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Which one of the following aqueous solutions will exhibit highest boiling point: (a) \(0.015 \mathrm{M}\) urea (b) \(0.01 \mathrm{M} \mathrm{KNO}_{3}\) (c) \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\) (d) \(0.015 \mathrm{M}\) glucose

Short Answer

Expert verified
The 0.10 M Na2SO4 solution will exhibit the highest boiling point as it produces more particles in the solution, leading to a greater boiling point elevation.

Step by step solution

01

Understand Boiling Point Elevation

The boiling point of an aqueous solution depends on the number of solute particles present in the solution because it is a colligative property. The more solute particles, the higher the boiling point will be due to the elevation of the boiling point, known as boiling point elevation.
02

Determine the Number of Particles in Solution

We need to determine the number of particles each solute produces in solution. Urea (NH2CONH2) and glucose (C6H12O6) are non-electrolytes, so they will not dissociate in solution and each will remain as one particle per formula unit. KNO3 dissociates into K+ and NO3-, two particles per formula unit. Na2SO4 dissociates into 2Na+ and SO4^2-, which yields three particles per formula unit.
03

Calculate the Total Number of Particles for Each Solution

For 0.015 M urea: total particles = 0.015 M * 1 particle/unit = 0.015. For 0.015 M glucose: total particles = 0.015 M * 1 particle/unit = 0.015. For 0.01 M KNO3: total particles = 0.01 M * 2 particles/unit = 0.02. For 0.10 M Na2SO4: total particles = 0.10 M * 3 particles/unit = 0.30.
04

Determine the Solution with the Highest Number of Particles

Comparing the total number of particles for each solution, we find that the solution of 0.10 M Na2SO4 has the highest number of particles per liter of solution.
05

Identify the Solution with the Highest Boiling Point

Since the number of particles determines the elevation in boiling point, the solution with the highest number of particles will exhibit the highest boiling point. Therefore, the 0.10 M Na2SO4 solution will have the highest boiling point among the given options.

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Key Concepts

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

Colligative Properties
Colligative properties are unique characteristics of solutions that depend solely on the number of solute particles in a given amount of solvent, regardless of the identity of the solute particles. Boiling point elevation is one such property, along with freezing point depression and osmotic pressure.

When a non-volatile solute is dissolved in a solvent, it disrupts the solvent's ability to evaporate by getting in between solvent molecules. This disruption requires additional energy to convert the solvent into a gas at the surface, thus raising the boiling point of the solution relative to the pure solvent. This is a direct result of the presence of solute particles, which is why colligative properties are such an important concept in chemistry, especially when discussing boiling point elevation.

Additionally, the effect on the boiling point is also influenced by the type of solute particles. This is where dissociation and non-dissociation of solutes come into play, affecting the extent of boiling point elevation.
Solution Concentration
The concentration of a solution is a measure of how much solute is dissolved in a given amount of solvent. It is often expressed in molarity (M), which is a ratio of moles of solute to liters of solution. The concentration plays a pivotal role in determining the extent of boiling point elevation.

The higher the concentration of solute particles, the more pronounced the colligative properties. In the context of boiling point elevation, the solution with a higher concentration of solute particles will have a higher boiling point because more energy is needed to allow the solvent to escape into a gaseous state. This fundamental concept helps us understand the outcome of the original exercise: a comparison of boiling points among different solutions.
Dissociation of Electrolytes
Electrolytes are substances that conduct electricity in solution due to the movement of ions. When an electrolyte dissolves in water, it dissociates into its constituent ions. For example, sodium chloride (NaCl) dissociates into Na+ and Cl- ions.

In the given exercise, KNO3 and Na2SO4 are examples of electrolytes that dissociate into multiple particles when dissolved. The number of ions produced is crucial because it affects the total number of solute particles in solution. This, in turn, impacts the boiling point elevation according to the colligative properties principle. When there are more dissolved particles, the greater the effect on the boiling point, explaining why a solution like Na2SO4 with a greater extent of dissociation might raise the boiling point more significantly than non-electrolytes or weak electrolytes with fewer particles in solution.
Non-electrolytes
Non-electrolytes, contrastingly, are substances that do not produce ions when dissolved in water. These include molecular compounds such as urea (NH2CONH2) and glucose (C6H12O6). These substances remain intact as molecules in solution, contributing relatively fewer particles compared to electrolytes that dissociate.

In terms of boiling point elevation, this means that solutions of non-electrolytes typically exhibit less of an increase in boiling point compared to solutions of electrolytes with the same molar concentration. This is because the number of solute particles is directly responsible for the boiling point elevation, and non-electrolytes contribute only one particle per formula unit, in contrast to electrolytes which can produce multiple particles per formula unit.

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Most popular questions from this chapter

An azeotropic mixture of two liquids has a boiling point higher than either of them when it: (a) shows positive deviation from Raoult's law (b) shows negative deviation from Raoult's law (c) shows ideal behaviour (d) is saturated

If \(1 \mathrm{~m}\) solution of benzoic acid in benzene has a freezing point depression of \(2.4^{\circ} \mathrm{C}\). \(\left(K_{f}=5.12^{\circ} \mathrm{C} \mathrm{mol}^{-1} \mathrm{~kg}\right)\) and boiling point elevation of \(2.53^{\circ} \mathrm{C}\left(K_{b}=2.53^{\circ} \mathrm{C} \mathrm{mol}^{-1} \mathrm{~kg}\right)\) then select the correct statement/s : Statement I : there is dimer formation when undergoing freezing Statement II : there is no change when undergoing boiling Statement III : reverse of I and II Statement IV : dimer formation in freezing and boiling state (a) I, II (b) II, III (c) III, I (d) only I

The osmotic pressures of \(0.010 \mathrm{M}\) solutions of \(\mathrm{KI}\) and of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) are \(0.432\) atm and \(0.24\) atm respectively. The van't Hoff factor for KI is : (a) \(1.80\) (b) \(0.80\) (c) \(1.2\) (d) \(1.0\)

One molal solution of a carboxylic acid in benzene shows the elevation of boiling point of \(1.518 \mathrm{~K}\). The degree of association for dimerization of the acid in benzene is \(\left(K_{b}\right.\) for benzene \(\left.=2.53 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\right):\) (a) \(60 \%\) (b) \(70 \%\) (c) \(75 \%\) (d) \(80 \%\)

The freezing point of a \(4 \%\) aqueous solution of ' \(A\) ' is equal to the freezing point of \(10 \%\) aqueous solution of ' \(B\) '. If the molecular weight of ' \(A\) ' is 60 , then the molecular weight of ' \(B\) ' will be: (a) 160 (b) 90 (c) 45 (d) 180

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