Question

A \(325-\mathrm{mL}\) sample of solution contains \(25.3 \mathrm{~g}\) of \(\mathrm{CaCl}_{2}\) (a) Calculate the molar concentration of \(\mathrm{Cl}^{-}\) in this solution. (b) How many grams of \(\mathrm{Cl}^{-}\) are in \(0.100 \mathrm{~L}\) of this solution?

Short answer

(a) The molar concentration of \(\mathrm{Cl}^{-}\) is approximately 1.40 M. (b) There are about 4.96 grams of \(\mathrm{Cl}^{-}\) in 0.100 L of the solution.

Step-by-step solution

Calculate Moles of CaCl2

First, we need to find the number of moles of \(\mathrm{CaCl}_{2}\) in the solution. Using the molar mass of \(\mathrm{CaCl}_{2}\), which is approximately \(110.98\,\text{g/mol}\), the formula for moles is given by: \( n = \frac{\text{mass}}{\text{molar mass}} \). Thus, the number of moles of \(\mathrm{CaCl}_{2}\) is: \[ n = \frac{25.3\,\text{g}}{110.98\,\text{g/mol}} \approx 0.228\,\text{mol}.\]

Relate Moles of CaCl2 to Moles of Cl-

In \(\mathrm{CaCl}_{2}\), each formula unit produces 2 ions of \(\mathrm{Cl}^{-}\). Therefore, the number of moles of \(\mathrm{Cl}^{-}\) is twice the number of moles of \(\mathrm{CaCl}_{2}\): \[ n_{\mathrm{Cl}^{-}} = 2 \times 0.228\,\text{mol} = 0.456\,\text{mol}.\]

Calculate Molar Concentration of Cl-

To find the molarity, or molar concentration, of \(\mathrm{Cl}^{-}\) ions, divide the moles of \(\mathrm{Cl}^{-}\) by the volume of the solution in liters. The given volume is \(325\,\text{mL}\), which is equivalent to \(0.325\,\text{L}\). Now, calculate the concentration: \[ [\mathrm{Cl}^{-}] = \frac{0.456\,\text{mol}}{0.325\,\text{L}} \approx 1.40\,\text{M}.\]

Calculate Grams of Cl- in 0.100 L of Solution

Since we have the concentration of \(\mathrm{Cl}^{-}\) as \(1.40\,\text{M}\), we can find out how many moles of \(\mathrm{Cl}^{-}\) are in \(0.100\,\text{L}\) of the solution using the formula: \[ \text{Moles of } \mathrm{Cl}^{-} = 1.40 \times 0.100 = 0.140\,\text{mol}.\]Now, convert these moles into grams using the molar mass of \(\mathrm{Cl}\) which is approximately \(35.45\,\text{g/mol}\):\[ \text{Mass of } \mathrm{Cl}^{-} = 0.140\,\text{mol} \times 35.45\,\text{g/mol} \approx 4.96\,\text{g}.\]

Key concepts

CaCl2

Calcium chloride, commonly written as \(\mathrm{CaCl}_{2}\), is an ionic compound comprised of calcium and chloride ions. It plays a significant role in many chemical and industrial applications because of its properties.

• The calcium ion carries a charge of +2, while each chloride ion carries a charge of -1.
• In the compound \(\mathrm{CaCl}_{2}\), there are two chloride ions for every calcium ion.
• This implies that the dissociation of \(\mathrm{CaCl}_{2}\) in water results in two chloride ions being released for each formula unit of calcium chloride.

Understanding this dissociation is crucial when calculating the number of moles of chloride ions in a solution. By knowing the amount of \(\mathrm{CaCl}_{2}\) present, one can determine how many chloride ions are present through simple multiplication.

moles

The concept of moles is a fundamental building block in chemistry. A mole is a unit used to express the amount of a chemical substance.

• The formula used to calculate moles is: \[ n = \frac{\text{mass}}{\text{molar mass}} \]
• In this exercise, the mass of \(\mathrm{CaCl}_{2}\) is given, and the molar mass (approximately \(110.98\,\text{g/mol}\)) is used to find the moles.
• Thus, the number of moles tells us how many units of \(\mathrm{CaCl}_{2}\) are present in terms of Avogadro's number, which is \(6.022 \times 10^{23}\) formula units per mole.

Calculating moles helps us convert given information into a form that can be easily manipulated to find concentrations and other required chemical quantities.

chloride ions

Chloride ions, \(\mathrm{Cl}^{-}\), are negatively charged ions that significantly affect the chemical behavior of a solution. When considering \(\mathrm{CaCl}_{2}\), it is important to determine how many chloride ions are released when the compound dissociates. Since one formula unit of \(\mathrm{CaCl}_{2}\) yields two chloride ions, one can quickly calculate the moles of chloride ions given the moles of \(\mathrm{CaCl}_{2}\).

• For every mole of \(\mathrm{CaCl}_{2}\), there are two moles of \(\mathrm{Cl}^{-}\).
• This stoichiometric relationship simplifies the calculation of chloride ion concentration, as evidenced by multiplying the moles of \(\mathrm{CaCl}_{2}\) by 2.

Knowing the moles of chloride ions aids in determining the molar concentration, which ultimately tells us how many of these ions are present in a given volume.

solution volume

Solution volume is an essential parameter in calculating concentrations. In chemistry, solution volumes are often expressed in liters, since molarity (molar concentration) is defined as moles of solute per liter of solution.

• In this exercise, the original volume of the solution is turned from \(325\,\text{mL}\) to \(0.325\,\text{L}\).
• Converting milliliters to liters simply involves dividing by 1000, as there are 1000 milliliters in a liter.
• The formula for molarity \([\mathrm{Cl}^{-}] = \frac{\text{moles of } \mathrm{Cl}^{-}}{\text{volume in liters}}\) revolves around precisely knowing the solution volume.

Accurate volume measurement ensures precise computation of ion concentrations, which is crucial for applications requiring a specific chemical balance. Such meticulous conversion and calculation allow accurate chemical analyses.