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The freezing point of equimolal aqueous solution will be highest for (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}^{+} \mathrm{H}_{3} \mathrm{Cl}^{-}\)(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)

Short Answer

Expert verified
The highest freezing point is for glucose (d) \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \).

Step by step solution

01

Understanding Depression of Freezing Point

The freezing point of a solution decreases from that of the pure solvent. For solutions, this depression in freezing point is dependent on the number of solute particles, given by the formula: \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor, \( K_f \) is the freezing point depression constant, and \( m \) is the molality. Thus, the higher the \( i \), the greater the freezing point depression and the lower the freezing point of the solution.
02

Calculating van't Hoff Factor

The van't Hoff factor \( i \) is determined by the number of ions each solute dissociates into: - (a) Aniline hydrochloride \( \mathrm{C}_{6} \mathrm{H}_{5} \mathrm{N}^{+} \mathrm{H}_{3} \mathrm{Cl}^{-} \): \( i = 2 \) - (b) \( \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2} \): \( i = 3 \) - (c) \( \mathrm{La}\left(\mathrm{NO}_{3}\right)_{3} \): \( i = 4 \) - (d) Glucose \( \mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6} \): \( i = 1 \), as it does not dissociate.
03

Identifying Solution with Highest Freezing Point

The solution with the lowest depression in freezing point (highest freezing point) will have the smallest \( i \). Among the options, glucose (d) with \( i = 1 \) undergoes no dissociation, leading to the least lowering of the freezing point.

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Key Concepts

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

Freezing Point Depression
Freezing point depression is a fascinating colligative property that occurs when a solute is added to a solvent. In simpler terms, it means that the solution will freeze at a lower temperature than the pure solvent alone. The phenomenon can be explained with the formula: \[ \Delta T_f = i \cdot K_f \cdot m \] where \( \Delta T_f \) is the change in the freezing point, \( i \) is the van't Hoff factor, \( K_f \) is the freezing point depression constant of the solvent, and \( m \) is the molality of the solution.To understand this further, let's break it down:
  • The van't Hoff factor \( i \): This denotes the number of particles the solute produces in solution. More particles mean a greater effect on freezing point depression.
  • The constant \( K_f \): This is specific to each solvent and indicates how much the freezing point depresses with the addition of solute. Pure water, for example, has a specific \( K_f \).
  • Molality \( m \): This refers to moles of solute per kilogram of solvent. More solute generally means more depression of the freezing point.
Understanding freezing point depression helps chemists predict the degree of cooling required for a solution to freeze.
Van't Hoff Factor
The van't Hoff factor, labeled as \( i \), plays a crucial role in determining the extent of colligative properties like freezing point depression. It tells us the number of individual particles formed in a solution from the dissociation of solute molecules. Knowing \( i \) allows predictions about how much a solution's freezing point will drop. The concept becomes even clearer with an example:
  • Consider **calcium nitrate**, \( \text{Ca(NO}_3\text{)}_2 \). It dissociates into 3 ions: 1 calcium ion \( \text{Ca}^{2+} \) and 2 nitrate ions \( \text{NO}_3^- \). Here, \( i = 3 \).
  • Contrast this with **glucose**, \( \text{C}_6 \text{H}_{12} \text{O}_6 \), a non-electrolyte. It does not dissociate in water, so \( i = 1 \).
Given solutions of equal molality, the one with a higher \( i \) will show greater freezing point depression. Understanding \( i \) is fundamental when predicting solution behavior.
Aqueous Solution
An aqueous solution is a specific type of chemical solution where the solvent is water. Water's unique properties make it an excellent solvent for many substances, assisting processes such as dissolution and reaction. Here's a simple breakdown of why aqueous solutions are important in this context:
  • Dissolution: Solutes dissolve in water due to its polar nature, enabling the solute molecules or ions to spread throughout the solution.
  • Colligative Properties: In learning about freezing point depression, water's solvent capabilities mean it will lower its freezing point based on the solute added.
  • Everyday relevance: Many everyday solutions, like saline or sugar water, demonstrate the principles of aqueous solutions and colligative properties.
The effects of solutes like glucose or ionic compounds can be studied in water as it provides a consistent and familiar medium to understand these properties.

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Most popular questions from this chapter

Osmotic pressure observed when benzoic acid is dissolved in benzene is less than that expected from theoretical considerations. This is because (a) benzoic acid is an organic solute (b) benzoic acid has higher molar mass than benzene (c) benzoic acid gets associated in benzene (d) benzoic acid gets dissociated in benzene

The amount of dissolved oxygen in 1 litre water in equilibrium with air at 1 atm pressure at \(25^{\circ} \mathrm{C}\) will be (assume that air contains 20 mole \% oxygen, Henry's constant \((\mathrm{kH})\) for oxygen is \(3.04 \times 10^{7} \mathrm{~mm} \mathrm{Hg}\) and density of \(\mathrm{H}_{2} \mathrm{O}\) at \(25^{\circ} \mathrm{C}\) is \(1 \mathrm{~g} / \mathrm{cc}\) ) (a) the mole fraction of \(\mathrm{O}_{2}\) in solution is \(5 \times 10^{-6}\) (b) the mole fraction of \(\mathrm{O}_{2}\) in solution is \(5 \times 10^{-7}\) (c) the molarity of \(\mathrm{O}_{2}\) in solution is \(2.77 \times 10^{-5} \mathrm{M}\) (d) the molarity of \(\mathrm{O}_{2}\) in solution is \(2.77 \times 10^{-4} \mathrm{M}\)

Two liquids having vapour pressures \(\mathrm{P}_{1}^{0}\) and \(\mathrm{P}_{2}^{0}\) in pure state in the ratio of \(2: 1\) are mixed in a molar ratio of \(1: 2\). The ratio of their moles in the vapour state can be (a) \(2: 1\) (b) \(1: 2\) (c) \(1: 1\) (d) \(3: 3\)

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

The vapour pressure of pure benzene and toluene are 160 and 60 torr respectively. The mole fraction of toluene in vapour phase in contact with equimolar solution of benzene and toluene is (a) \(0.6\) (b) \(0.50\) (c) \(0.27\) (d) \(0.74\)

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