A sugar syrup solution contains \(15.0 \%\) sugar, \(\mathrm{C}_{12} \mathrm{H}_{22} \mathrm{O}_{11}\), by mass and has a density of \(1.06 \mathrm{~g} / \mathrm{mL}\). (a) How many grams of sugar are in \(1.0 \mathrm{~L}\) of this syrup? (b) What is the molarity of this solution? (c) What is the molality of this solution?

Molarity is a measure of the concentration of a solute in a solution. It is defined as the number of moles of solute per liter of solution. This is represented by the symbol M. When we say a solution is '1 M,' it means there is 1 mole of solute in every liter of solution.

For example, in the given problem, we calculated the number of moles of sugar (solute) in 1 liter of sugar syrup (solution) as 0.464 moles. Therefore, the molarity of the sugar syrup is 0.464 M. This helps us determine the concentration of the sugar in the solution in a straightforward manner.

To recap, molarity (M) can be calculated using the formula: \( M = \frac{\text{moles of solute}}{\text{liters of solution}} \). This concept is fundamental in understanding the concentration of solutions in chemistry.

While molarity measures the concentration based on the volume of the solution, molality focuses on the mass of the solvent. Molality (m) is defined as the number of moles of solute per kilogram of solvent.

In the given problem, after determining the mass of water (solvent) in the syrup as 901 g (or 0.901 kg), we calculated the molality of the solution. With 0.464 moles of sugar (solute) and 0.901 kg of water (solvent), the molality is: \( m = \frac{0.464 \, \text{mol}}{0.901 \, \text{kg}} \approx 0.515 \, \text{m} \). This measure helps when the temperature and pressure impact the volume but not the mass, making it useful for comparing concentrations under different conditions.

Density is a physical property defined as mass per unit volume. It can be represented by the formula: \( \text{Density} = \frac{\text{mass}}{\text{volume}} \).

In the problem, the density of the sugar syrup is given as 1.06 g/mL. Using the density, we can calculate the mass of 1 liter (which is 1000 mL) of the syrup: \( \text{Mass} = 1.06 \, \frac{\text{g}}{\text{mL}} \times 1000 \, \text{mL} = 1060 \, \text{g} \).

This calculation is crucial because it helps us understand how much sugar and water are in the solution's mass. Knowing the density allows us to move between volume and mass with ease, which is essential when dealing with solution concentrations in chemistry.

Mass percentage is a way to express the concentration of a solute in a solution. It is given by the formula: \( \text{Mass percentage} = \frac{\text{mass of solute}}{\text{total mass of solution}} \times 100 \).

In this problem, the sugar syrup contains 15% sugar by mass. Given the total mass of the solution (1060 g), we calculated the mass of sugar: \( \text{Mass of sugar} = 0.15 \times 1060 \, \text{g} = 159 \, \text{g} \).

Knowing the mass percentage allows us to determine the mass of the solute directly related to the total mass of the solution, giving a clear picture of the composition of the solution.

Molar mass is the mass of one mole of a substance. It is expressed in grams per mole (g/mol). To calculate the molar mass of a compound, you sum up the atomic masses of all the atoms in the molecule.

For sugar (\(\mathrm{C}_{12}\mathrm{H}_{22}\mathrm{O}_{11}\)), the molar mass is calculated as follows: \( 12 \times 12.01 + 22 \times 1.01 + 11 \times 16.00 = 342.34 \, \text{g/mol} \).

In the problem, we used this molar mass to convert the given mass of sugar (159 g) into moles: \( \text{Moles of sugar} = \frac{159 \, \text{g}}{342.34 \, \text{g/mol}} \approx 0.464 \, \text{mol} \).

Understanding molar mass is crucial for converting between mass and moles, which is fundamental in stoichiometry and solution concentration calculations.