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Chapter 13: Problem 98

Acetonitrile \(\left(\mathrm{CH}_{3} \mathrm{CN}\right)\) is a polar organic solvent that dissolves a wide range of solutes, including many salts. The density of a \(1.80 \mathrm{M}\) LiBr solution in acetonitrile is \(0.826 \mathrm{~g} / \mathrm{cm}^{3} .\) Calculate the concentration of the solution in (a) molality, (b) mole fraction of \(\mathrm{LiBr},(\mathrm{c})\) mass percentage of \(\mathrm{CH}_{3} \mathrm{CN}\).

Short Answer

Expert verified
The concentration of the LiBr solution in acetonitrile is: (a) \(2.69 \, mol/kg\) (molality) (b) \(0.0994\) (mole fraction of LiBr) (c) \(81.08\%\) (mass percentage of CH3CN).

Step by step solution

01

Determine the mass of LiBr and CH3CN in the solution

First, we need to find the mass of each component in the solution. We are given that the molarity of the solution is 1.80 M, which means there are 1.80 moles of LiBr in every 1 liter of the solution. We also know the density of the solution is 0.826 g/cm³. We can use these values to find the total mass of solution in 1 liter: 1 liter = 1000 cm³ Total mass of solution = Volume × Density = (1000 cm³) × (0.826 g/cm³) = 826 g Now we need to find the mass of LiBr in 1 liter of the solution: Molar mass of LiBr = 6.94 g/mol (Li) + 79.9 g/mol (Br) = 86.84 g/mol Mass of LiBr = Moles × Molar mass = (1.80 moles) × (86.84 g/mol) = 156.31 g To find the mass of CH3CN in 1 liter of the solution, we can subtract the mass of LiBr from the total mass of the solution: Mass of CH3CN = Total mass of solution - Mass of LiBr = 826 g - 156.31 g = 669.69 g
02

Calculate the molality of the solution

Now that we have the mass of LiBr and CH3CN, we can calculate the molality of the solution. Molality is defined as the moles of solute (LiBr) per kilogram of solvent (CH3CN). First, we need to convert the mass of CH3CN to kilograms: Mass of CH3CN = 669.69 g × (1 kg/1000 g) = 0.66969 kg Now we can calculate the molality: Molality = Moles of LiBr / Mass of CH3CN in kg = (1.80 moles) / (0.66969 kg) = 2.69 mol/kg
03

Determine the mole fraction of LiBr

Next, we will calculate the mole fraction of LiBr in the solution. To do this, we need to find the moles of CH3CN first. Molar mass of CH3CN = 12.01 g/mol (C) + 1.01 g/mol (H) * 3 + 14.01 g/mol (N) = 41.05 g/mol Moles of CH3CN = Mass of CH3CN / Molar mass of CH3CN = 669.69 g / 41.05 g/mol = 16.31 moles We can now find the mole fraction of LiBr: Mole fraction of LiBr = Moles of LiBr / (Moles of LiBr + Moles of CH3CN) = 1.80 moles / (1.80 moles + 16.31 moles) = 0.0994
04

Calculate the mass percentage of CH3CN

Finally, we will calculate the mass percentage of CH3CN in the solution. Mass percentage of CH3CN = (Mass of CH3CN / Total mass of solution) × 100 = (669.69 g / 826 g) × 100 = 81.08 % So, the concentration of the LiBr solution in acetonitrile is: (a) 2.69 mol/kg (molality) (b) 0.0994 (mole fraction of LiBr) (c) 81.08 % (mass percentage of CH3CN).

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Key Concepts

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

Understanding Molality
Molality is a measure of the concentration of a solution in terms of the amount of a substance (the solute) dissolved in a specific mass of the solvent. Unlike molarity, molality is not affected by changes in temperature, which makes it particularly useful for studying solutions where temperatures can fluctuate.

Molality is calculated with the formula:
\[ \text{molality} = \frac{\text{moles of solute}}{\text{kilograms of solvent}} \]
It's essential to take note that when converting grams to kilograms in the solvent's mass, a kilogram equates to 1,000 grams. Helpful tip: Always double-check your units to make sure you convert from grams to kilograms when necessary to calculate molality correctly.
Deciphering Mole Fraction
Mole fraction is a ratio representing the number of moles of a component relative to the total number of moles in a mixture. The beauty of mole fraction is that it's a dimensionless quantity—that is, it has no units.

To compute the mole fraction, use the formula:
\[ \text{mole fraction} = \frac{\text{moles of component}}{\text{total moles in mixture}} \]
  • The sum of all mole fractions in a solution always equals 1.
  • Mole fraction is particularly useful when dealing with gas mixtures and solutions where the precise composition is required.
Remember to find the total moles by adding the moles of the solute and the solvent together before calculating the mole fraction.
Mass Percentage Made Simple
When speaking of mass percentage (or weight percentage), we're looking at the concentration of a component represented by the mass of that component relative to the total mass of the mixture, expressed as a percentage.

The formula to get the mass percentage is:
\[ \text{mass percentage} = \left(\frac{\text{mass of component}}{\text{total mass of solution}}\right) \times 100\% \]
There are a few pointers to keep in mind with mass percentage:
  • There are no unit conversions required—the result is always a percentage.
  • It’s an easily understood concentration metric because it directly tells you ‘how much’ of the substance is in the solution compared to the whole.
Ensuring that the mass of the solution includes all its components will enable accurate calculation of mass percentages.
Mastering Molarity
Molarity, often denoted as M, is the number of moles of solute per liter of solution. It's one of the most common units that you'll encounter when dealing with chemical solutions.

To calculate molarity:
\[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]
Here are some considerations for molarity:
  • Volume is temperature-dependent, meaning molarity can change with temperature as the solution expands or contracts.
  • In the context of textbook solutions, it's often given that you'll need to work backwards using molarity to find moles or liters if not provided.
Remember that when you are given density and molarity, like in the example problem, you can find the total mass of the solution by multiplying the volume by the density, and then work further to find the masses of individual components if needed.

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Most popular questions from this chapter

The solubility of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) in water at \(20^{\circ} \mathrm{C}\) is \(70 \mathrm{~g}\) per \(100 \mathrm{~mL}\) of water. (a) Is a \(1.22 \mathrm{M}\) solution of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) in water at \(20^{\circ} \mathrm{C}\) saturated, supersaturated, or unsaturated? (b) Given a solution of \(\mathrm{MnSO}_{4} \cdot \mathrm{H}_{2} \mathrm{O}\) of unknown concentration, what experiment could you perform to determine whether the new solution is saturated, supersaturated, or unsaturated?

By referring to Figure 13.18 , determine whether the addition of \(40.0 \mathrm{~g}\) of each of the following ionic solids to \(100 \mathrm{~g}\) of water at \(40^{\circ} \mathrm{C}\) will lead to a saturated solution: (a) \(\mathrm{NaNO}_{3},\) (b) KCl, (c) \(\mathrm{K}_{2} \mathrm{Cr}_{2} \mathrm{O}_{7}\) (d) \(\mathrm{Pb}\left(\mathrm{NO}_{3}\right)_{2}\)

How does the lattice energy of an ionic solid affect its solubility in water? [Section 13.1\(]\)

List four properties of a solution that depend on the total concentration but not the type of particle or particles present as solute. Write the mathematical expression that describes how each of these properties depends on concentration.

(a) What is the molality of a solution formed by dissolving \(1.12 \mathrm{~mol}\) of \(\mathrm{KCl}\) in \(16.0 \mathrm{~mol}\) of water? (b) How many grams of sulfur \(\left(\mathrm{S}_{8}\right)\) must be dissolved in \(100.0 \mathrm{~g}\) naphthalene \(\left(\mathrm{C}_{10} \mathrm{H}_{8}\right)\) to make a \(0.12 \mathrm{~m}\) solution?
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