For which of the following compounds is it possible to make...

For which of the following compounds is it possible to make a $1.0 M$ solution at $20^{\circ} \mathrm{C} ?$ a. $\mathrm{CuSO}_{4},$ solubility $=32.0 \mathrm{g} / 100 \mathrm{mL}$ b. $\mathrm{Ba}(\mathrm{OH})_{2},$ solubility $=3.9 \mathrm{g} / 100 \mathrm{mL}$ c. $\mathrm{FeCl}_{2}$, solubility $=68.5 \mathrm{g} / 100 \mathrm{mL}$ d. $\mathrm{Ca}(\mathrm{OH})_{2},$ solubility $=0.173 \mathrm{g} / 100 \mathrm{mL}$

For which of the following compounds is it possible to make
a $1.0 M$ solution at $20^{\circ} \mathrm{C} ?$
a. $\mathrm{CuSO}_{4},$ solubility $=32.0 \mathrm{g} / 100 \mathrm{mL}$
b. $\mathrm{Ba}(\mathrm{OH})_{2},$ solubility $=3.9 \mathrm{g} / 100 \mathrm{mL}$
c. $\mathrm{FeCl}_{2}$, solubility $=68.5 \mathrm{g} / 100 \mathrm{mL}$
d. $\mathrm{Ca}(\mathrm{OH})_{2},$ solubility $=0.173 \mathrm{g} / 100 \mathrm{mL}$

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Key Concepts

Molar Mass Calculations

Calculating the molar mass of a compound allows the conversion from mass (in grams) to moles. This is crucial in evaluating if the available solubility in grams per liter is sufficient to achieve the desired molarity for the solution.

Unit Conversions

Converting units (e.g., grams per 100 mL to grams per liter) is fundamental when relating solubility data to molarity. This step ensures that all measurements are in compatible units so that calculations involving molar mass and volume are accurate.

Molarity

Molarity is a concentration unit that represents the number of moles of solute per liter of solution. In problems involving solution concentration, it is critical to convert between the mass of a compound dissolved and the molar amount based on its molar mass.

Solubility

Solubility indicates how much solute can dissolve in a given amount of solvent at a specific temperature, often expressed in grams per 100 mL. Understanding solubility is essential for determining whether a required concentration, such as 1.0 M, is achievable under given conditions.

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