Question

What volume of a \(0.580-M\) solution of \(\mathrm{CaCl}_{2}\) contains \(1.28 \mathrm{~g}\) solute?

Short answer

The volume of a \(0.580-M\) solution of \(CaCl_{2}\) that contains \(1.28 g\) solute is approximately \(2.02 mL\).

Step-by-step solution

Calculate moles of solute

First, we need to convert the mass of the solute (1.28 g) to moles. For this, we have to find the molar mass of CaCl2. The molar mass of Ca (calcium) is 40.08 g/mol, and that of Cl (chlorine) is 35.45 g/mol. Molar mass of CaCl2 = 40.08 + (2 * 35.45) = 40.08 + 70.9 = 110.98 g/mol Now divide the mass of solute by the calculated molar mass: moles of solute = mass of solute / molar mass of CaCl2 moles = 1.28 g / 110.98 g/mol

Calculate volume of the solution

Now, we can use the provided molarity of the solution to find the volume: Molarity = moles of solute / volume of solution 0.580 M = (1.28 g / 110.98 g/mol) / volume Volume = (1.28 g / 110.98 g/mol) / 0.580 M After calculating, we get: Volume ≈ 0.00202 L Now convert the volume to milliliters: Volume ≈ 0.00202 L × (1000 mL / 1 L) ≈ 2.02 mL

Report the result

So, the volume of a 0.580-M solution of CaCl2 that contains 1.28 g solute is approximately 2.02 mL.

Key concepts

Understanding Moles of Solute

To grasp the concept of molarity, it's essential to understand the idea of moles first. A mole is a quantity unit in chemistry that measures the amount of a substance. It's similar to how a dozen means twelve items. However, in moles, we deal with an incredibly large number called Avogadro's number: approximately \(6.022 \times 10^{23}\).

When we talk about moles of solute, we're referring to the number of moles of the substance that has been dissolved in a solvent. In the example exercise, we need to find the number of moles of calcium chloride (\(\text{CaCl}_2\)) from a given mass.

This is done by dividing the mass of the solute by the molar mass to convert the mass into moles. For 1.28 grams of \(\text{CaCl}_2\), use the molar mass of \(110.98 \text{ g/mol}\) to find the moles. In equation form, it looks like this:

• Moles of solute = \( \frac{1.28\, \text{g}}{110.98\, \text{g/mol}} \)

This conversion is crucial as it sets the foundation for further calculations involving concentration and volume.

Unlocking the Molar Mass

Molar mass is another fundamental concept key to chemistry. It gives us the mass of one mole of a substance, generally expressed in grams per mole (g/mol). The molar mass is calculated by adding up the atomic masses of each element in the compound from the periodic table.

In the context of \(\text{CaCl}_2\), this requires understanding the atomic masses of calcium and chlorine. From the periodic table, calcium has an atomic mass of \(40.08 \text{ g/mol}\), and for chlorine, it's \(35.45 \text{ g/mol}\). Since \(\text{CaCl}_2\) has one calcium atom and two chlorine atoms, you calculate its molar mass as follows:

• \(\text{Molar mass of CaCl}_2 = 40.08 + (2 \times 35.45) = 110.98 \text{ g/mol} \)

Molar mass is a bridge between the macroscopic world (mass we can measure) and the microscopic world (moles that represent a count of atoms or molecules), enabling conversions that are central to stoichiometric calculations.

Exploring Volume of Solution

Volume of solution is another pivotal concept when dealing with solutions in chemistry. It refers to the total space that the solution occupies, and is usually measured in liters or milliliters. To calculate it from a given mass of solute and molarity, we need to manipulate the formula for molarity.

Molarity (\(M\)) is defined as the ratio of moles of solute to the volume of solution in liters. Therefore, we can rearrange the formula to solve for the volume of solution:

• \(\text{Volume of solution} = \frac{\text{Moles of solute}}{\text{Molarity}}\)

Using the given molarity \(0.580 \text{ M}\), and moles of \(\text{CaCl}_2\) calculated earlier, the volume can be derived. We end up with a solution volume of approximately \(0.00202 \text{ L}\) or \(2.02 \text{ mL}\) once converted.

Understanding volume calculations is invaluable for preparing chemical solutions accurately in a laboratory setting, ensuring precision across various experiments.