Calculate the molality of each of these solutions: (a) \(14.3 \mathrm{~g}\) of sucrose \(\left(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\right)\) in \(676 \mathrm{~g}\) of water, (b) 7.20 moles of ethylene \(\operatorname{glycol}\left(\mathrm{C}_{2} \mathrm{H}_{6} \mathrm{O}_{2}\right)\) in \(3546 \mathrm{~g}\) of water.

Understanding the concept of molar mass is fundamental when tackling chemistry problems, especially those involving molality calculations. Molar mass is defined as the mass of one mole of a particular substance, measured in grams per mole (g/mol). It is essentially the sum of the atomic masses of every atom within a molecule. For instance, in the case of sucrose (C₁₂H₂₂O₁₁), we must calculate the molar mass by adding up the masses of each carbon (C), hydrogen (H), and oxygen (O) atom, considering the number of times each appears in the molecule.

To calculate the molar mass of sucrose, you would multiply the atomic mass of carbon by 12 (since there are 12 carbon atoms), add it to the product of the atomic mass of hydrogen and 22 (the number of hydrogen atoms), and then add this to the product of the atomic mass of oxygen and 11 (the number of oxygen atoms). This gives us the molar mass of sucrose that is used in molality calculations. Knowing the molar mass allows us to convert between grams of a substance and moles, a conversion that is necessary to find the molality of a solution.

It's important to ensure you are using the correct atomic masses for these calculations and to be accurate in your addition, as small differences can lead to significant errors in chemistry.

The mole concept is a bridge between the microscopic world of atoms and molecules and the macroscopic world we observe. One mole equals Avogadro's number (approximately 6.022 x 10²³) of particles, which is a foundational concept in chemistry representing a standard quantity of chemical units. When we speak of moles of solute in the realm of solution concentration, we're counting the number of moles of a substance dissolved in a given volume or mass of solvent.

For example, in our exercise where molality must be calculated, step one requires finding the number of moles of sucrose that can be found in 14.3 grams. Using the molar mass of sucrose calculated prior, you can convert the gram measurement into moles. Similarly, having moles of a solute, such as the 7.20 moles of ethylene glycol given directly in the exercise, simplifies subsequent calculations. Remember, the moles of solute are a central part of the formula used to calculate molality, and accurate determination of this quantity is pivotal for obtaining the correct solution concentration.

Solution concentration refers to the amount of a solute that is dissolved in a given quantity of solvent and is a key concept in various sectors, including chemistry, biology, and environmental sciences. There are multiple ways to express solution concentration, including molarity, molality, and percent composition, among others. In this problem, we're focusing on molality, which is defined as the moles of solute per kilogram of solvent (mol/kg).

Molality is particularly useful when dealing with temperature changes, as it does not change with temperature. This is because, unlike volume, mass does not expand or contract with temperature fluctuations. To calculate the molality of a solution, it's crucial to precisely determine the number of moles of the solute, as covered in the previous section, and the mass of the solvent converted into kilograms.

Following the exercise steps, after finding the moles of solute and converting the solvent's mass from grams to kilograms, you can divide the moles of solute by the kilograms of solvent to find the molality. As an example, for a sucrose solution with 0.0418 moles of sucrose in 0.676 kilograms of water, the concentration, expressed in molality, would be 0.0618 mol/kg or 0.0618m.