Problem 25 - Chemistry: An Atoms-Focused Approach

Question

For which of the following compounds is it possible to make a \(1.0 M\) solution at \(0^{\circ} \mathrm{C} ?\) a. \(\mathrm{CuSO}_{4} \cdot 5 \mathrm{H}_{2} \mathrm{O},\) solubility \(=23.1 \mathrm{g} / 100 \mathrm{mL}\) b. \(A g N O_{3},\) solubility \(=122\) g/ 100 mL c. \(\mathrm{Fe}\left(\mathrm{NO}_{3}\right)_{2} \cdot 6 \mathrm{H}_{2} \mathrm{O},\) solubility \(=113 \mathrm{g} / 100 \mathrm{mL}\) d. \(\mathrm{Ca}(\mathrm{OH})_{2}\), solubility \(=0.185 \mathrm{g} / 100 \mathrm{mL}\)

Step-by-Step Solution

Answer

a. CuSO4·5H2O b. AgNO3 c. Fe(NO3)2·6H2O d. Ca(OH)2

Step 1: Calculate the molar mass of each compound

To calculate the molar mass of each compound, first, determine the atomic mass of each element in the compound and then sum them up. For simplicity, we will not show the full calculation for each compound. Using a periodic table, this is done as follows: a. CuSO4·5H2O: Molar mass \(\approx 249.7 \mathrm{g/mol}\). b. AgNO3: Molar mass \(\approx 169.9 \mathrm{g/mol}\). c. Fe(NO3)2·6H2O: Molar mass \(\approx 179.9 \mathrm{g/mol}\). d. Ca(OH)2: Molar mass \(\approx 74.1 \mathrm{g/mol}\).

Step 2: Calculate the maximum concentration of each compound at \(0^{\circ}\mathrm{C}\)"

To do this, divide the solubility by the molar mass and then convert the result from moles per 100 mL of solution to moles per 1 L (moles/L or M): a. CuSO4·5H2O: \(\text{Maximum Concentration} = \dfrac{23.1 \mathrm{g}}{249.7 \mathrm{g/mol}} \cdot \dfrac{1000 \mathrm{mL}}{100 \mathrm{mL}} = \dfrac{23.1}{249.7} * 10 \approx 0.93 \text{ M}\). b. AgNO3: \(\text{Maximum Concentration} = \dfrac{122 \mathrm{g}}{169.9 \mathrm{g/mol}} \cdot \dfrac{1000 \mathrm{mL}}{100 \mathrm{mL}} = \dfrac{122}{169.9} * 10 \approx 7.18 \text{ M}\). c. Fe(NO3)2·6H2O: \(\text{Maximum Concentration} = \dfrac{113 \mathrm{g}}{179.9 \mathrm{g/mol}} \cdot \dfrac{1000 \mathrm{mL}}{100 \mathrm{mL}} = \dfrac{113}{179.9} * 10 \approx 6.28 \text{ M}\). d. Ca(OH)2: \(\text{Maximum Concentration} = \dfrac{0.185 \mathrm{g}}{74.1 \mathrm{g/mol}} \cdot \dfrac{1000 \mathrm{mL}}{100 \mathrm{mL}} = \dfrac{0.185}{74.1} * 10 \approx 0.025 \text{ M}\).

Step 3: Compare the maximum concentration to the desired 1.0 M concentration

We now compare the maximum concentration of each compound (calculated in Step 2) to the desired concentration of \(1.0\,\text{M}\). If the maximum concentration is equal to or greater than 1.0, it is possible to make a \(1.0\) M solution at \(0^{\circ}\mathrm{C}\). Otherwise, it is not possible. a. CuSO4·5H2O: Maximum concentration: \(0.93 \text{ M}\). Not possible. b. AgNO3: Maximum concentration: \(7.18 \text{ M}\). Possible. c. Fe(NO3)2·6H2O: Maximum concentration: \(6.28 \text{ M}\). Possible. d. Ca(OH)2: Maximum concentration: \(0.025 \text{ M}\). Not possible. In conclusion, it is possible to make a \(1.0\,\text{M}\) solution at \(0^{\circ}\mathrm{C}\) for the compounds b (AgNO3) and c (Fe(NO3)2·6H2O).

Key Concepts

Molar Mass Calculation

Understanding how to calculate the molar mass of a compound is crucial in determining how much of a compound is needed to achieve a certain concentration. Each element has a specific atomic mass, which can be found on the periodic table. To determine the molar mass of a compound, you sum up the atomic masses of all elements in that compound, taking into account the number of atoms of each element in its chemical formula.

• For example, in the compound CuSO₄·5H₂O, we need the atomic masses of Cu, S, O, and H. Calculate it as: Cu (63.55 g/mol) + S (32.07 g/mol) + O₄ (4 x 16.00 g/mol) + 5H₂O (5 x (2 x 1.01 + 16.00) g/mol).
• Once calculated, the molar mass of CuSO₄·5H₂O is approximately 249.7 g/mol.

This calculated molar mass is essential for converting the amount of solubility from grams to moles, which is the next step in concentration calculations.

Maximum Concentration Calculation

Once the molar mass of a compound is known, we can determine the maximum concentration of that compound in a given volume of solvent at a specific temperature. This involves using the solubility of the compound, usually given in grams per 100 mL, and converting it to moles per liter (Molarity, M).

• First, the solubility in grams is divided by the molar mass to convert to moles.
• Then, convert the volume from 100 mL to 1 liter by multiplying the result by 10 (since 1000 mL = 1 L).

For example, if AgNO₃ has a solubility of 122 g/100 mL, its molar mass is 169.9 g/mol. The calculation for maximum concentration is: \(\frac{122}{169.9} \times 10\), which results in roughly 7.18 M.

Chemical Solutions

In chemistry, a solution is a homogenous mixture where solutes are dissolved in a solvent, like water. Understanding solutions goes beyond knowing that they are mixtures; it involves comprehending the behavior of solutes and solvents at a molecular level.

• Concentration is key, quantified as Molarity (M), defined as moles of solute per liter of solution.
• Compounds dissolve at certain rates depending on properties like temperature and pressure.
• Not all compounds have the same solubility, which influences how solutions are prepared and utilized in various applications.

Solutions are foundational in chemical reactions, with their properties significantly affecting reaction rates and outcomes.

Solubility at Specific Temperatures

Solubility is the maximum amount of solute that can dissolve in a solvent at a given temperature and pressure. The solubility often depends heavily on temperature; a compound may be much more soluble in a warm solution than in a cold one.

• It's measured as grams of solute per 100 mL of solvent, simplified further into molarity when needed.
• Understanding solubility can indicate which solution concentrations are feasible at certain temperatures, such as in the exercise where measurements are at \(0^{\circ}C\).

At \(0^{\circ}C\), some compounds dissolve sufficiently to create concentrated solutions, while others may not reach desired concentrations without temperature adjustments.