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Which one of the following aqueous solutions will exhibit highest boiling point? (a) \(0.05 \mathrm{M}\) glucose (b) \(0.01 \mathrm{M} \mathrm{KNO}_{3}\) (c) \(0.015 \mathrm{M}\) urea (d) \(0.01 \mathrm{M} \mathrm{Na}_{2} \mathrm{SO}_{4}\)

Short Answer

Expert verified
The 0.05 M glucose solution has the highest boiling point elevation.

Step by step solution

01

Understand Boiling Point Elevation

The boiling point of a solution increases when a solute is dissolved in it, a phenomenon known as boiling point elevation. The increase in boiling point is directly proportional to the van't Hoff factor and the molality of the solution: \[\Delta T_b = i \cdot K_b \cdot m\]where \(\Delta T_b\) is the change in boiling point, \(i\) is the van't Hoff factor, \(K_b\) is the ebullioscopic constant, and \(m\) is the molality of the solution.
02

Identify Van't Hoff Factors

Determine the van't Hoff factor \(i\) for each solute:- Glucose: \(i = 1\) because it is a non-electrolyte.- KNO鈧: \(i = 2\) because it dissociates into two ions: K鈦 and NO鈧冣伝.- Urea: \(i = 1\) because it is a non-electrolyte.- Na鈧係O鈧: \(i = 3\) because it dissociates into three ions: 2 Na鈦 and SO鈧劼测伝.
03

Calculate Molality for Each Solution

For each solution, the molality \(m\) is given as follows (since they already list the molarity, we will approximate that it's equal to molality here due to typically low concentration in dilute solutions):- Glucose: \(m = 0.05\, \text{mol/kg}\)- KNO鈧: \(m = 0.01\, \text{mol/kg}\)- Urea: \(m = 0.015\, \text{mol/kg}\)- Na鈧係O鈧: \(m = 0.01\, \text{mol/kg}\)
04

Calculate Boiling Point Elevation for Each Solution

Use the formula \(\Delta T_b = i \cdot m\) to calculate the boiling point elevation for each solution:- Glucose: \(\Delta T_b = 1 \times 0.05 = 0.05\)- KNO鈧: \(\Delta T_b = 2 \times 0.01 = 0.02\)- Urea: \(\Delta T_b = 1 \times 0.015 = 0.015\)- Na鈧係O鈧: \(\Delta T_b = 3 \times 0.01 = 0.03\)
05

Determine Highest Boiling Point Elevation

Compare the boiling point elevation calculated for each solution:- Glucose: \(0.05\)- KNO鈧: \(0.02\)- Urea: \(0.015\)- Na鈧係O鈧: \(0.03\)The highest increase in boiling point is with glucose at \(0.05\).

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Key Concepts

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

van't Hoff factor
The van鈥檛 Hoff factor, often represented by the symbol \(i\), is a key component in the study of collegiate chemistry, particularly when examining colligative properties such as boiling point elevation. This factor relates to how many particles a compound turns into when dissolved in a solution.

For non-electrolytes like glucose and urea, \(i = 1\), since they do not dissociate into ions. On the other hand, electrolytes like potassium nitrate (KNO鈧) and sodium sulfate (Na鈧係O鈧) dissociate into multiple ions. For KNO鈧, \(i = 2\) because it dissociates into two ions: K鈦 and NO鈧冣伝. Meanwhile, Na鈧係O鈧 splits into three ions: 2 Na鈦 and SO鈧劼测伝, giving it a van鈥檛 Hoff factor of \(i = 3\).

Understanding the van鈥檛 Hoff factor is essential because it directly impacts the magnitude of colligative properties. Higher values of \(i\) indicate more particles, which leads to greater changes in properties like boiling point elevation.
molality
In collegiate chemistry, molality is a crucial concept when discussing solutions and their properties. Molality (\(m\)) is a measure of the concentration of solute in a solution and is defined as the number of moles of solute per kilogram of solvent.

Unlike molarity, which depends on the volume of the solution, molality is dependent only on the mass of the solvent. This makes it particularly useful in colligative property calculations, like boiling point elevation, because it remains unaffected by changes in temperature or pressure. Given that a difference in temperature can influence volume, using molality provides a more stable basis for calculating changes like boiling point elevations.

In the provided exercise, you can notice that molality was approximated to be equal to molarity due to the assumption of dilute solutions, where the impact of the minor differences often can be negligible.
collegiate chemistry
Collegiate chemistry involves the study of complex concepts and principles that help explain the physical and chemical properties of substances and mixtures. Topics like boiling point elevation, van鈥檛 Hoff factor, and molality are fundamental to understanding the nature of solutions.

These concepts are part of what we call 'colligative properties'. Colligative properties depend on the number of solute particles in a solution rather than the type of particles. This means regardless of whether the solute is a sugar or salt, its ability to affect boiling points hinges on how many particles result from its dissolution.

Understanding these properties helps students predict important behaviors of solutions. It becomes clear why solutions like Na鈧係O鈧, with more particles released upon dissolution, might be expected to cause a greater boiling point elevation, even when its molality is the same as simpler non-dissociating solutes such as glucose.

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Most popular questions from this chapter

Which one of the statements given below concerning properties of solutions, describe a colligative effect? (a) boiling point of pure water decreases by the addition of ethanol (b) vapour pressure of pure water decreases by the addition of nitric acid (c) vapour pressure of pure benzene decreases by the addition of naphthalene (d) boiling point of pure benzene increases by the addition of toluene

During depression in freezing point in a solution, the following are in equilibrium (a) liquid solvent, solid solvent (b) liquid solvent, solid solute (c) liquid solute, solid solute (d) liquid solute, solid solvent

Ethylene glycol is used as antifreeze in a cold climate. Mass of ethylene glycol which should be added to \(4 \mathrm{~kg}\) of water to prevent it form freezing at \(-6^{\circ} \mathrm{C}\) will be: \(\left(\mathrm{K}_{f}\right.\) for water \(=1.86 \mathrm{~kg} \mathrm{~mol}^{-1}\), and molar mass of ethylene glycol \(=62 \mathrm{~g} \mathrm{~mol}^{-1}\) ) (a) \(204.11 \mathrm{~g}\) (b) \(804.32 \mathrm{~g}\) (c) \(600.20 \mathrm{~g}\) (d) \(302.40 \mathrm{~g}\)

If liquids A and B form an ideal solution, the [2003] (a) enthalpy of mixing is zero(b) entropy of mixing is zero (c) free energy of mixing is zero (d) free energy as well as the entropy of mixing are each zero

A \(0.2\) molal aqueous solution of a weak acid \((\mathrm{HX})\) is \(20 \%\) ionized. The freezing point of this solution is (Given \(\mathrm{K}_{\mathrm{f}}=1.86^{\circ} \mathrm{C} \mathrm{kg} \mathrm{mol}^{-1}\) for water \()\) (a) \(-0.45^{\circ} \mathrm{C}\) (b) \(-0.90^{\circ} \mathrm{C}\) (c) \(-0.21^{\circ} \mathrm{C}\) (d) \(-0.43^{\circ} \mathrm{C}\)

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