/*! 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 93 An aqueous solution of urea \((\... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 6: Problem 93

An aqueous solution of urea \((\mathrm{MW}=60.06)\) freezes at \(-4.6^{\circ} \mathrm{C}\) and 1 atm. Estimate the normal boiling point of the solution; then calculate the mass of urea (grams) that would have to be added to \(1.00 \mathrm{kg}\) of solution to raise the normal boiling point by \(3^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The normal boiling point of the solution is 101.26掳C. To raise the boiling point by 3掳C, you need to add 353 g of urea.

Step by step solution

01

Calculating the Normal Boiling Point

Let's start by using the formula for freezing point depression: \[\Delta T_f = K_f \cdot m \cdot i\]where \(K_f\) is the cryoscopic constant for water equal to 1.86掳C/molality, \(m\) is the molality of the solution, and \(i\) is the number of particles per formula unit for the solute. Since urea is a non-volatile, non-electrolyte compound, we can consider \(i\) as 1. Let's rearrange the formula to find \(m\):\[m = \Delta T_f / (K_f \cdot i) = (-4.6掳C) / (1.86掳C/mol) = -2.47 mol/kg \]Now, let's calculate the mole fraction (X) of urea. \[X_{urea} = \frac{mol_{urea}}{mol_{urea} + mol_{water}}\]Assuming 1 kg of water, the molality is roughly equal to the mole fraction of urea since 1 kg of water is about 55.5 moles. Now, we can find the boiling point elevation using the formula:\[\Delta T_b = K_b \cdot m \cdot i\]where \(K_b\) is the ebullioscopic constant for water equal to 0.51掳C/molality. By placing the found molality value into the equation, the normal boiling point elevation caused by the urea can be calculated. Add this to the normal boiling point of water (100掳C) to find the normal boiling point of the solution.
02

Finding Mass of Urea to Raise Boiling Point

The next part of the problem asks how much additional urea is needed to further raise the boiling point by 3掳C. Convert this temperature change to a molality increase using the formula for boiling point elevation:\[m = \Delta T_b / (K_b \cdot i) = 3掳C / (0.51掳C/mol) = 5.88 mol/kg\]We know how much the molality needs to increase, multiply this by the molar mass of urea (60.06 g/mol) to convert this into grams per kilogram of water.

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 91影视!

Key Concepts

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

Freezing Point Depression
Understanding the concept of freezing point depression is vital in explaining why certain solutions freeze at lower temperatures than their pure solvents. This phenomenon occurs when a solute is dissolved in a solvent, resulting in a decrease in the temperature at which the solution becomes solid.

When a solute, such as urea, is added to a solvent like water, the solute particles disrupt the formation of the structured solid phase (ice) of the solvent. This interruption requires the temperature to be decreased further before the solvent can successfully transition into its solid state. The extent of freezing point depression can be quantified by the formula:
\(\Delta T_f = K_f \cdot m \cdot i\)
where \(\Delta T_f\) represents the change in freezing point, \(K_f\) is the cryoscopic constant (specific to the solvent), \(m\) is the molality of the solution, and \(i\) is the van 't Hoff factor, indicating the number of particles the solute dissociates into. For nonelectrolytes like urea, which do not dissociate, \(i\) is equal to 1.

The relationship between these variables shows that the decrease in the freezing point of the solution is directly proportional to the molality of the solute.
Boiling Point Elevation
In contrast to freezing point depression, boiling point elevation describes how the boiling point of a liquid solvent increases upon the addition of a solute. This increase happens because the presence of solute particles in the solvent makes it more difficult for the solvent molecules to escape into the gaseous phase, which is how boiling occurs.

The amount by which the boiling point increases can be determined by the equation:
\(\Delta T_b = K_b \cdot m \cdot i\)
Here, \(\Delta T_b\) is the boiling point elevation, \(K_b\) is the ebullioscopic constant (specific to the solvent), \(m\) is the molality of the solution, and \(i\) is once again the van 't Hoff factor. Similar to freezing point depression, for non-dissociating solutes like urea, \(i\) equals 1.

The practical application of this principle is seen when calculating how much a solution's boiling point will increase for a given amount of solute (such as urea), as well as determining how much additional solute is required to achieve a desired increase in boiling point.
Molality
Molality, a measure of the concentration of a solution, is defined as the number of moles of solute per kilogram of solvent. Unlike molarity, which is the number of moles of solute per liter of solution, molality is not affected by changes in temperature or pressure because it is based on mass, not volume.

The formula for molality is expressed as:
\(m = \frac{moles_{solute}}{mass_{solvent}(kg)}\)
This concentration measurement is particularly useful in colligative properties calculations such as freezing point depression and boiling point elevation, where the effect on the solvent's physical properties depends directly on the number of solute particles, regardless of their individual properties. The utilization of molality ensures accuracy in these calculations since it is independent of the solution's volume changes with temperature.

When solving problems that involve adjusting the freezing point or boiling point of a solution, determining the correct molality based on the amount of sold solute and mass of the solvent is a crucial step.

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 fuel gas containing methane and ethane is burned with air in a furnace, producing a stack gas at \(300^{\circ} \mathrm{C}\) and \(105 \mathrm{kPa}\) (absolute). You analyze the stack gas and find that it contains no unburned hydrocarbons, oxygen, or carbon monoxide. You also determine the dew-point temperature.(a) Estimate the range of possible dew-point temperatures by determining the dew points when the feed is either pure methane or pure ethane. (b) Estimate the fraction of the feed that is methane if the measured dew- point temperature is \(59.5^{\circ} \mathrm{C}\). (c) What range of measured dew point temperatures would lead to calculated methane mole fractions within 5\% of the value determined in Part (b)?

An ore containing \(90 \mathrm{wt} \% \mathrm{MgSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) and the balance insoluble minerals is fed to a dissolution tank at a rate of \(60,000 \mathrm{lb}_{\mathrm{m}} / \mathrm{h}\) along with fresh water and a recycle stream. The tank contents are heated to \(120^{\circ} \mathrm{F}\), causing all of the magnesium sulfate monohydrate in the ore to dissolve, forming a solution 10^0 F above saturation. The resulting slurry of the insoluble minerals in MgSO_solution is pumped to a heated filter, where a wet filter cake is separated from a solids-free filtrate. The filter cake retains \(5 \mathrm{lb}_{\mathrm{m}}\) of solution per \(100 \mathrm{lb}_{\mathrm{m}}\) of solids. The filtrate is sent to a crystallizer in which the temperature is reduced to \(50^{\circ} \mathrm{F},\) producing a slurry of \(\mathrm{MgSO}_{4} \cdot 7 \mathrm{H}_{2} \mathrm{O}\) crystals in a saturated solution that is sent to another filler. The product filter cake contains all of the prea entrained solution in a ratio of \(5 \mathrm{Ib}_{\mathrm{m}}\) solution per \(100 \mathrm{lb}_{\mathrm{m}}\) crystals. The filtrate from this filter is returned to the dissolution tank as the recycle stream.Solubility data: Saturated magnesium sulfate solutions at \(110^{\circ} \mathrm{F}\) and \(50^{\circ} \mathrm{F}\) contain \(32 \mathrm{wt} \%\) \(\mathrm{MgSO}_{4}\) and \(23 \mathrm{wt} \% \mathrm{MgSO}_{4},\) respectively.(a) Explain why the solution is first heated (in the dissolution tank) and filtered and then cooled (in the crystallizer) and filtered. (b) Calculate the production rate of crystals and the required feed rate of fresh water to the dissolution tank. (Note: Don't forget to include water of hydration when you write a mass balance on water.)(c) Calculate the ratio \(\mathrm{lb}_{\mathrm{m}}\) recycle/lb \(_{\mathrm{m}}\) makeup water.

A storage tank for liquid \(n\) -octane has a diameter of \(30 \mathrm{ft}\) and a height of \(20 \mathrm{ft}\). During a typical \(24-\mathrm{h}\) period the level of liquid octane falls from 18 ft to 8 ft, after which fresh octane is pumped into the tank to return the level to \(18 \mathrm{ft}\). As the level in the tank falls, nitrogen is fed into the free space to maintain the pressure at 16 psia; when the tank is being refilled, the pressure is maintained at 16 psia by discharging gas from the vapor space to the environment. The nitrogen in the tank may be considered saturated with octane vapor at all times. The average tank temperature is \(90^{\circ} \mathrm{F}\). (a) What is the daily rate, in gallons and \(1 \mathrm{b}_{\mathrm{m}}\), at which octane is used? (b) What is the variation in absolute pressure at the bottom of the tank in inches of mercury? (c) How much octane is lost to the environment during a 24 -h period? (d) Why is nitrogen used in the vapor space of the tank when air would be cheaper? (e) Suggest a means by which the octane can be recovered from the gas stream discharged to the atmosphere.

You were recently hired as a process engineer by a pulp and paper manufacturing firm. Your new boss calls you in and tells you about a pulp dryer designed to reduce the moisture content of \(1500 \mathrm{kg} / \mathrm{min}\) of wet pulp from \(0.9 \mathrm{kg} \mathrm{H}_{2} \mathrm{O} / \mathrm{kg}\) dry pulp to \(0.15 \mathrm{wt} \% \mathrm{H}_{2} \mathrm{O}\). The design called for drawing atmospheric air at \(90 \%\) relative humidity, \(25^{\circ} \mathrm{C}, 760 \mathrm{mm}\) Hg into a blower that forces the air through a heater and into the dryer. When the operation was put into service, weather conditions were exactly as assumed in the design, and measurements showed that the air leaving the dryer was at \(80^{\circ} \mathrm{C}\) and a gauge pressure of \(10 \mathrm{mm}\) Hg. However, there was no way to check the operation of the blower to see if it was delivering the specified volumetric flow rate of air. Your boss wants to check that value and asks you to devise a method for doing so. You go back to your office, sketch the process, and determine that you can estimate the air flow rate from the given information if you also know the moisture content of the air leaving the dryer.(a) Propose a method to estimate the moisture content of the exit air. (b) Suppose your measurement is carried out and you learn that the exit air at \(10 \mathrm{mm}\) Hg gauge has a dew point of \(40^{\circ} \mathrm{C}\). Use that information and the mass of water removed from the wet pulp to determine the volumetric flow rate ( \(\mathrm{m}^{3} / \mathrm{min}\) ) of air entering the system.

A \(50.0-\mathrm{L}\) tank contains an air-carbon tetrachloride gas mixture at an absolute pressure of \(1 \mathrm{atm}, \mathrm{a}\) temperature of \(34^{\circ} \mathrm{C},\) and a relative saturation of \(30 \% .\) Activated carbon is added to the tank to remove the \(\mathrm{CCl}_{4}\) from the gas by adsorption and the tank is then sealed. The volume of added activated carbon may be assumed negligible in comparison to the tank volume.(a) Calculate \(p_{\mathrm{CCl}_{4}}\) at the moment the tank is sealed, assuming ideal-gas behavior and neglecting adsorption that occurs prior to sealing. (b) Calculate the total pressure in the tank and the partial pressure of carbon tetrachloride at a point when half of the CCl_ initially in the tank has been adsorbed. Note: It was shown in Example \(6.7-1\) that at \(34^{\circ} \mathrm{C}\).$$X^{*}\left(\frac{\mathrm{g} \mathrm{CCl}_{4} \text { adsorbed }}{\mathrm{g} \text { carbon }}\right)=\frac{0.0762 p_{\mathrm{CCl}_{4}}}{1+0.096 p_{\mathrm{CCl}_{4}}}$$ where \(p_{\mathrm{CCl}_{4}}\) is the partial pressure (in \(\mathrm{mm} \mathrm{Hg}\) ) of carbon tetrachloride in the gas contacting the carbon.(c) How much activated carbon must be added to the tank to reduce the mole fraction of \(\mathrm{CCl}_{4}\) in the gas to 0.001?
See all solutions

Recommended explanations on Chemistry Textbooks

Kinetics

Read Explanation

The Earths Atmosphere

Read Explanation

Chemical Reactions

Read Explanation

Physical Chemistry

Read Explanation

Chemical Analysis

Read Explanation

Nuclear Chemistry

Read Explanation
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.