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Chapter 16: Problem 5

Elevation in the boiling point for 1 molal solution of glucose is \(2 \mathrm{~K}\). The depression in the freezing point for 2 molal solution of glucose in the same solvent is \(2 \mathrm{~K}\). The relation between \(\mathrm{K}_{\mathrm{b}}\) and \(\mathrm{K}_{\mathrm{f}}\) is: (a) \(\mathrm{K}_{\mathrm{b}}=1.5 \mathrm{~K}_{\mathrm{f}}\) (b) \(\mathrm{K}_{\mathrm{b}}=\mathrm{K}_{\mathrm{f}}\) (c) \(\mathrm{K}_{\mathrm{b}}=0.5 \mathrm{~K}_{\mathrm{f}}\) (d) \(\mathrm{K}_{\mathrm{b}}=2 \mathrm{~K}_{\mathrm{f}}\)

Short Answer

Expert verified
\( K_b = 2 K_f \), option (d).

Step by step solution

01

Understanding the Problem

The problem involves the colligative properties of solutions: boiling point elevation and freezing point depression. Given data include the boiling point elevation for a 1 molal solution and the freezing point depression for a 2 molal solution, both being 2 K, and we are to find the relation between the boiling point elevation constant \( K_b \) and the freezing point depression constant \( K_f \).
02

Boiling Point Elevation Equation

The boiling point elevation equation is given by \( \Delta T_b = K_b \cdot m \), where \( \Delta T_b = 2 \text{ K} \) and \( m = 1 \text{ molal} \). From this, we can calculate \( K_b \) as: \[ K_b = \frac{2}{1} = 2 \text{ K} \].
03

Freezing Point Depression Equation

The freezing point depression equation is given by \( \Delta T_f = K_f \cdot m \), where \( \Delta T_f = 2 \text{ K} \) and \( m = 2 \text{ molal} \). From this, we can calculate \( K_f \) as: \[ K_f = \frac{2}{2} = 1 \text{ K} \].
04

Relating \(K_b\) and \(K_f\)

Now, we know \( K_b = 2 \text{ K} \) and \( K_f = 1 \text{ K} \). To find the relation, we compare the two constants: \( K_b = 2 \cdot K_f \). Thus, \( K_b = 2 K_f \).
05

Conclusion

The correct relation between \( K_b \) and \( K_f \) is \( K_b = 2 K_f \), which matches option \( (d) \). Thus, option \( (d) \) is correct.

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Key Concepts

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

Boiling Point Elevation
Boiling point elevation is a fascinating concept that explains how the boiling point of a solvent increases when a solute is added. This is one of the colligative properties, which depend only on the number of solute particles in the solution, not the type of particles.
Let's break it down:
  • The increase in the boiling point is directly proportional to the molality of the solution, which is the number of moles of solute per kilogram of solvent.
  • The formula for boiling point elevation is \( \Delta T_b = K_b \cdot m \), where \( \Delta T_b \) represents the change in boiling point, \( K_b \) is the ebullioscopic constant specific to each solvent, and \( m \) is the molality.
In the exercise, a 1 molal glucose solution causes a 2 K elevation. Therefore, \( \Delta T_b = 2 \); and, as calculated, \( K_b = 2 \ \, \text{K} \). Understanding how the boiling point increases with solute addition helps us grasp real-world applications like in cooking, where salt in water raises the boiling point to cook food more efficiently.
Freezing Point Depression
Freezing point depression is another intriguing colligative property, closely related to boiling point elevation. It describes how the freezing point of a solvent decreases when a solute is dissolved in it.
The basic principles are:
  • The freezing point decreases in proportion to the molality of the solution.
  • Its formula is given by \( \Delta T_f = K_f \cdot m \), where \( \Delta T_f \) is the decrease in freezing point, \( K_f \) is the cryoscopic constant of the solvent, and \( m \) is the molality.
In the provided exercise, a 2 molal glucose solution lowers the freezing point by the same amount鈥�2 K. By applying the formula, we find \( K_f = 1 \ \, \text{K} \). This concept is often applied in real-life scenarios like in adding antifreeze to car radiators to lower the freezing point of the engine's coolant, thus preventing damage in cold weather.
Molality
Molality is a measure of the concentration of a solution. It is used specifically in colligative properties due to its temperature-independent nature. Let's delve into its specifics:
Molality refers to:
  • The number of moles of solute per kilogram of solvent, denoted by \( m \).
  • Unlike molarity, which involves volume and can vary with temperature, molality is unaffected by temperature changes, making it reliable for studies involving boiling point elevation and freezing point depression.
In the solved exercise, molality helps describe how the given concentration of glucose solutions affects the boiling and freezing points.Thus, molality serves as a crucial parameter in understanding colligative properties, allowing for predictable and quantifiable changes in physical properties of the solution just by dissolving more or fewer particles.

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Most popular questions from this chapter

What is the molarity and molality of a \(13 \%\) solution (by weight) of sulphuric acid with a density of \(1.02 \mathrm{~g} / \mathrm{mL} ?\) To what volume should \(100 \mathrm{~mL}\) of this acid be diluted in order to prepare a \(1.5 \mathrm{~N}\) solution?

The vapour pressure of pure benzene at a certain temperature is 640 \(\mathrm{mm}\) Hg. A non-volatile non-electrolyte solid weighing \(2.175 \mathrm{~g}\) is added to \(39.0 \mathrm{~g}\) of benzene. The vapour pressure of the solution is 600 \(\mathrm{mm} \mathrm{Hg}\). What is the molecular weight of the solid substance?

A set of solutions is prepared using \(180 \mathrm{~g}\) of water as a solvent and \(10 \mathrm{~g}\) of different non-volatile solutes \(\mathrm{A}, \mathrm{B}\) and \(\mathrm{C}\). The relative lowering of vapour pressure in the presence of these solutes are in the order [Given, molar mass of \(\mathrm{A}=100 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{B}=200 \mathrm{~g} \mathrm{~mol}^{-1} ; \mathrm{C}=10,000 \mathrm{~g}\) \(\left.\mathrm{mol}^{-1}\right]\) (a) \(\mathrm{B}>\mathrm{C}>\mathrm{A}\) (b) \(\mathrm{C}>\mathrm{B}>\mathrm{A}\) (c) \(\mathrm{A}>\mathrm{B}>\mathrm{C}\) (d) \(\mathrm{A}>\mathrm{C}>\mathrm{B}\)

Properties such as boiling point, freezing point and vapour pressure of a pure solvent change when solute molecules are added to get homogeneous solution. These are called colligative properties. Application of colligative properties are very useful in day-to-day life. One of its example is the use of ethylene glycol and water mixture as anti-freezing liquid in the radiator of automobiles. A solution \(\mathrm{M}\) is prepared by mixing ethanol and water. The mole fraction of ethanol in the mixture is \(0.9\) Given : Freezing point depression constant of water \(\left(K_{f}^{\text {water }}\right)\) $$ =1.86 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$ Freezing point depression constant of ethanol ( \(\left.K_{f}{ }^{\text {ethanol }}\right)\) $$ =2.0 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1} $$ Boiling point elevation constant of water \(\left(K_{b}^{\text {water }}\right)\) \(=0.52 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) Boiling point elevation constant of ethanol \(\left(K_{b}^{\text {ethanol }}\right)=1.2 \mathrm{~K} \mathrm{~kg} \mathrm{~mol}^{-1}\) Standard freezing point of water \(=273 \mathrm{~K}\) Standard freezing point of ethanol \(=155.7 \mathrm{~K}\) Standard boiling point of water \(=373 \mathrm{~K}\) Standard boiling point of ethanol \(=351.5 \mathrm{~K}\) Vapour pressure of pure water \(=32.8 \mathrm{~mm} \mathrm{Hg}\) Vapour pressure of pure ethanol \(=40 \mathrm{~mm} \mathrm{Hg}\) Molecular weight of water \(=18 \mathrm{~g} \mathrm{~mol}^{-1}\) Molecular weight of ethanol \(=46 \mathrm{~g} \mathrm{~mol}^{-1}\) In answering the following questions, consider the solution to be ideal dilute solutions and solutes to be non-volatile and non-dissociative. The freezing point of the solution \(\mathrm{M}\) is (a) \(268.7 \mathrm{~K}\) (b) \(268.5 \mathrm{~K}\) (c) \(234.2 \mathrm{~K}\) (d) \(150.9 \mathrm{~K}\)

The freezing point of equimolal aqueous solutions will be highest for : (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{NH}_{3} \mathrm{Cl}\) (aniline hydrochloride) (b) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) (c) \(\mathrm{La}\left(\mathrm{NO}_{3}\right)_{3}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (glucose)
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