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Chapter 5: Problem 54

Which of the following solutions are isotonic with respect to one another? 1\. \(0.15 \mathrm{M}\) urea 2\. \(0.05 \mathrm{M} \mathrm{CaCl}_{2}\) 3\. \(0.1 \mathrm{M} \mathrm{MgSO}_{4}\) 4\. \(0.15 \mathrm{M}\) glucose Select the correct answer using the codes given below: (a) 1 and 4 (b) 1,2 and 4 (c) 2 and 3 (d) 2,3 and 4

Short Answer

Expert verified
Option (b) 1, 2, and 4 are isotonic.

Step by step solution

01

Understand Isotonic Solutions

Isotonic solutions have the same osmotic pressure, which means they have the same concentration of solute particles in solution. This is achieved when their molar concentrations of solute particles are equal after considering dissociation.
02

Consider the Dissociation of Compounds

To determine isotonicity, consider the dissociation of ionic compounds. \( \mathrm{CaCl}_{2} \) dissociates into three ions (1 \(\mathrm{Ca}^{2+}\) and 2 \(\mathrm{Cl}^{-}\)), so a \(0.05 \mathrm{M} \mathrm{CaCl}_{2}\) solution has an effective concentration of \(3 \times 0.05 = 0.15 \mathrm{M}\). \( \mathrm{MgSO}_{4} \) dissociates into two ions (\(\mathrm{Mg}^{2+}\) and \(\mathrm{SO}_4^{2-}\)), resulting in a \(0.1 \mathrm{M} \mathrm{MgSO}_{4}\) solution having an effective concentration of \(2 \times 0.1 = 0.2 \mathrm{M}\).
03

Calculate Effective Concentrations for Molecular Solutes

Molecular solutes like urea and glucose don't dissociate into ions. Therefore, a \(0.15 \mathrm{M}\) solution of urea and a \(0.15 \mathrm{M}\) solution of glucose both have effective concentrations of \(0.15 \mathrm{M}\).
04

Compare Effective Concentrations for Isotonicity

Compare the effective concentrations: Urea (0.15 M), CaCl2 (0.15 M), MgSO4 (0.2 M), and glucose (0.15 M). The solutions with the same effective concentrations are urea, CaCl2, and glucose, all at 0.15 M, making them isotonic with each other.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Osmotic Pressure
Osmotic pressure is a fundamental concept when understanding isotonic solutions. It refers to the pressure required to stop the flow of water across a semipermeable membrane. When two solutions have the same osmotic pressure, they are considered isotonic, meaning there is no net movement of water across the membrane between them.
  • Osmotic pressure depends on the number of solute particles in a solution rather than their identity.
  • Solutions with higher concentrations of solute particles will have higher osmotic pressures.
  • Isotonic solutions are essential in medical applications, such as intravenous solutions, to prevent cell shrinkage or swelling.
Once you understand osmotic pressure, you will find it easier to grasp why isotonic solutions are important and how they behave in biological systems.
Ionic Dissociation
Ionic dissociation is the process of breaking down ionic compounds into their individual ions when dissolved in water. This process is crucial for determining the effective concentration of solutes in a solution.
  • Ionic compounds like CaCl2 and MgSO4 dissociate into multiple ions, thereby affecting osmotic pressure.
  • CaCl2 dissociates into three ions: one Ca虏鈦� and two Cl鈦�.
  • MgSO鈧� dissociates into two ions: Mg虏鈦� and SO鈧劼测伝.
Considering ionic dissociation helps in calculating the total concentration of solute particles when dealing with solutions that contain ionic compounds. This calculation is vital for comparing solutions to determine isotonicity.
Molar Concentration
Molar concentration, or molarity, is the amount of solute (in moles) present in one liter of solution. This is an essential measurement when calculating isotonic solutions, as it allows us to determine the number of solute particles present.
  • Molarity is expressed in moles per liter (M).
  • For molecular solutes like urea and glucose, the molarity directly represents the number of solute particles because they do not dissociate.
  • For ionic compounds, the effective molarity is greater due to dissociation into multiple ions.
By understanding molarity, you will be more prepared to calculate effective concentrations and compare them for isotonicity.
Solute Particles
Solute particles are the individual units in a solution that contribute to osmotic pressure. These particles can be ions or molecules, depending on whether the solute is ionic or molecular.
  • The total number of solute particles in a solution determines its effective concentration.
  • Molecular solutes contribute one particle per molecule (e.g., urea, glucose).
  • Ionic solutes contribute multiple particles per formula unit due to dissociation (e.g., CaCl鈧� dissociates into three particles).
Understanding solute particles and their effects on osmotic pressure is vital for determining if solutions are isotonic, as it helps quantify how many particles are actually present in the solution.

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Most popular questions from this chapter

The vapour pressure of two liquids ' \(\mathrm{P}\) ' and ' \(\mathrm{Q}\) ' are 80 and 60 torr respectively. The total vapour pressure of solution obtained by mixing 3 mole of \(\mathrm{P}\) and \(2 \mathrm{~mol}\) of Q would be (a) 20 torr (b) 72 torr (c) 68 torr (d) 140 torr

Benzoic acid undergoes dimerization in benzene solution, the van't Hoff factor 'i' is related to the degree of association ' \(\mathrm{x}\) ' of to the acid as (a) \(\mathrm{i}=(1+\mathrm{x})\) (b) \(\mathrm{i}=(1-\mathrm{x})\) (c) \(\mathrm{i}=(1-\mathrm{x} / 2)\) (d) \(\mathrm{i}=(1+\mathrm{x} / 2)\)

Which of the following is correct for a solution showing positive deviations from Raoult's law? (a) \(\Delta \mathrm{V}=+\mathrm{ve}, \Delta \mathrm{H}=+\mathrm{ve}\) (b) \(\Delta \mathrm{V}=-\mathrm{ve}, \Delta \mathrm{H}=+\mathrm{ve}\) (c) \(\Delta \mathrm{V}=+\mathrm{ve}, \Delta \mathrm{H}=-\mathrm{ve}\) (d) \(\Delta \mathrm{V}=-\mathrm{ve}, \Delta \mathrm{H}=-\mathrm{ve}\)

A binary liquid solution is prepared by mixing \(n\)-heptane and ethanol. Which on of the following statement is correct regarding the behavior of the solution? (a) The solution in non-ideal, showing +ve deviation from Raoult's Law. (b) The solution in non-ideal, showing \(-\) ve deviation from Raoult's Law. (c) n-heptane shows tre deviation while ethanol shows -ve deviation from Raoult's Law. (d) The solution formed is an ideal solution.

The rise in the boiling point of a solution containing \(1.8 \mathrm{~g}\) of glucose in \(100 \mathrm{~g}\) of solvent is \(0.1^{\circ} \mathrm{C}\). The molal elevation constant of the liquid is (a) \(1 \mathrm{~K} / \mathrm{m}\) (b) \(0.1 \mathrm{~K} / \mathrm{m}\) (c) \(0.01 \mathrm{~K} / \mathrm{m}\) (d) \(10 \mathrm{~K} / \mathrm{m}\)
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