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Chapter 12: Problem 17

Calculate the molality of each of the following solutions: (a) \(14.3 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) in \(676 \mathrm{~g}\) of water, (b) 7.20 moles of ethylene glycol \(\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) in \(3546 \mathrm{~g}\) of water.

Short Answer

Expert verified
(a) The molality of the sucrose solution is 0.0618 m. (b) The molality of the ethylene glycol solution is 2.03 m.

Step by step solution

01

Calculation for (a)

First, calculate the molar mass of sucrose \(C_{12}H_{22}O_{11}\). This is done by adding the atomic masses of each atom within the formula. That results in \(12(12.01 g/mol) + 22(1.008 g/mol) + 11(16.00 g/mol) = 342.3 g/mol\). Now, calculate the number of moles in 14.3 g of sucrose using this molar mass: \((14.3 g) / (342.3 g/mol) = 0.0418 moles.\) Next, convert the weight of the water from grams to kilograms, which gives: \(676 g = 0.676 kg\). The molality can then be calculated using the formula: molality = moles of solute / kg of solvent, thus resulting in, molality = \(0.0418 moles / 0.676 kg = 0.0618 mol/kg = 0.0618 m\).
02

Calculation for (b)

In this scenario, the number of moles of ethylene glycol is already given as 7.20 moles. Convert the weight of the water from grams to kilograms, which gives: \(3546 g = 3.546 kg\). The molality can then be calculated using the formula: molality = moles of solute / kg of solvent, thus resulting in, molality = \(7.20 moles / 3.546 kg = 2.03 m\).

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Key Concepts

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

Understanding Molality
Molality, a measure of concentration, is defined as the number of moles of solute per kilogram of solvent. It's represented with the symbol 'm' and calculated with the formula:
\[\text{molality (m)} = \frac{\text{moles of solute}}{\text{kg of solvent}}\]
This measure is particularly useful because it remains constant regardless of temperature changes, unlike molarity, which varies as the volume of the solution changes with temperature. In practice, molality is frequently used in boiling point elevation and freezing point depression calculations.
Molar Mass
The molar mass of a compound is the mass of one mole of that substance. It's the sum of the atomic masses of all atoms in the molecular formula, usually expressed in grams per mole (g/mol). For example, to calculate the molar mass of sucrose, each atom within the molecular formula (\(C_{12}H_{22}O_{11}\)) contributes to the total weight:
\[\text{Molar mass of sucrose} = 12(\text{Carbon}) + 22(\text{Hydrogen}) + 11(\text{Oxygen})\]
Identifying the molar mass is a fundamental step in molality calculations because you need to convert the mass of the solute to moles.
Calculating Moles of Solute
To find moles of solute, you divide the mass of the solute by its molar mass:
\[\text{moles of solute} = \frac{\text{mass of solute(g)}}{\text{molar mass(g/mol)}}\]
For a given mass, this equation enables the conversion from grams to moles, providing a measure of the number of particles in the sample, which is essential for quantifying the concentration of a solution.
Sucrose Solubility
Sucrose, or table sugar, is highly soluble in water. Solubility refers to the maximum amount of solute that can dissolve in a solvent at a given temperature. Knowing the solubility of sucrose is important when preparing solutions to avoid supersaturation, where excess sugar cannot dissolve. In the context of studying properties like osmosis, diffusion, and crystallization, understanding sucrose solubility is crucial.
Ethylene Glycol Properties
Ethylene glycol (\(C_2H_6O_2\)) is a colorless, odorless liquid often used as an automotive antifreeze and coolant. It has a higher boiling point and lower freezing point than water, which allows it to regulate engine temperatures effectively. In chemistry, the knowledge of ethylene glycol's properties can inform decisions about its use in solutions, particularly when preparing mixtures that require specific physical properties.

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Most popular questions from this chapter

Calculate the amount of water (in grams) that must be added to (a) \(5.00 \mathrm{~g}\) of urea \(\left(\mathrm{NH}_{2}\right)_{2} \mathrm{CO}\) in the preparation of a 16.2 percent by mass solution, and (b) \(26.2 \mathrm{~g}\) of \(\mathrm{MgCl}_{2}\) in the preparation of a 1.5 percent by mass solution.

The vapor pressure of ethanol \(\left(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}\right)\) at \(20^{\circ} \mathrm{C}\) is \(44 \mathrm{mmHg},\) and the vapor pressure of methanol \(\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) at the same temperature is \(94 \mathrm{mmHg} . \mathrm{A}\) mixture of \(30.0 \mathrm{~g}\) of methanol and \(45.0 \mathrm{~g}\) of ethanol is prepared (and can be assumed to behave as an ideal solution). (a) Calculate the vapor pressure of methanol and ethanol above this solution at \(20^{\circ} \mathrm{C}\). (b) Calculate the mole fraction of methanol and ethanol in the vapor above this solution at \(20^{\circ} \mathrm{C}\). (c) Suggest a method for separating the two components of the solution.

How does each of the following affect the solubility of an ionic compound? (a) lattice energy, (b) solvent (polar versus nonpolar), (c) enthalpies of hydration of cation and anion

Describe the fractional crystallization process and its application.

Basing your answer on intermolecular force considerations, explain what "like dissolves like" means.
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