Molality: Videos & Practice Problems

Molality, represented by the lowercase letter m, is defined as the number of moles of solute per kilogram of solvent. This concept is crucial in understanding solutions, as it highlights the relationship between the solute and the solvent within a solution. Specifically, molality is calculated using the formula:

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

In contrast, molarity, denoted by the capital letter M, also uses moles of solute in the numerator, but its denominator is the volume of the solution in liters. This distinction is important, as it allows for conversions between the two measurements when necessary.

For example, if you have 0.30 moles of sodium chloride (NaCl), this indicates that you have 0.30 moles of the solute dissolved in 1 kilogram of solvent, typically water. Similarly, if you encounter a solution that is 0.25 molar (0.25 M) in glucose, this means there are 0.25 moles of glucose in 1 liter of solution, which is often water as the solvent.

Understanding these definitions and calculations is essential, especially when converting between molality, molarity, mole fraction, or mass percent. Remember, when given the molality of a solution, it directly translates to the number of moles of solute per kilogram of solvent, a key concept for further studies in chemistry.

Ionic molality, also known as osmolality, refers to the concentration of dissolved ions in a solution. To calculate the osmolality of a solution, it is essential to consider the dissociation of the solute into its constituent ions. For instance, when sodium chloride (NaCl) is dissolved in water, it dissociates into sodium ions (Na⁺) and chloride ions (Cl^-), resulting in a total of two ions per formula unit.

To find the osmolality, the formula used is:

Osmolality = Number of ions × Molality

In the case of a 0.30 molal solution of sodium chloride, the calculation would be:

Osmolality = 2 × 0.30 molal = 0.60 osmolal

This means that the ionic molality or osmolality of the sodium chloride solution is 0.60 osmolal. It is crucial to remember that the osmolality reflects the total concentration of all ions in the solution, which is determined by the number of ions produced upon dissociation. To reinforce your understanding, try the example provided at the bottom of the page and compare your results.

To calculate the molality of potassium chlorate in a solution, we start by determining the number of moles of potassium chlorate (KClO₃) present in the solution. Given that 43 grams of KClO₃ is dissolved, we can convert this mass into moles using its molecular weight, which is 122.55 grams per mole. The conversion is done as follows:

Number of moles = \(\frac{\text{mass (g)}}{\text{molecular weight (g/mol)}} = \frac{43 \, \text{g}}{122.55 \, \text{g/mol}} \approx 0.3509 \, \text{moles}\).

Next, we need to find the mass of the solvent (water) in kilograms. We know the total volume of the solution is 100 mL and the density of the solution is 1.760 grams per milliliter. We can calculate the total mass of the solution using the formula:

Total mass of solution = density × volume = \(1.760 \, \text{g/mL} \times 100 \, \text{mL} = 176 \, \text{g}\).

Since the solution consists of both solute (KClO₃) 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 = Total mass of solution - Mass of solute = \(176 \, \text{g} - 43 \, \text{g} = 133 \, \text{g}\).

To convert this mass into kilograms, we divide by 1000:

Mass of solvent in kg = \(\frac{133 \, \text{g}}{1000} = 0.133 \, \text{kg}\).

Now that we have both the number of moles of KClO₃ and the mass of the solvent in kilograms, we can calculate the molality (m) of the solution using the formula:

Molality (m) = \(\frac{\text{moles of solute}}{\text{kilograms of solvent}} = \frac{0.3509 \, \text{moles}}{0.133 \, \text{kg}} \approx 2.64 \, \text{mol/kg}\).

This calculation shows that the molality of the potassium chlorate solution is approximately 2.64 mol/kg. Understanding how to manipulate the relationships between mass, volume, and moles is crucial for solving problems related to solution concentration, particularly when determining molality.

To convert the molality of glucose in an aqueous solution to molarity, we start with the given molality of 2.56 molal, which indicates there are 2.56 moles of glucose per kilogram of solvent (water). Additionally, the density of the solution is provided as 1.530 grams per milliliter.

First, we calculate the mass of glucose in grams. The molar mass of glucose (C₆H₁₂O₆) is approximately 180.156 grams per mole. Therefore, the mass of glucose in 2.56 moles is:

Mass of glucose = 2.56 moles × 180.156 g/mole = 461.119 grams.

Next, we need to find the total mass of the solution, which consists of the mass of the solute (glucose) and the mass of the solvent (water). Since we have 1 kilogram (1000 grams) of water, the total mass of the solution is:

Total mass of solution = mass of glucose + mass of water = 461.119 grams + 1000 grams = 1461.119 grams.

Using the density of the solution, we can convert the mass of the solution to volume in milliliters:

Volume of solution (mL) = Total mass of solution / Density = 1461.119 grams / 1.530 g/mL ≈ 955.032 mL.

To convert milliliters to liters, we divide by 1000:

Volume of solution (L) = 955.032 mL / 1000 = 0.955032 L.

Now, we can calculate the molarity (M), which is defined as the number of moles of solute per liter of solution:

Molarity (M) = moles of solute / liters of solution = 2.56 moles / 0.955032 L ≈ 2.68 M.

This calculation demonstrates how to transition from molality to molarity using the density of the solution, highlighting the relationship between these two concentration measures in solution chemistry.

To determine the ionic molality of nitrate ions in a solution of lead(IV) nitrate, we start by understanding the dissociation of the compound in solution. Lead(IV) nitrate, represented chemically as Pb(NO₃)₄, dissociates into one lead ion (Pb⁴⁺) and four nitrate ions (NO₃^-). This means that for every mole of lead(IV) nitrate, there are a total of five ions produced: one lead ion and four nitrate ions.

The ionic molality of a specific ion can be calculated by multiplying the number of that ion produced during dissociation by the molality of the compound. In this case, the molality of lead(IV) nitrate is given as 0.305 molal. Therefore, the ionic molality of the nitrate ions is calculated as follows:

Ionic Molality of Nitrate Ions:
Ionic molality = Number of nitrate ions × Molality of lead(IV) nitrate
Ionic molality = 4 × 0.305 molal = 1.22 molal

It is important to note that if asked for the total ionic molality, which accounts for all ions in solution, you would multiply the total number of ions (5) by the molality of the compound:

Total Ionic Molality:
Total ionic molality = Total number of ions × Molality of lead(IV) nitrate
Total ionic molality = 5 × 0.305 molal = 1.525 molal

In summary, when calculating ionic molality, always consider the dissociation of the compound and the number of ions produced to find the molality of individual ions or the total ionic molality as needed.