Problem 54
Question
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
Step-by-Step Solution
Verified Answer
Option (b) 1, 2, and 4 are isotonic.
1Step 1: 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.
2Step 2: 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 \(2 \times 0.1 = 0.2 \mathrm{M}\).
3Step 3: 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 \(0.15 \mathrm{M}\).
4Step 4: 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.
Key Concepts
Osmotic PressureIonic DissociationMolar ConcentrationSolute Particles
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.
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 Ca²⁺ and two Cl⁻.
- MgSO₄ dissociates into two ions: Mg²⁺ and SO₄²⁻.
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.
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., CaCl₂ dissociates into three particles).
Other exercises in this chapter
Problem 51
The vapours pressure of water at \(23^{\circ} \mathrm{C}\) is \(19.8 \mathrm{~mm}\). of Hg. \(0.1\) mole of glucose is dissolved in \(178.2 \mathrm{~g}\) of wat
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\(0.5 \mathrm{M}\) of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) is diluted from 1 litre to 10 litre, normaliy of the resulting solution is (a) \(1 \mathrm{~N}\) (b) \(
View solution Problem 56
\(50 \mathrm{~mL}\) of \(10 \mathrm{~N} \mathrm{H}_{2} \mathrm{SO}_{4}, 25 \mathrm{~mL}\) of \(12 \mathrm{~N} \mathrm{HCl}\) and \(40 \mathrm{~mL}\) of \(5 \mat
View solution Problem 58
The elevation in boiling point for \(13.44 \mathrm{~g}\) of \(\mathrm{CuCl}_{2}\) dissolved in \(1 \mathrm{~kg}\) of water as solvent will be \(\left(\mathrm{K}
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