/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 76 What is the freezing point of an... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What is the freezing point of an aqueous solution that boils at \(105.0^{\circ} \mathrm{C}\) ?

Short Answer

Expert verified
The freezing point of the aqueous solution is \(-18.186^{\circ}\mathrm{C}\).

Step by step solution

01

Calculate boiling point elevation

Since we know the boiling point of the aqueous solution (\(105.0^{\circ}\mathrm{C}\)) and the normal boiling point of water (\(100.0^{\circ}\mathrm{C}\)), we can calculate the boiling point elevation: \(\Delta T_b = T_{b, solution} - T_{b, pure \, water}\) \(\Delta T_b = 105.0 - 100.0\) \(\Delta T_b = 5.0^{\circ} \mathrm{C}\)
02

Calculate the molality of the solution

Use the boiling point elevation formula to calculate the molality of the solution: \(\Delta T_b = K_b \cdot m\) \(5.0 = 0.512 \cdot m\) \(m = \frac{5.0}{0.512}\) \(m = 9.766\, \mathrm{mol/kg}\)
03

Calculate the freezing point depression

Use the molality and the cryoscopic constant of water to calculate the freezing point depression: \(\Delta T_f = K_f \cdot m\) \(\Delta T_f = 1.86 \cdot 9.766\) \(\Delta T_f = 18.186^{\circ} \mathrm{C}\)
04

Calculate the freezing point of the aqueous solution

Subtract the freezing point depression from the normal freezing point of water to determine the freezing point of the aqueous solution: \(T_{f, solution} = T_{f, pure \, water} - \Delta T_f\) \(T_{f, solution} = 0.0^{\circ}\mathrm{C} - 18.186^{\circ}\mathrm{C}\) \(T_{f, solution} = -18.186^{\circ}\mathrm{C}\) So the freezing point of the aqueous solution is \(-18.186^{\circ}\mathrm{C}\).

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Over 30 million students worldwide already upgrade their learning with 91Ó°ÊÓ!

Key Concepts

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

Boiling Point Elevation
Boiling point elevation is a colligative property, meaning it depends on the number of solute particles in a solution and not on their identity. When you dissolve a nonvolatile solute into a solvent, the boiling point of the resulting solution is higher than that of the pure solvent. This happens because the solute particles disrupt the formation of vapor at the surface of the liquid, requiring more heat energy to reach the point of boiling.

This relationship can be quantified using the formula for boiling point elevation:
  • \[\Delta T_b = K_b \cdot m\]
  • Where \(\Delta T_b\) is the change in boiling point, \(K_b\) is the ebullioscopic constant specific to the solvent (for water it's 0.512 \(\text{°C/m}\)), and \(m\) is the molality of the solution.
In our exercise, the boiling point of the aqueous solution was observed at 105.0 °C, which is 5.0 °C above the normal boiling point of water at 100.0 °C. By using the formula, we calculated the molality, which helps to further determine changes in the freezing point, illustrating the connection between boiling point elevation and freezing point depression.
Freezing Point Depression
Freezing point depression, another important colligative property, occurs when the freezing point of a solution is lower than that of the pure solvent. Just like boiling point elevation, this effect is due to the presence of solute particles. These particles disrupt the orderly crystal formation of the solid phase, requiring a lower temperature to freeze the solution.

To model this concept, we use the formula:
  • \[\Delta T_f = K_f \cdot m\]
  • Here, \(\Delta T_f\) is the freezing point depression, \(K_f\) is the cryoscopic constant (for water it's 1.86 \(\text{°C/m}\)), and \(m\) is the molality of the solution.
In the given problem, after determining the molality from boiling point elevation, we used it to find the freezing point depression. The aqueous solution's freezing point was calculated to drop by 18.186 °C, leading to a new freezing point of -18.186 °C. This showcases how changes in solute concentration can significantly alter the physical properties of a solution.
Molality
Molality is a measure of the concentration of a solute in a solution, expressed in moles of solute per kilogram of solvent. Unlike molarity, molality remains unaffected by changes in temperature because it depends on the mass of the solvent, not the volume of the entire solution.

To find molality, you use the formula:
  • \[m = \frac{\text{moles of solute}}{\text{kg of solvent}}\]
In the context of colligative properties like boiling point elevation and freezing point depression, molality is crucial because these effects depend on solute concentration.
In our exercise, the molality was calculated using the boiling point elevation data. It was found to be 9.766 mol/kg, indicating a relatively concentrated solution. This concentration directly affected both the increase in boiling point and the decrease in freezing point, underlining how molality is a key factor in manipulating the colligative properties of solutions.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

(a) Does a \(0.10 \mathrm{~m}\) aqueous solution of \(\mathrm{NaCl}\) have a higher boiling point, a lower boiling point, or the same boiling point as a \(0.10 \mathrm{~m}\) aqueous solution of \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) ? (b) The experimental boiling point of the \(\mathrm{NaCl}\) solution is lower than that calculated assuming that \(\mathrm{NaCl}\) is completely dissociated in solution. Why is this the case?

A solution is made containing \(20.8 \mathrm{~g}\) of phenol \(\left(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{OH}\right)\) in \(425 \mathrm{~g}\) of ethanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\right)\). Calculate (a) the mole fraction of phenol, (b) the mass percent of phenol, (c) the molality of phenol.

Indicate the type of solute-solvent interaction (Section 11.2) that should be most important in each of the following solutions: (a) \(\mathrm{CCl}_{4}\) in benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\), (b) methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) in water, (c) \(\mathrm{KBr}\) in water, (d) \(\mathrm{HCl}\) in acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\).

(a) A sample of hydrogen gas is generated in a closed container by reacting \(2.050 \mathrm{~g}\) of zinc metal with \(15.0 \mathrm{~mL}\) of \(1.00 \mathrm{M}\) sulfuric acid. Write the balanced equation for the reaction, and calculate the number of moles of hydrogen formed, assuming that the reaction is complete. (b) The volume over the solution in the container is \(122 \mathrm{~mL}\). Calculate the partial pressure of the hydrogen gas in this volume at \(25^{\circ} \mathrm{C}\), ignoring any solubility of the gas in the solution. (c) The Henry's law constant for hydrogen in water at \(25^{\circ} \mathrm{C}\) is \(7.8 \times 10^{-4} \mathrm{~mol} / \mathrm{L}\)-atm. Estimate the number of moles of hydrogen gas that remain dissolved in the solution. What fraction of the gas molecules in the system is dissolved in the solution? Was it reasonable to ignore any dissolved hydrogen in part (b)? [13.111] The following table presents the solubilities of several gases in water at \(25^{\circ} \mathrm{C}\) under a total pressure of gas and water vapor of \(1 \mathrm{~atm}\). (a) What volume of \(\mathrm{CH}_{4}(\mathrm{~g})\) under standard conditions of temperature and pressure is contained in \(4.0 \mathrm{~L}\) of a saturated solution at \(25^{\circ} \mathrm{C}\) ? (b) Explain the variation in solubility among the hydrocarbons listed (the first three compounds), based on their molecular structures and intermolecular forces. (c) Compare the solubilities of \(\mathrm{O}_{2}, \mathrm{~N}_{2}\), and \(\mathrm{NO}\), and account for the variations based on molecular structures and intermolecular forces. (d) Account for the much larger values observed for \(\mathrm{H}_{2} \mathrm{~S}\) and \(\mathrm{SO}_{2}\) as compared with the other gases listed. (e) Find several pairs of substances with the same or nearly the same molecular masses (for example, \(\mathrm{C}_{2} \mathrm{H}_{4}\) and \(\mathrm{N}_{2}\) ), and use intermolecular interactions to explain the differences in their solubilities.

Calculate the molality of each of the following solutions: (a) \(8.66 \mathrm{~g}\) of benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) dissolved in \(23.6 \mathrm{~g}\) of carbon tetrachloride \(\left(\mathrm{CCl}_{4}\right)\), (b) \(4.80 \mathrm{~g}\) of \(\mathrm{NaCl}\) dissolved in \(0.350 \mathrm{~L}\) of water.

See all solutions

Recommended explanations on Chemistry Textbooks

View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.