/*! 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 67 What mass of glycerin \(\left(\m... [FREE SOLUTION] | 91Ó°ÊÓ

91Ó°ÊÓ

What mass of glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right),\) a nonelectrolyte, must be dissolved in \(200.0 \mathrm{g}\) water to give a solution with a freezing point of \(-1.50^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
To obtain a solution with a freezing point of -1.50ËšC, approximately 14.84 g of glycerin \(\left(\mathrm{C}_{3} \mathrm{H}_{8} \mathrm{O}_{3}\right)\) must be dissolved in 200.0 g of water.

Step by step solution

01

Calculate the change in freezing point (∆Tf)

The change in freezing point is the difference between the freezing point of the solution and the freezing point of the pure solvent. For this problem, the freezing point of the solution is given as -1.50˚C and the freezing point of pure water is 0˚C. ∆Tf = (Freezing point of pure solvent) - (Freezing point of the solution) ∆Tf = 0 - (-1.50˚C) ∆Tf = 1.50˚C
02

Calculate the molality of the glycerin solution

Now that we have the change in freezing point and the freezing point constant for water, we can use the freezing point depression formula to find the molality of the glycerin solution. The formula is: ∆Tf = Kf * molality Here, ∆Tf = 1.50˚C and Kf = 1.86 ˚C/m 1.50˚C = 1.86˚C/m * molality Molality = 1.50˚C / 1.86˚C/m Molality ≈ 0.806 moles glycerin/kg water
03

Calculate the moles of glycerin needed

Now we know the molality of the glycerin solution, we can compute the number of moles of glycerin needed to be dissolved in 200 g water. moles of glycerin = molality * mass of water (in kg) moles of glycerin = 0.806 mol/kg * 0.200 kg moles of glycerin ≈ 0.1612 mol
04

Calculate the mass of glycerin needed

Now we know the moles of glycerin, we can calculate their mass using the formula: mass = moles * molar mass The molar mass of glycerin (C₃H₈O₃) is Molar mass = (3 * 12.01 g/mol) + (8 * 1.01 g/mol) + (3 * 16.00 g/mol) ≈ 92.09 g/mol mass of glycerin = 0.1612 mol * 92.09 g/mol mass of glycerin ≈ 14.84 g So, to get a solution with a freezing point of -1.50˚C, approximately 14.84 g of glycerin must be dissolved in 200.0 g 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 91Ó°ÊÓ!

Key Concepts

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

Colligative Properties
Colligative properties are characteristics of solutions that depend solely on the number of dissolved particles in the solution, not on the identity of those particles. Examples of colligative properties include boiling point elevation, freezing point depression, vapor pressure lowering, and osmotic pressure. Freezing point depression, the focus of our textbook exercise, is particularly important in many practical applications, such as anti-freeze solutions for cars and the salting of roads in winter.

When a non-volatile solute like glycerin is dissolved in a solvent like water, it disrupts the formation of ice at 0°C, causing the solution to freeze at a lower temperature. This is because the solute particles interfere with the ability of the solvent molecules to form a crystalline solid (ice) by getting in between them, requiring a lower temperature to achieve the same level of solid formation.
Molality
Molality is a measure of the concentration of a solution that expresses the moles of solute per kilogram of solvent. It is distinct from molarity, which is the moles of solute per liter of solution. Molality is particularly useful in the study of colligative properties because it is not affected by temperature changes, unlike molarity. This is because the molality is based on mass, a property that doesn't change with temperature, while volume, used in molarity, can expand or contract with temperature.

To calculate molality, you divide the number of moles of solute by the mass of the solvent in kilograms. In our exercise, we calculated the molality to determine the amount of glycerin needed to lower the freezing point of water. This concept is crucial for understanding how solutes affect the physical properties of solutions in a temperature-dependent manner.
Solution Concentration
Solution concentration is a quantitative measure of how much solute is present in a given quantity of solvent or solution. It can be expressed in various ways, including molality, molarity, parts per million, and weight percent. Understanding concentration is fundamental in chemistry because it affects reaction rates, chemical equilibria, and physical properties of solutions.

In the context of our textbook problem, we focused on the concentration expressed as molality to utilize the linear relationship between the freezing point depression and the molality of the solution. Knowing the concentration allows us to predict and control the freezing point of the solution to suit particular needs, such as creating a solution with a specific freezing point.
Molar Mass Calculation
Molar mass is the mass of one mole of a substance, typically expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all the atoms in a single molecule of the substance. The molar mass is an important factor when converting between the mass of a substance and the number of moles of that substance.

In our solution, we used the molar mass of glycerin to find the mass required to achieve the desired freezing point depression. The molar mass enabled us to convert moles (calculated from molality and mass of the solvent) into grams of glycerin needed. A correct molar mass is essential for accurately determining the amount of solute in a solution and for ensuring desired outcomes in chemical reactions and processes.

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

Patients undergoing an upper gastrointestinal tract laboratory test are typically given an X-ray contrast agent that aids with the radiologic imaging of the anatomy. One such contrast agent is sodium diatrizoate, a nonvolatile water-soluble compound. A \(0.378-m\) solution is prepared by dissolving 38.4 g sodium diatrizoate (NaDTZ) in \(1.60 \times 10^{2} \mathrm{mL}\) water at \(31.2^{\circ} \mathrm{C}\) (the density of water at \(31.2^{\circ} \mathrm{C}\) is \(0.995 \mathrm{g} / \mathrm{cm}^{3}\) ). What is the molar mass of sodium diatrizoate? What is the vapor pressure of this solution if the vapor pressure of pure water at \(31.2^{\circ} \mathrm{C}\) is 34.1 torr?

You have read that adding a solute to a solvent can both increase the boiling point and decrease the freezing point. A friend of yours explains it to you like this: "The solute and solvent can be like salt in water. The salt gets in the way of freezing in that it blocks the water molecules from joining together. The salt acts like a strong bond holding the water molecules together so that it is harder to boil." What do you say to your friend?

In the winter of \(1994,\) record low temperatures were registered throughout the United States. For example, in Champaign, Illinois, a record low of \(-29^{\circ} \mathrm{F}\) was registered. At this temperature can salting icy roads with \(\mathrm{CaCl}_{2}\) be effective in melting the ice? a. Assume \(i=3.00\) for \(\mathrm{CaCl}_{2}\) b. Assume the average value of \(i\) from Exercise 87 (The solubility of \(\mathrm{CaCl}_{2}\) in cold water is \(74.5 \mathrm{g}\) per \(100.0 \mathrm{g}\) water.

Calculate the solubility of \(\mathrm{O}_{2}\) in water at a partial pressure of \(\mathrm{O}_{2}\) of 120 torr at \(25^{\circ} \mathrm{C}\). The Henry's law constant for \(\mathrm{O}_{2}\) is \(1.3 \times 10^{-3} \mathrm{mol} / \mathrm{L} \cdot \mathrm{atm}\) for Henry's law in the form \(C=k P\) where \(C\) is the gas concentration (mol/L).

A solution is prepared by mixing 1.000 mole of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) and 3.18 moles of propanol \(\left(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{CH}_{2} \mathrm{OH}\right) .\) What is the composition of the vapor (in mole fractions) at \(40^{\circ} \mathrm{C} ?\) At \(40^{\circ} \mathrm{C},\) the vapor pressure of pure methanol is 303 torr, and the vapor pressure of pure propanol is 44.6 torr.

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.