Chemical Concentrations: Videos & Practice Problems

In the study of solutions, understanding the properties and behaviors of solutes and solvents is crucial. Solutions are defined as homogeneous mixtures where the solvent dissolves a solute to a significant extent. The concentration of these solutions can be quantified using two primary units: molarity and molality.

Molarity, denoted as M, is calculated by the formula:

M = \frac{n}{V}

where n represents the number of moles of solute, and V is the volume of the solution in liters. In this context, the solute is the smaller component of the solution, while the solvent is the larger component.

Molality, on the other hand, is defined as:

m = \frac{n}{m_{solvent}}

Here, n again refers to the moles of solute, but m_solvent is the mass of the solvent in kilograms. Although molarity and molality differ in their denominators, they both rely on the moles of solute, which is a key concept in both general and analytical chemistry.

Recognizing these definitions and their applications is essential for solving problems related to solution concentrations, particularly as we delve deeper into analytical chemistry. The knowledge gained from general chemistry regarding colligative properties and solution behaviors will be beneficial as we tackle more complex examples in future discussions.

To determine the molarity of a cadmium chloride solution, we start by identifying the key components of the problem. Molarity (M) is defined as the number of moles of solute per liter of solution. In this case, cadmium chloride (CdCl₂) is the solute, and water is the solvent.

We are given 20 grams of cadmium chloride and 80 grams of water. The molecular weight of cadmium chloride is 183.317 grams per mole. To find the moles of cadmium chloride, we use the formula:

moles of solute = \(\frac{\text{mass of solute (g)}}{\text{molar mass (g/mol)}}\)

Substituting the values, we have:

moles of CdCl₂ = \(\frac{20 \text{ g}}{183.317 \text{ g/mol}} \approx 0.1091 \text{ moles}\)

Next, we need to calculate the total volume of the solution in liters. The total mass of the solution is the sum of the mass of the solute and the solvent, which is 20 grams + 80 grams = 100 grams. The density of the solution is given as 1.1988 grams per cubic centimeter (g/cm³), which can also be expressed as grams per milliliter (g/mL). To find the volume in liters, we use the formula:

volume (L) = \(\frac{\text{mass (g)}}{\text{density (g/mL)}}\)

Calculating the volume gives us:

volume = \(\frac{100 \text{ g}}{1.1988 \text{ g/mL}} \approx 83.417 \text{ mL} = 0.083417 \text{ L}\)

Now, we can calculate the molarity of the cadmium chloride solution:

Molarity (M) = \(\frac{\text{moles of solute}}{\text{liters of solution}} = \frac{0.1091 \text{ moles}}{0.083417 \text{ L}} \approx 1.3079 \text{ M}\)

Considering significant figures, the values provided (20.00 g, 80.00 g, and 1.1988 g/mL) suggest that our final answer should also reflect four significant figures. Rounding 1.3079 M gives us:

Molarity = 1.308 M

This calculation illustrates the process of determining the molarity of a solution by converting grams of solute to moles, calculating the total volume of the solution, and applying the molarity formula. Understanding these steps is crucial for solving similar problems in chemistry.

To sterilize drinking water, chlorine is commonly added, typically in the form of the diatomic molecule Cl₂. In a park's water fountain, the chlorine concentration is measured at 185 parts per million (PPM). To convert this concentration into molarity and subsequently into micromolar, we start by understanding that 1 PPM is equivalent to 1 milligram of solute per liter of solution. Therefore, 185 PPM indicates that there are 185 milligrams of Cl₂ in 1 liter of water.

First, we convert milligrams to grams: 185 mg is equal to 0.185 grams (since 1 mg = 10^-3 g). Next, we need to calculate the number of moles of Cl₂ using its molecular weight. The atomic weight of chlorine is approximately 35.453 g/mol, so the molecular weight of Cl₂ is:

$2^{\mathrm{Cl}}=235.453=70.906g/mol$

Now, we can find the number of moles of Cl₂ in 0.185 grams:

$\frac{0.185}{70.906}=\mathrm{2.61\; \backslash times\; 10^\{-3\}}moles$

Since this amount is in 1 liter of solution, the molarity (M) of the solution is:

$M\; =\; \backslash frac\{2.61\; \backslash times\; 10^\{-3\}\; \backslash text\{\; moles\}\}\{1\; \backslash text\{\; L\}\}\; =\; 2.61\; \backslash times\; 10^\{-3\}\; \backslash text\{\; M\}$

To convert molarity to micromolar (µM), we recognize that 1 M = 10⁶ µM. Thus, we multiply the molarity by 10⁶:

$2.61\; \backslash times\; 10^\{-3\}\; \backslash text\{\; M\}\; \backslash times\; 10^\{6\}\; =\; 2610\; \backslash text\{\; \mu M\}$

However, since we need to express this with the correct number of significant figures based on the original measurement (185 PPM has three significant figures), we round it to:

$2.61\; \backslash times\; 10^\{3\}\; \backslash text\{\; \mu M\}$

This process illustrates the conversion between parts per million, molarity, and micromolar concentrations, emphasizing the importance of understanding solution concentrations in various contexts, including molarity and molality.

In this discussion, we explore the concepts of molarity and molality, focusing on how to calculate the molality of a solution. To illustrate this, consider a solution prepared by dissolving 41.33 grams of nitric acid (HNO₃) in enough water to make 100 mL of solution, with a solution density of 1.380 grams per milliliter.

To find the molality (m), we use the formula:

m = \(\frac{\text{moles of solute}}{\text{kilograms of solvent}}\)

First, we need to convert grams of nitric acid to moles. The molecular mass of nitric acid is 63.018 grams per mole. Thus, the number of moles can be calculated as follows:

moles of HNO₃ = \(\frac{41.33 \text{ g}}{63.018 \text{ g/mol}} \approx 0.6558 \text{ moles}\)

Next, we determine the mass of the solution. Using the density, we can find the total mass of the solution:

mass of solution = volume × density = 100 \text{ mL} × 1.380 \text{ g/mL} = 138 \text{ g}

Since the solution consists of both solute (nitric acid) and solvent (water), we can find the mass of the solvent by subtracting the mass of the solute from the total mass of the solution:

mass of solvent = mass of solution - mass of solute = 138 \text{ g} - 41.33 \text{ g} = 96.67 \text{ g}

To convert grams of solvent to kilograms, we divide by 1000:

mass of solvent = \(\frac{96.67 \text{ g}}{1000} = 0.09667 \text{ kg}\)

Now, we can substitute the values into the molality formula:

m = \(\frac{0.6558 \text{ moles}}{0.09667 \text{ kg}} \approx 6.784 \text{ mol/kg}\)

Considering significant figures, since the values used in the calculation have four significant figures, we round the final answer to four significant figures:

molality of the nitric acid solution = 6.784 mol/kg.

This example illustrates the process of calculating molality, emphasizing the importance of understanding the relationship between solute, solvent, and solution properties. In further exercises, students can practice calculating both molality and molarity to reinforce these concepts.

To determine the molality and molarity of an ethanol solution with a mole fraction of 0.090 and a density of 1.35 g/mL, we start by calculating the molality. Molality (m) is defined as the number of moles of solute (ethanol) per kilogram of solvent (water). Given the mole fraction, we know that the mole fraction of ethanol (x) is the ratio of moles of ethanol to the total moles of the solution. Thus, if we have 0.090 moles of ethanol, the total moles of the solution can be expressed as:

$$ x = \frac{n_{ethanol}}{n_{solution}} = \frac{0.090}{1} $$

This implies that the moles of solvent (water) can be calculated as:

$$ n_{water} = n_{solution} - n_{ethanol} = 1 - 0.090 = 0.910 $$

Next, we convert the moles of water to grams using the molar mass of water (18.0153 g/mol):

$$ \text{mass of water} = n_{water} \times \text{molar mass of water} = 0.910 \, \text{mol} \times 18.0153 \, \text{g/mol} = 16.394 \, \text{g} $$

To convert grams to kilograms, we divide by 1000:

$$ \text{mass of water in kg} = \frac{16.394 \, \text{g}}{1000} = 0.016394 \, \text{kg} $$

Now, we can calculate the molality:

$$ m = \frac{n_{ethanol}}{mass_{water}} = \frac{0.090 \, \text{mol}}{0.016394 \, \text{kg}} \approx 5.49 \, \text{mol/kg} $$

Rounding to two significant figures, the molality is approximately 5.5 molal.

Next, we calculate the molarity (M), which is defined as the number of moles of solute per liter of solution. We already have the moles of ethanol (0.090 moles). To find the volume of the solution in liters, we first need to determine the total mass of the solution. The total mass is the sum of the mass of ethanol and the mass of water. The molar mass of ethanol (C₂H₅OH) is approximately 46.0684 g/mol, so:

$$ \text{mass of ethanol} = n_{ethanol} \times \text{molar mass of ethanol} = 0.090 \, \text{mol} \times 46.0684 \, \text{g/mol} \approx 4.146 \, \text{g} $$

Now, we can find the total mass of the solution:

$$ \text{mass of solution} = \text{mass of water} + \text{mass of ethanol} = 16.394 \, \text{g} + 4.146 \, \text{g} \approx 20.54 \, \text{g} $$

Using the density of the solution, we can find the volume in milliliters:

$$ \text{volume of solution} = \frac{\text{mass of solution}}{\text{density}} = \frac{20.54 \, \text{g}}{1.35 \, \text{g/mL}} \approx 15.2 \, \text{mL} $$

To convert milliliters to liters:

$$ \text{volume in liters} = \frac{15.2 \, \text{mL}}{1000} = 0.0152 \, \text{L} $$

Finally, we can calculate the molarity:

$$ M = \frac{n_{ethanol}}{V_{solution}} = \frac{0.090 \, \text{mol}}{0.0152 \, \text{L}} \approx 5.92 \, \text{mol/L} $$

Rounding to two significant figures, the molarity is approximately 5.9 M. In summary, the molality of the ethanol solution is 5.5 molal, and the molarity is 5.9 M.