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Chapter 5: Problem 159

Density of a \(2.05 \mathrm{M}\) solution of acetic acid in water is \(1.02 \mathrm{~g} / \mathrm{mL}\). The molality of the solution is (a) \(1.14 \mathrm{~mol} \mathrm{~kg}^{-1}\) (b) \(3.28 \mathrm{~mol} \mathrm{~kg}^{-1}\) (c) \(2.28 \mathrm{~mol} \mathrm{~kg}^{-1}\) (d) \(0.44 \mathrm{~mol} \mathrm{~kg}^{-1}\)

Short Answer

Expert verified
The molality of the solution is \(2.28 \mathrm{~mol} \mathrm{~kg}^{-1}\). Option (c).

Step by step solution

01

Understand Density and Molarity

We know the molarity of acetic acid is given as \(2.05 \mathrm{M}\). This means there are \(2.05\) moles of acetic acid in \(1\) liter of solution. The density of the solution is \(1.02 \mathrm{~g/mL}\), which can also be expressed as \(1.02 \times 10^3 \mathrm{~g/L}\). This means \(1\) liter of solution weighs \(1020 \mathrm{~g}\).
02

Calculate Mass of Solvent

To find the mass of the solvent, we first calculate the mass of acetic acid in \(1\) liter of solution. The molecular weight of acetic acid \((\mathrm{CH}_3\mathrm{COOH})\) is approximately \(60 \mathrm{~g/mol}\). Therefore, the mass of acetic acid is calculated as follows:\[ \text{Mass of acetic acid} = 2.05 \times 60 = 123 \mathrm{~g} \]The mass of the solvent (water) is the total mass of the solution minus the mass of acetic acid:\[ \text{Mass of water} = 1020 - 123 = 897 \mathrm{~g} \] or \(0.897 \mathrm{~kg}\).
03

Calculate Molality

Molality is defined as the number of moles of solute divided by the mass (in kg) of the solvent:\[ \text{Molality} = \frac{2.05 \text{ moles}}{0.897 \text{ kg}} = 2.28 \mathrm{~mol} \mathrm{~kg}^{-1} \]
04

Choose the Correct Option

The calculated molality is \(2.28 \mathrm{~mol} \mathrm{~kg}^{-1}\), which matches option \(c\). Therefore, the correct answer is \(c\).

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Key Concepts

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

Molarity
Molarity is a fundamental concept in chemistry, important for understanding the concentration of solutions in terms of volume. When we talk about molarity, represented by the symbol \( M \), we refer to the number of moles of solute present in one liter of solution.
  • In the context of acetic acid in water, a \( 2.05 \mathrm{M} \) solution means there are \( 2.05 \) moles of acetic acid dissolved per liter.
  • Molarity serves as a convenient way to express concentration, particularly for reactions in aqueous solutions.
  • It helps in calculating how much of a substance is needed or produced during chemical reactions.
To find the molarity, one typically divides the number of moles of the solute by the total volume of the solution in liters. For our exercise, the molarity was utilized to determine how much acetic acid was present in a defined volume, progressing our understanding of the solution's characteristics. Overall, molarity is a key tool in quantifying chemical solutions, beneficial for analytical and practical applications.
Acetic Acid
Acetic acid, chemically represented as \( \mathrm{CH}_3\mathrm{COOH} \), is a simple carboxylic acid known for its distinct acidic properties and vinegary smell. It's a significant industrial chemical and an essential component in vinegar.
  • Its molecular weight is approximately \( 60 \mathrm{~g/mol} \).
  • In our exercise, knowing this molecular weight helped in calculating the mass of acetic acid present in the solution.
  • Acetic acid is a weak acid, partially dissociating in water, which means its molecules do not completely split into ions.
  • This weak acidic nature impacts the calculations of reacting species in solutions.
Acetic acid has numerous applications, such as in the production of acetate salts, vinyl acetate monomer, and as a solvent in the chemical industry. Understanding acetic acid's role and characteristics is crucial for both industrial processes and laboratory analyses, including the calculations of concentration and the preparation of solutions.
Density of Solutions
Density is a measure of how much mass is contained within a given volume. For solutions, knowing the density helps to convert between mass and volume, offering a bridge between weight-based and volume-based concentrations.
  • The density of the acetic acid solution given in the exercise is \( 1.02 \mathrm{~g/mL} \).
  • This was used to determine that \( 1 \) liter (\( 1000 \mathrm{~mL} \)) of solution has a mass of \( 1020 \mathrm{~g} \).
  • Density reflects the compactness of molecules within the solution, important for both chemical and physical processes.
Calculating with density allows us to translate answers more adeptly between moles, masses, and volumes. In many chemical calculations, having density information is crucial when dealing with physical properties alongside chemical properties, enabling a clearer understanding of the composition of solutions.
Mass of Solvent
The mass of the solvent in any solution plays a key role in determining various concentration measures, like molality. In our exercise, the solvent was water, considered when calculating the molality of the acetic acid solution.
  • To find the mass of the solvent, subtract the mass of the solute from the total mass of the solution.
  • In this case, \( 1020 \mathrm{~g} \) was the mass of the entire solution, while \( 123 \mathrm{~g} \) was the mass of acetic acid.
  • The remaining \( 897 \mathrm{~g} \) or \( 0.897 \mathrm{~kg} \) represents the water mass, serving as the solvent.
Understanding the mass of the solvent is crucial for correctly using formulas like molality, as molality depends on the mass of the solvent in kilograms. This knowledge helps in achieving accurate results when working with different solutions and their various properties, ensuring precision in experimental chemistry.

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Most popular questions from this chapter

Which pair of the following will not form an ideal solution? (a) \(\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{Br}+\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{I}\) (b) \(\mathrm{H}_{2} \mathrm{O}+\mathrm{C}_{4} \mathrm{H}_{9} \mathrm{OH}\) (c) \(\mathrm{CCl}_{4}+\mathrm{SiCl}_{4}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{14}+\mathrm{C}_{7} \mathrm{H}_{16}\)

The colligative properties of electrolytes require a slightly different approach than the one used for the colligative properties of non- electrolytes. The electrolytes dissociate into ions in solution. It is the number of solute particles that determine the colligative properties of a solution. The electrolyte solutions, therefore show abnormal colligative properties. To account for this effect we define a quantity; called the van't Hoff factor which is given by [solution] \(i=\) \(\frac{\text { Actual number of particles in solution after dissociation }}{\text { Number of formula units initally dissolved in solution }}\) \(\mathrm{i}=1\) (for non - electrolytes); \(\mathrm{i}>1\) (for electrolytes, undergoing dissociation) \(\mathrm{i}<1\) (for solute, undergoing association) \(0.1 \mathrm{M} \mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) is \(60 \%\) ionized. What will be its van't Hoff factor? (a) \(3.4\) (b) \(1.7\) (c) \(2.4\) (d) \(2.2\)

The beans are cooked sooner in a pressure cooker, because (a) boiling point increase with increasing pressure (b) boiling point decrease with increasing pressure (c) extra pressure of pressure cooker, softens the beans (d) internal energy is not lost while cooking in pressure cooker.

Ethylene glycol is used as antifreeze in a cold climate. Mass of ethylene glycol which should be added to \(4 \mathrm{~kg}\) of water to prevent it form freezing at \(-6^{\circ} \mathrm{C}\) will be: \(\left(\mathrm{K}_{f}\right.\) for water \(=1.86 \mathrm{~kg} \mathrm{~mol}^{-1}\), and molar mass of ethylene glycol \(=62 \mathrm{~g} \mathrm{~mol}^{-1}\) ) (a) \(204.11 \mathrm{~g}\) (b) \(804.32 \mathrm{~g}\) (c) \(600.20 \mathrm{~g}\) (d) \(302.40 \mathrm{~g}\)

The freezing point of equimolal aqueous solution will be highest for (a) \(\mathrm{C}_{6} \mathrm{H}_{5} \mathrm{~N}^{+} \mathrm{H}_{3} \mathrm{Cl}^{-}\)(aniline hydrochloride) (b) \(\mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) (c) \(\mathrm{La}\left(\mathrm{NO}_{3}\right)_{3}\) (d) \(\mathrm{C}_{6} \mathrm{H}_{12} \mathrm{O}_{6}\) (glucose)
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