a. Suppose you wanted to produce 1.00 \(\mathrm{L}\) of a 3.50 \(\mathrm{M}\) aqueous solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (1) What is the solute? (2) What is the solvent? (3) How many grams of solute are needed to make this solution? b. How many grams of solute are needed to make 2.50 L of a 1.75 \(\mathrm{M}\) solution of \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} ?\)

In any solution, the solute is the substance that is dissolved within another substance. Generally, the solute is present in a smaller quantity compared to the solvent. For instance, in the exercise, \(\mathrm{H}_{2} \mathrm{SO}_{4}\) (sulfuric acid) acts as the solute in the aqueous solution prepared.
The quality of a solute directly affects the characteristics of the solution it is a part of, including its molarity and other properties such as boiling and freezing points.

The solvent is the component in which the solute dissolves. It is usually present in a larger amount than the solute. Water is the most common solvent used in many solutions, particularly those referred to as aqueous solutions. In the given exercise, the solvent is water (\(\mathrm{H}_2 \mathrm{O}\)).
Understanding the role of a solvent is crucial, as it determines the nature of the solution, including its ability to conduct electricity, its overall stability, and its reaction environment.

Molarity (\(M\)) is a way to express the concentration of a solution. It is defined as the number of moles of solute per liter of solution. The formula for molarity is:
\[ M = \frac{n}{V} \]
Where \(M\) is molarity, \(n\) is the number of moles of the solute, and \(V\) is the volume of the solution in liters.
For example, in part (a) of the exercise, a 3.50 M solution of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) means there are 3.50 moles of sulfuric acid dissolved in each liter of the solution.

The concept of moles is fundamental in chemistry for quantifying the amount of a substance. One mole is \(6.022 \times 10^{23}\) entities of any substance (Avogadro's number). This quantity helps convert between the mass of a substance and the number of its constituent particles.
For instance, to find the moles of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) needed to prepare a given solution, use the molarity formula rearranged for moles: \[ n = M \times V \]
Given a 3.50 M solution and 1.00 L volume, the calculation is: \[ n = 3.50 \text{ mol/L} \times 1.00 \text{ L} = 3.50 \text{ mol} \]

Molar mass is the mass of one mole of a substance, expressed in grams per mole (\text{g/mol}). It is the sum of the atomic masses of each atom in a molecule.
As an example, the molar mass of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) can be calculated as:
\[ (2 \times 1.01) + 32.07 + (4 \times 16.00) = 98.09 \text{ g/mol} \]
This molar mass is then used to convert moles into grams when determining the quantity of \(\mathrm{H}_{2} \mathrm{SO}_{4}\) needed to make the solution:
\[ \text{Mass} = 3.50 \text{ mol} \times 98.09 \text{ g/mol} = 343.315 \text{ g} \]
Similarly, for \(\mathrm{Ba}(\mathrm{NO}_3)_2\), its molar mass is calculated and used to find the required mass in the solution.