Problem 6.80QP

Interpretation Introduction

Interpretation:

0.10 M glucose and 0.10 M KCl are separated by a semi-permeable membrane and the direction of net flow of water has to be predicted.

Concept introduction:

The passage of solvent molecule occurs across the semi-permeable membrane due to the difference in osmotic pressure of solution and pure solvent. Osmotic pressure is directly proportional to the concentration of the solute. That is, whenever there is a concentration gradient setup between the semi-permeable membrane, the solvent will move from its higher concentration to the lower.

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Students have asked these similar questions

Identifying the major species in weak acid or weak base equilibria The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that HF is a weak acid. acids: 0.2 mol of KOH is added to 1.0 L of a 0.5 M HF solution. bases: Х other: ☐ acids: 0.10 mol of HI is added to 1.0 L of a solution that is 1.4M in both HF and NaF. bases: other: ☐ 0,0,... ด ? 18 Ar

Identifying the major species in weak acid or weak base equilibria The preparations of two aqueous solutions are described in the table below. For each solution, write the chemical formulas of the major species present at equilibrium. You can leave out water itself. Write the chemical formulas of the species that will act as acids in the 'acids' row, the formulas of the species that will act as bases in the 'bases' row, and the formulas of the species that will act as neither acids nor bases in the 'other' row. You will find it useful to keep in mind that NH3 is a weak base. acids: ☐ 1.8 mol of HCl is added to 1.0 L of a 1.0M NH3 bases: ☐ solution. other: ☐ 0.18 mol of HNO3 is added to 1.0 L of a solution that is 1.4M in both NH3 and NH₁Br. acids: bases: ☐ other: ☐ 0,0,... ? 000 18 Ar B 1

Using reaction free energy to predict equilibrium composition Consider the following equilibrium: 2NH3 (g) = N2 (g) +3H₂ —N2 (g) AGº = 34. kJ Now suppose a reaction vessel is filled with 4.19 atm of ammonia (NH3) and 9.94 atm of nitrogen (N2) at 378. °C. Answer the following questions about this system: rise Under these conditions, will the pressure of NH 3 tend to rise or fall? ☐ x10 fall Х Is it possible to reverse this tendency by adding H₂? In other words, if you said the pressure of NH 3 will tend to rise, can that be changed to a tendency to fall by adding H₂? Similarly, if you said the pressure of NH3 will tend to fall, can that be changed to a tendency to rise by adding H₂? If you said the tendency can be reversed in the second question, calculate the minimum pressure of H₂ needed to reverse it. Round your answer to 2 significant digits. yes no atm 00. 18 Ar 무ㅎ ?