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Chapter 13: Problem 64

List these aqueous solutions in order of decreasing freezing point. (a) \(0.10 \mathrm{~mol}\) methanol/kg (b) \(0.10 \mathrm{~mol} \mathrm{KCl} / \mathrm{kg}\) (c) \(0.080 \mathrm{~mol} \mathrm{BaCl}_{2} / \mathrm{kg}\) (d) \(0.040 \mathrm{~mol} \mathrm{Na}_{2} \mathrm{SO}_{4} / \mathrm{kg}\) (Assume that all of the salts dissociate completely into their ions in solution.)

Short Answer

Expert verified
Order: (a) methanol, (d) Na鈧係O鈧�, (b) KCl, (c) BaCl鈧�.

Step by step solution

01

Understand Freezing Point Depression

The freezing point depression of a solution is given by the formula \( \Delta T_f = i \cdot K_f \cdot m \), where \( \Delta T_f \) is the decrease in freezing temperature, \( K_f \) is the freezing point depression constant, \( m \) is the molality, and \( i \) is the van 't Hoff factor representing the number of particles the solute breaks into in solution.
02

Calculate van 't Hoff Factor for Each Solute

(a) Methanol is a nonelectrolyte, so \( i = 1 \).(b) KCl dissociates into 2 ions, so \( i = 2 \).(c) BaCl鈧� dissociates into 3 ions (Ba虏鈦� and 2Cl鈦�), so \( i = 3 \).(d) Na鈧係O鈧� dissociates into 3 ions (2Na鈦� and SO鈧劼测伝), so \( i = 3 \).
03

Calculate Freezing Point Depression for Each Solution

Using the formula \( \Delta T_f = i \cdot K_f \cdot m \), and assuming \( K_f \) is the same for all solutions, we compare \( i \cdot m \) for each:(a) \( i = 1, m = 0.10 \) -> \( i \cdot m = 0.10 \)(b) \( i = 2, m = 0.10 \) -> \( i \cdot m = 0.20 \)(c) \( i = 3, m = 0.080 \) -> \( i \cdot m = 0.24 \)(d) \( i = 3, m = 0.040 \) -> \( i \cdot m = 0.12 \)
04

List Solutions in Decreasing Order of Freezing Point

Higher \( i \cdot m \) values mean a lower freezing point. Thus, the order of decreasing freezing point is:1. Solution (a) Methanol: \( 0.10 \)2. Solution (d) Na鈧係O鈧�: \( 0.12 \)3. Solution (b) KCl: \( 0.20 \)4. Solution (c) BaCl鈧�: \( 0.24 \)

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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 't Hoff factor, represented by the symbol \( i \), is an important concept in understanding colligative properties such as freezing point depression. It expresses the number of particles a compound dissociates into when dissolved in a solution. For example, in the case of electrolytes like salts:
  • Methanol, being a nonelectrolyte, does not dissociate and thus has an \( i \) value of 1.
  • Potassium chloride (KCl) dissociates into two ions (K+ and Cl-), resulting in an \( i \) of 2.
  • Barium chloride (BaCl鈧�) separates into three ions (one Ba2+ and two Cl-), giving an \( i \) of 3.
  • Sodium sulfate (Na鈧係O鈧�) also dissociates into three ions (two Na+ and one SO鈧�2-), leading to an \( i \) of 3.
Understanding the van 't Hoff factor is crucial for predicting how a solute will affect the colligative properties of a solution, like decreasing the freezing point.
molality
Molality, indicated by \( m \), is a measure of concentration defined as the number of moles of solute per kilogram of solvent. It differs from molarity, which measures concentration per liter of solution. This distinction is significant especially in colligative properties because:
  • Molality is unaffected by changes in temperature and pressure, making it more reliable for calculations involving changes in physical states.
  • The freezing point depression, a colligative property, is calculated using molality in the formula \( \Delta T_f = i \cdot K_f \cdot m \).
  • In our exercise, each solution's molality was already provided, such as \(0.10\, \text{mol} / \text{kg}\) for KCl.
Considering molality ensures more accurate insights into how solute particles influence a solution under varying environmental conditions.
aqueous solutions
Aqueous solutions play a significant role in chemistry, referring simply to any solution where water acts as the solvent. Understanding their character is crucial when dealing with colligative properties like freezing point depression:
  • Water, as a universal solvent, can dissolve a variety of solutes including salts and other electrolytes.
  • Aqueous solutions offer a medium where ionic compounds dissociate into ions, impacting the solution's properties.
  • In the context of freezing point depression, it鈥檚 important to remember that the nature of water allows for significant interactions and changes in properties such as boiling points and freezing points.
Through complete dissociation of solutes in aqueous solutions, students can observe the real-world application of the concepts like the van 't Hoff factor and molality, providing a concrete learning experience in chemical studies.

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Most popular questions from this chapter

Describe what happens when an ionic solid dissolves in water. Sketch an illustration that includes at least three positive ions, three negative ions, and a dozen or so water molecules in the vicinity of the ions.

Anthracene, a hydrocarbon obtained from coal, has an empirical formula of \(\mathrm{C}_{7} \mathrm{H}_{5} .\) To find its molecular formula you dissolve \(0.500 \mathrm{~g}\) anthracene in \(30.0 \mathrm{~g}\) henzene. The boiling point of the solution is \(80.34{ }^{\circ} \mathrm{C}\). Determine the molar mass and molecular formula of anthracene.

In chemical research, newly synthesized compounds are often sent to commercial laboratories for analysis that determines the weight percent of \(\mathrm{C}\) and \(\mathrm{H}\) by burning the compound and collecting the evolved \(\mathrm{CO}_{2}\) and \(\mathrm{H}_{2} \mathrm{O}\) ( Sec. \(3-9\) ). The molar mass is determined by measuring the osmotic pressure of a solution of the compound. Calculate the empirical and molecular formulas of a compound, \(\mathrm{C}_{x} \mathrm{H}_{y} \mathrm{Cr}\), given this information: (a) The compound contains \(73.94 \% \mathrm{C}\) and \(8.27 \% \mathrm{H} ;\) the remainder is chromium. (b) At \(25^{\circ} \mathrm{C}\), the osmotic pressure of \(5.00 \mathrm{mg}\) of the unknown dissolved in \(100 . \mathrm{mL}\) of chloroform solution is \(3.17 \mathrm{mmHg}\)

Simple acids such as formic acid, \(\mathrm{HCOOH},\) and acetic acid, \(\mathrm{CH}_{3} \mathrm{COOH},\) are very soluble in water; however, fatty acids such as stearic acid, \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{16} \mathrm{COOH},\) and palmitic acid, \(\mathrm{CH}_{3}\left(\mathrm{CH}_{2}\right)_{14} \mathrm{COOH},\) are water- insoluble. Based on what you know about the solubility of alcohols, explain the solubility of these organic acids.

A solution, prepared by dissolving \(9.41 \mathrm{~g} \mathrm{NaHSO}_{3}\) in \(1.00 \mathrm{~kg}\) water, freezes at \(-0.33{ }^{\circ} \mathrm{C}\). From these data, decide which of these equations is the correct expression for the dissociation of the salt. (a) \(\mathrm{NaHSO}_{3}(\mathrm{aq}) \longrightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{HSO}_{3}^{-}(\mathrm{aq})\) (b) \(\mathrm{NaHSO}_{3}(\mathrm{aq}) \longrightarrow \mathrm{Na}^{+}(\mathrm{aq})+\mathrm{H}^{+}(\mathrm{aq})+\mathrm{SO}_{3}^{2-}(\mathrm{aq})\)
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