Question

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 \(\mathrm{Ba}\left(\mathrm{NO}_{3}\right)_{2} ?\)

Short answer

(a) 343.315 g of \(\mathrm{H}_2 \mathrm{SO}_4\). (b) 1143.373 g of \(\mathrm{Ba}(\mathrm{NO}_3)_2\).

Step-by-step solution

Title - Identify the Solute

The solute is the substance that is dissolved in the solvent. For part (a), it is stated that the solution is of \(\mathrm{H}_2 \mathrm{SO}_4\). Therefore, the solute is \(\mathrm{H}_2 \mathrm{SO}_4\).

Title - Identify the Solvent

The solvent is the substance in which the solute is dissolved. It is given that the solution is aqueous, meaning the solvent is water (\(\mathrm{H}_2\mathrm{O}\)).

Title - Calculate the Moles of Solute for 1.00 L of 3.50 M Solution

The formula for molarity (\(M\)) is \[ M = \frac{n}{V}\] where \(n\) is the number of moles and \(V\) is the volume in liters. Rearrange the formula to solve for \(n\): \[ n = M \times V \] Given \(M = 3.50\, \mathrm{M}\) and \(V = 1.00\, \mathrm{L}\): \[ n = 3.50\, \mathrm{mol/L} \times 1.00\, \mathrm{L} = 3.50\, \mathrm{mol} \]Therefore, 3.50 moles of \(\mathrm{H}_2 \mathrm{SO}_4\) are needed.

Title - Calculate the Mass of the Solute for \(\mathrm{H}_2 \mathrm{SO}_4\)

Use the molar mass of \(\mathrm{H}_2 \mathrm{SO}_4\) to convert moles to grams. The molar mass of \(\mathrm{H}_2 \mathrm{SO}_4\) is \(2 \times 1.01 + 32.07 + 4 \times 16.00 = 98.09 \, \mathrm{g/mol}\). \[ \text{Mass} = n \times \text{molar mass} \] \[ \text{Mass} = 3.50\, \mathrm{mol} \times 98.09 \, \mathrm{g/mol} = 343.315\, \mathrm{g} \] Therefore, 343.315 grams of \(\mathrm{H}_2 \mathrm{SO}_4\) are needed.

Title - Calculate the Moles of Solute for 2.50 L of 1.75 M Solution

Using the formula for molarity again: \[ n = M \times V \] Given \(M = 1.75\, \mathrm{M}\) and \(V = 2.50\, \mathrm{L}\): \[ n = 1.75\, \mathrm{mol/L} \times 2.50\, \mathrm{L} = 4.375\, \mathrm{mol} \] Therefore, 4.375 moles of \(\mathrm{Ba}(\mathrm{NO}_3)_2\) are needed.

Title - Calculate the Mass of the Solute for \(\mathrm{Ba}(\mathrm{NO}_3)_2\)

Use the molar mass of \(\mathrm{Ba}(\mathrm{NO}_3)_2\) to convert moles to grams. The molar mass of \(\mathrm{Ba}(\mathrm{NO}_3)_2\) is \[137.33 + 2 \times (14.01 + 3 \times 16.00) = 261.34\, \mathrm{g/mol}\]. \[ \text{Mass} = n \times \text{molar mass} \] \[ \text{Mass} = 4.375\, \mathrm{mol} \times 261.34\, \mathrm{g/mol} = 1143.373\, \mathrm{g} \] Therefore, 1143.373 grams of \(\mathrm{Ba}(\mathrm{NO}_3)_2\) are needed.

Key concepts

Solute

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.

Solvent

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

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.

Moles

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

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.