Question

A chemist wants to make 5.5 \(\mathrm{L}\) of a 0.300 \(\mathrm{M} \mathrm{CaCl}_{2}\) solution. What mass of \(\mathrm{CaCl}_{2}(\mathrm{in} \mathrm{g})\) should the chemist use?

Short answer

The chemist should use 110.5 g of \(\mathrm{CaCl}_{2}\).

Step-by-step solution

Determine the amount of substance needed

First, calculate the moles of \(\mathrm{CaCl}_{2}\) needed by using the molarity (M) formula, which is Molarity = moles of solute / liters of solution. Rearrange the formula to find the moles of solute, which gives us moles = Molarity \(\times\) Volume. For a 0.300 M solution in 5.5 L, multiply 0.300 moles/L by 5.5 L.

Calculate the molar mass of \(\mathrm{CaCl}_{2}\)

Find the molar mass of \(\mathrm{CaCl}_{2}\) by adding the atomic masses of calcium (Ca) and chlorine (Cl). Calcium has an atomic mass of approximately 40.08 g/mol, and chlorine has an atomic mass of approximately 35.45 g/mol. Since there are two chlorine atoms, the formula is 40.08 g/mol for Ca + 2(35.45 g/mol for Cl).

Calculate the mass of \(\mathrm{CaCl}_{2}\) needed

Multiply the moles of \(\mathrm{CaCl}_{2}\) calculated in Step 1 by the molar mass of \(\mathrm{CaCl}_{2}\) calculated in Step 2 to find the mass in grams needed to make the solution.

Key concepts

Chemical Molarity

Molarity is a crucial concept in chemistry, as it measures the concentration of a solution. It's defined as the number of moles of solute (the substance being dissolved) per liter of solution. The molarity formula is simple:
\( Molarity (M) = \frac{moles of solute}{liters of solution} \).
Understanding molarity is key for many laboratory tasks, including the preparation of chemical solutions with precise concentrations. When a chemist is asked to prepare a specified molarity, they're essentially determining how much of a substance to dissolve in a certain volume of solvent to achieve the desired concentration. Students often struggle with visualizing molarity, but it can be helpful to think of it as the 'strength' of the solution - the higher the molarity, the more 'potent' it is with the solute. To aid comprehension, imagine adding a spoonful of salt to a glass of water. The amount of water and salt will define the molarity of your saltwater solution.

Molar Mass Calculation

Molar mass is essentially the weight of one mole of a substance and is expressed in grams per mole (g/mol). It is calculated by summing the atomic masses of all atoms in a molecule, as found on the periodic table of elements.
To calculate molar mass, each element's atomic mass must be multiplied by the number of times it occurs in the molecule, and these are then added together. For example, calcium chloride (\(\mathrm{CaCl}_2\)) contains one calcium atom and two chlorine atoms. Therefore, the molar mass is the atomic mass of calcium plus twice the atomic mass of chlorine. In the case of \(\mathrm{CaCl}_2\), this is \(40.08\,\mathrm{g/mol} + 2 \times 35.45\,\mathrm{g/mol}\).
Understanding the concept of molar mass allows students to convert between the mass of a substance and the number of moles—a critical step in many chemical calculations, including reaction stoichiometry and solution preparation.

Solution Preparation

Preparing a chemical solution of a specific molarity requires careful calculation and precise measurement. The process can be summarized in three primary steps: determining the desired moles of solute, finding the molar mass of the solute, and measuring out the corresponding mass of solute to dissolve in solvent.
For instance, to prepare a 0.300 M \(\mathrm{CaCl}_2\) solution in 5.5 L, one would first calculate the required moles of \(\mathrm{CaCl}_2\), which in this case, is the molarity multiplied by the volume. Next, one would calculate the molar mass of \(\mathrm{CaCl}_2\), and finally, multiply the moles by the molar mass to get the required mass in grams. This mass of \(\mathrm{CaCl}_2\) would then be precisely weighed and dissolved in enough water to make 5.5 liters of solution.
Students should remember to account for significant figures and use clean measuring tools to ensure precision. A common pitfall in solution preparation is inaccurately measuring volumes or masses, which can lead to a solution's concentration being off from the intended molarity. Further, it's important to dissolve the solute completely before making up the final volume, as undissolved particles can lead to inaccuracies.