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(a) Which solution is expected to have the higher boiling point: \(0.20 \mathrm{m}\) KBr or \(0.30 \mathrm{m}\) sugar? (b) Which aqueous solution has the lower freezing point: \(0.12 \mathrm{m} \mathrm{NH}_{4} \mathrm{NO}_{3}\) or \(0.10 \mathrm{m} \mathrm{Na}_{2} \mathrm{CO}_{3} ?\)

Short Answer

Expert verified
(a) 0.20 m KBr has the higher boiling point. (b) 0.12 m NH鈧凬O鈧 has the lower freezing point.

Step by step solution

01

Define the boiling point elevation formula

Boiling point elevation is described by the formula \( \Delta T_b = i \cdot K_b \cdot m \), where \( i \) is the van't Hoff factor, \( K_b \) is the ebullioscopic constant, and \( m \) is the molality. A higher \( \Delta T_b \) indicates a higher boiling point.
02

Calculate boiling point elevation for KBr solution

KBr dissociates into two ions: K鈦 and Br鈦, so \( i = 2 \). The molality \( m = 0.20 \mathrm{m} \) for KBr. Substitute these into \( \Delta T_b = i \cdot K_b \cdot m \) to get \( \Delta T_{b,\mathrm{KBr}} = 2 \cdot K_b \cdot 0.20 = 0.40 \cdot K_b \).
03

Calculate boiling point elevation for the sugar solution

Sugar does not dissociate in solution, so \( i = 1 \). The molality \( m = 0.30 \mathrm{m} \). Substitute into the formula: \( \Delta T_{b,\mathrm{sugar}} = 1 \cdot K_b \cdot 0.30 = 0.30 \cdot K_b \).
04

Compare boiling point elevations

Since \( \Delta T_{b,\mathrm{KBr}} = 0.40 \cdot K_b \) is greater than \( \Delta T_{b,\mathrm{sugar}} = 0.30 \cdot K_b \), the KBr solution has a higher boiling point.
05

Define the freezing point depression formula

Freezing point depression is described by \( \Delta T_f = i \cdot K_f \cdot m \), where \( i \) is the van't Hoff factor, \( K_f \) is the cryoscopic constant, and \( m \) is the molality. A lower \( \Delta T_f \) results in a lower freezing point.
06

Calculate freezing point depression for NH鈧凬O鈧 solution

NH鈧凬O鈧 dissociates into two ions: NH鈧勨伜 and NO鈧冣伝, so \( i = 2 \). The molality \( m = 0.12 \mathrm{m} \). Substitute into the formula: \( \Delta T_{f,\mathrm{NH}_4 ext{NO}_3} = 2 \cdot K_f \cdot 0.12 = 0.24 \cdot K_f \).
07

Calculate freezing point depression for Na鈧侰O鈧 solution

Na鈧侰O鈧 dissociates into three ions: 2 Na鈦 and CO鈧兟测伝, so \( i = 3 \). The molality \( m = 0.10 \mathrm{m} \). Substitute into the formula: \( \Delta T_{f,\mathrm{Na}_2 ext{CO}_3} = 3 \cdot K_f \cdot 0.10 = 0.30 \cdot K_f \).
08

Compare freezing point depressions

\( \Delta T_{f,\mathrm{NH}_4 ext{NO}_3} = 0.24 \cdot K_f \) is less than \( \Delta T_{f,\mathrm{Na}_2 ext{CO}_3} = 0.30 \cdot K_f \). Therefore, the NH鈧凬O鈧 solution has a lower freezing point.

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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 property of solutions. When you dissolve a solute in a solvent, the boiling point of the solution becomes higher than that of the pure solvent. This occurs because the presence of solute particles causes the solvent molecules to require more energy to escape into the vapor phase.
This effect is quantitatively described by the formula \( \Delta T_b = i \cdot K_b \cdot m \).
  • \( \Delta T_b \) is the increase in boiling point.
  • \( i \) is the van't Hoff factor, representing the number of particles the solute dissociates into.
  • \( K_b \) is the ebullioscopic constant, specific to each solvent.
  • \( m \) is the molality of the solution.
For instance, in the problem given, when comparing a \(0.20 \mathrm{m}\) KBr solution to a \(0.30 \mathrm{m}\) sugar solution, we observe that KBr dissociates into two ions, contributing to a higher van't Hoff factor of 2. This results in a greater boiling point elevation for KBr, demonstrating how the increase in boiling point depends not only on the concentration of the solute but also on its ability to dissociate.
Freezing Point Depression
Freezing point depression occurs when a solute is added to a solvent, causing the freezing point of the solution to be lower than that of the pure solvent. In simple terms, it takes a lower temperature for the solution to solidify.
This phenomenon is described by the formula \( \Delta T_f = i \cdot K_f \cdot m \).
  • \( \Delta T_f \) is the decrease in freezing point.
  • \( i \) is the van't Hoff factor, indicating the number of particles the solute breaks into.
  • \( K_f \) is the cryoscopic constant.
  • \( m \) is the solution's molality.
In the example provided, \(0.12 \mathrm{m}\) \(\mathrm{NH}_{4}\mathrm{NO}_{3}\) solution with a van't Hoff factor of 2 results in a smaller decrease in temperature compared to \(0.10 \mathrm{m}\) \(\mathrm{Na}_{2}\mathrm{CO}_{3}\) with a factor of 3. This demonstrates how substances that dissociate into more ions tend to create a more significant reduction in freezing point, explaining why the \(\mathrm{NH}_{4}\mathrm{NO}_{3}\) solution ends up with a lower freezing point.
Van't Hoff Factor
The van't Hoff factor, denoted by \( i \), is a crucial part of colligative properties such as boiling point elevation and freezing point depression. The van't Hoff factor measures the effect of a solute on the colligative properties of a solution. More specifically, it accounts for the number of particles into which a solute dissociates in solution. The more particles that are produced, the greater the effect on colligative properties.
To determine \( i \):
  • A non-dissociating solute, like sugar, has \( i = 1 \).
  • Ionic compounds dissociate into their ions; for example, \(\mathrm{KBr}\) dissociates into \(\mathrm{K}^+\) and \(\mathrm{Br}^-\), thus \( i = 2 \).
  • Compounds like \(\mathrm{Na}_2\mathrm{CO}_3\) split into three ions (2 \(\mathrm{Na}^+\) and \(\mathrm{CO}_3^{2-}\)), giving \( i = 3 \).
The van't Hoff factor directly influences the magnitude of both boiling point elevation and freezing point depression. Consequently, deeper understanding and accurate calculation of this factor are imperative for analyzing and predicting changes in solutions' properties.

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Most popular questions from this chapter

To make homemade ice cream, you cool the milk and cream by immersing the container in ice and a concentrated solution of rock salt (NaCl) in water. If you want to have a water-salt solution that freezes at \(-10 .^{\circ} \mathrm{C},\) what mass of NaCl must you add to \(3.0 \mathrm{kg}\) of water? (Assume the van't Hoff factor, \(i,\) for \(\mathrm{NaCl}\) is \(1.85 .\) )

A \(35.0-\mathrm{g}\) sample of ethylene glycol, \(\mathrm{HOCH}_{2} \mathrm{CH}_{2} \mathrm{OH},\) is dissolved in \(500.0 \mathrm{g}\) of water. The vapor pressure of water at \(32^{\circ} \mathrm{C}\) is \(35.7 \mathrm{mm}\) Hg. What is the vapor pressure of the water-ethylene glycol solution at \(32^{\circ} \mathrm{C} ?\) (Ethylene glycol is nonvolatile.)

Account for the fact that alcohols such as methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) and ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) are quite miscible with water, whereas an alcohol with a long-carbon chain, such as octanol \(\left(\mathrm{C}_{8} \mathrm{H}_{17} \mathrm{OH}\right),\) is poorly soluble in water.

Dimethylglyoxime [DMG, (CH \(\left._{3} \mathrm{CNOH} \text { ) }_{2}\right]\) is used as a reagent to precipitate nickel ion. Assume that \(53.0 \mathrm{g}\) of DMG has been dissolved in \(525 \mathrm{g}\) of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) (IMAGE CAN'T COPY) The red, insoluble compound formed between nickel(II) ion and dimethylglyoxime (DMG) is precipitated when DMG is added to a basic solution of \(\mathrm{Ni}^{2+}(\mathrm{aq})\) (a) What is the mole fraction of DMG? (b) What is the molality of the solution? (c) What is the vapor pressure of the ethanol over the solution at ethanol's normal boiling point of \(78.4^{\circ} \mathrm{C} ?\) (d) What is the boiling point of the solution? (DMG does not produce ions in solution.) \(\left(K_{\mathrm{bp}} \text { for ethanol }=\right.\) \(\left.+1.22^{\circ} \mathrm{C} / m\right)\)

A compound is known to be a potassium halide, KX. If \(4.00 \mathrm{g}\) of the salt is dissolved in exactly \(100 \mathrm{g}\) of water, the solution freezes at \(-1.28^{\circ} \mathrm{C} .\) Identify the halide ion in this formula.

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