Calculate the molality of each of the following solutions: (a) \(2.0 \mathrm{~mol} \mathrm{HCl}\) in \(175 \mathrm{~g}\) water (b) \(14.5 \mathrm{~g} \mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\) in \(550.0 \mathrm{~g}\) water (c) \(25.2 \mathrm{~mL}\) methanol, \(\mathrm{CH}_{3} \mathrm{OH}(d=0.791 \mathrm{~g} / \mathrm{mL})\) in \(595 \mathrm{~g}\) ethanol, \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{OH}\)

Understanding the concept of molality is essential for calculations involving the concentration of a solution. It is a measure of the amount of a solute present in a given mass of solvent. Unlike molarity, which depends on the volume of the solution, molality is based solely on mass, making it unaffected by temperature changes.

With the formula for molality, \( m = \frac{mol_{solute}}{kg_{solvent}} \), you can determine how concentrated a solution is. This formula requires knowing the number of moles of the solute and the mass of the solvent in kilograms. Let's make this practical. If a student has a sample with 2.0 moles of hydrochloric acid in 175 grams of water, they would calculate the molality by converting the mass of water to kilograms (175g / 1000 = 0.175kg) and then use the formula: \( m = \frac{2.0 \, \text{mol}}{0.175 \, \text{kg}} \), yielding the answer in mol/kg, which is the unit for molality.

Molar mass is a cornerstone concept in chemistry, representing the mass of one mole of a substance. It is typically expressed in grams per mole (g/mol) and can be calculated by summing the atomic masses of all the atoms in a chemical formula.

To demonstrate its application, consider a student tasked with determining the molality of a sucrose solution. They have 14.5 grams of sucrose, which has a molar mass of 342.3 g/mol. To find the moles of sucrose, the student would divide the mass by the molar mass: \( \text{moles} = \frac{14.5 \, \text{g}}{342.3 \, \text{g/mol}} \). This step is crucial because molality calculations require the moles of solute. As science educators, we emphasize accuracy in identifying the correct molar mass to avoid errors and secure a strong foundation for further calculations.

Distinguishing between solute and solvent is essential when discussing solutions. The solute is the substance that is dissolved, while the solvent is the substance in which the solute dissolves. Typically, the solvent is present in a greater amount than the solute.

In the context of our exercise, hydrochloric acid (HCl) is the solute and water is the solvent for part (a). In part (b), sucrose (C12H22O11) is the solute and water again functions as the solvent. For part (c), methanol (CH3OH) is the solute and ethanol (CH3CH2OH) is the solvent, demonstrating that even liquids can act as solutes. Understanding the roles of solute and solvent not only aids in molality calculations but also in anticipating physical and chemical properties of solutions.

The concentration of a solution is a measure of how much solute is present in a given amount of solvent. In chemistry, there are several ways to express concentration, including molality, molarity, and percent composition.

For our exercises, molality is the chosen method, providing insight into the solution’s makeup without the influence of temperature. To calculate the concentration for a tricky scenario, such as part (c) with methanol in ethanol, the mass of methanol is first obtained using its volume and density, followed by conversion to moles using the molar mass. The molality is then calculated by dividing the moles of methanol by the kilogram of ethanol.

For students, grasping these concepts leads to a better understanding of solution chemistry and reactions in varying conditions, making it invaluable for applications in both laboratory and industry settings.