Question

Write two pairs of unit factors for the following aqueous solutions given the molar concentration. (a) \(0.100 M\) LiI (b) \(0.100 \mathrm{M} \mathrm{NaNO}_{3}\) (c) \(0.500 \mathrm{MK}_{2} \mathrm{CrO}_{4}\) (d) \(0.500 \mathrm{M} \mathrm{ZnSO}_{4}\)

Short answer

Each molar concentration has two unit factors: moles per liter of solution and liters of solution per moles.

Step-by-step solution

Understanding Molar Concentration

Molarity (M) is defined as the number of moles of solute per liter of solution. For each given concentration, such as 0.100 M LiI, this means there are 0.100 moles of LiI in every 1 liter of solution.

Constructing Unit Factors - Part (a)

For 0.100 M LiI, the unit factors can be expressed as: 1) \( \frac{0.100 \, ext{mol LiI}}{1 \, ext{L solution}} \) 2) \( \frac{1 \, ext{L solution}}{0.100 \, ext{mol LiI}} \).

Constructing Unit Factors - Part (b)

For 0.100 M NaNO₃, the unit factors can be expressed as: 1) \( \frac{0.100 \, ext{mol NaNO}_3}{1 \, ext{L solution}} \) 2) \( \frac{1 \, ext{L solution}}{0.100 \, ext{mol NaNO}_3} \).

Constructing Unit Factors - Part (c)

For 0.500 M K₂CrO₄, the unit factors can be expressed as: 1) \( \frac{0.500 \, ext{mol K}_2 ext{CrO}_4}{1 \, ext{L solution}} \) 2) \( \frac{1 \, ext{L solution}}{0.500 \, ext{mol K}_2 ext{CrO}_4} \).

Constructing Unit Factors - Part (d)

For 0.500 M ZnSO₄, the unit factors can be expressed as: 1) \( \frac{0.500 \, ext{mol ZnSO}_4}{1 \, ext{L solution}} \) 2) \( \frac{1 \, ext{L solution}}{0.500 \, ext{mol ZnSO}_4} \).

Key concepts

Understanding Unit Factors

Unit factors are a fundamental concept in chemistry that can help you convert between different units. They are especially useful when working with solutions and concentrations. In the context of molarity, unit factors allow you to convert between moles of solute and liters of solution. This method helps in performing stoichiometric calculations by relating two quantities through a conversion factor.
For example, if you have a solution with a concentration of 0.100 M, meaning it has 0.100 moles of solute per liter, your unit factors would look like this:

• \( \frac{0.100 \, \text{mol}}{1 \, \text{L}} \) means there are 0.100 moles of a solute in every liter of solution.
• \( \frac{1 \, \text{L}}{0.100 \, \text{mol}} \) allows you to convert from moles to liters if needed.

This approach is flexible and widely used for calculations in both academic and practical chemistry applications.

Exploring Molar Concentration

Molarity, denoted by the unit M, refers to molar concentration. It describes how much solute is present in a given volume of solution. Specifically, it is defined as the number of moles of solute per liter of solution.
To calculate molarity, you would use the formula: \[ \text{Molarity (M)} = \frac{\text{moles of solute}}{\text{liters of solution}} \]Understanding molarity is crucial as it directly impacts the reactivity and interactions within a solution. For instance, a 0.500 M solution of \( K_2CrO_4 \) indicates that in every liter of this solution, there are 0.500 moles of \( K_2CrO_4 \).
This concept is foundational in titrations, buffer preparations, and various chemical analyses. Molarity not only tells us the concentration but aids in predicting how reactions will proceed under given conditions.

Defining Aqueous Solutions

An aqueous solution is a solution where water is the solvent. In chemistry, these solutions are pivotal as many reactions occur in water. A solute, like \( NaNO_3 \) or \( ZnSO_4 \), is dissolved in water to form the solution. The prefix 'aqueous' (\( \text{aq} \)) indicates that the solvent involved is water.
These solutions exhibit various properties, like conductivity and pH, influenced by the solute's nature. For example, ionic compounds like \( LiI \) when dissolved, dissociate into ions, which can conduct electricity, making them conductive.

• An aqueous solution of \( 0.500 \, M \, ZnSO_4 \) means it has 0.500 moles of \( ZnSO_4 \) dissociated in each liter of water.
• These solutions play a critical role in biological systems and industrial processes due to water's unique properties as a solvent.

Understanding aqueous solutions is essential for chemists who analyze reaction mechanisms and material properties.